type-theory/resources/MayConcise/ConciseRevised.tex

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\title{A Concise Course in Algebraic Topology}

\author{J. P. May}

\begin{document}

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\maketitle

\tableofcontents

\clearpage

\thispagestyle{empty}

\chapter*{Introduction}

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The first year graduate program in mathematics at the University of Chicago 
consists of three three-quarter courses, in analysis, algebra, and topology.
The first two quarters of the topology sequence focus on manifold theory and
differential geometry, including differential forms and, usually, a glimpse 
of de Rham cohomology. The third quarter focuses on algebraic topology. I have 
been teaching the third quarter off and on since around 1970. Before that, the 
topologists, including me, thought that it would be impossible to squeeze a 
serious introduction to algebraic topology into a one quarter course, but we 
were overruled by the analysts and algebraists, who felt that it was unacceptable
for graduate students to obtain their PhDs without having some contact with
algebraic topology. 

This raises a conundrum. A large number of students at 
Chicago go into topology, algebraic and geometric. The introductory course
should lay the foundations for their later work, but it should also be 
viable as an introduction to the subject suitable for those going into other 
branches of mathematics. These notes reflect my efforts to organize the
foundations of algebraic topology in a way that caters to both pedagogical 
goals. There are evident defects from both points of view. A treatment more 
closely attuned to the needs of algebraic geometers and analysts would include 
\v{C}ech cohomology on the one hand and de Rham cohomology and perhaps Morse 
homology on the other. A treatment more closely attuned to the needs of 
algebraic topologists would include spectral sequences and an array of 
calculations with them. In the end, the overriding pedagogical goal has
been the introduction of basic ideas and methods of thought. 

Our understanding of the foundations of algebraic topology has undergone subtle but 
serious changes since I began teaching this course. These changes reflect in part an
enormous internal development of algebraic topology over this period, one which
is largely unknown to most other mathematicians,  even those working in such 
closely related fields as geometric topology and algebraic geometry. Moreover,
this development is poorly reflected in the textbooks that have appeared over 
this period.

Let me give a small but technically important example. The study of 
generalized homology and cohomology theories pervades modern algebraic topology. 
These theories satisfy the excision axiom. One constructs most such theories homotopically, 
by constructing representing objects called spectra, and one must then prove that 
excision holds. There is a way to do this in general that is no more difficult 
than the standard verification for singular homology and cohomology. I find this proof 
far more conceptual and illuminating than the standard one even when specialized to 
singular homology and cohomology. (It is based on the approximation of excisive triads 
by weakly equivalent CW triads.) This should by now be a standard approach. However, 
to the best of my knowledge, there exists no rigorous exposition of this approach in 
the literature, at any level.

More centrally, there now exist axiomatic treatments of large swaths of homotopy
theory based on Quillen's theory of closed model categories. While I do not think
that a first course should introduce such abstractions, I do think that the exposition
should give emphasis to those features that the axiomatic approach shows to be
fundamental. For example, this is one of the reasons, although by no means the only one, 
that I have dealt with cofibrations, fibrations, and weak equivalences much more thoroughly 
than is usual in an introductory course. 

Some parts of the theory are dealt with quite classically. The theory of fundamental
groups and covering spaces is one of the few parts of algebraic topology that has probably 
reached definitive form, and it is well treated in many sources. Nevertheless, this material
is far too important to all branches of mathematics to be omitted from a first course. For
variety, I have made more use of the fundamental groupoid than in standard 
treatments,\footnote{But see R. Brown's book cited in \S2 of the suggestions for further reading.}
and my use of it has some novel features. For conceptual interest, I have emphasized different 
categorical ways of modeling the topological situation algebraically, and I have taken the 
opportunity to introduce some ideas that are central to equivariant algebraic topology.  

Poincar\'e duality is also too fundamental to omit. There are more elegant ways to treat 
this topic than the classical one given here, but I have preferred to give the theory in 
a quick and standard fashion that reaches the desired conclusions in an economical way. 
Thus here I have not presented the truly modern approach that applies to generalized homology 
and cohomology theories.\footnote{That approach derives Poincar\'e duality as a consequence of 
Spanier-Whitehead and Atiyah duality, via the Thom isomorphism for oriented vector bundles.}

The reader is warned that this book is not designed as a textbook, although it could 
be used as one in exceptionally strong graduate programs. Even then, it would be 
impossible to cover all of the material in detail in a quarter, or even in a year.
There are sections that should be omitted on a first reading and others that are 
intended to whet the student's appetite for further developments. In practice, when 
teaching, my lectures are regularly interrupted by (purposeful) digressions, most often 
directly prompted by the questions of students. These introduce more advanced topics that 
are not part of the formal introductory course: cohomology operations, characteristic 
classes, $K$-theory, cobordism, etc., are often first introduced earlier in the lectures 
than a linear development of the subject would dictate.

These digressions have been expanded and written up here as sketches
without complete proofs, in a logically coherent order, in the last four chapters. These are 
topics that I feel must be introduced in some fashion in any serious graduate level introduction to 
algebraic topology. A defect of nearly all existing texts is that they do not go far enough into 
the subject to give a feel for really substantial applications: the reader sees
spheres and projective spaces, maybe lens spaces, and applications accessible with knowledge of
the homology and cohomology of such spaces. That is not enough to give a real feeling for the
subject. I am aware that this treatment suffers the same defect, at least before its sketchy
last chapters.

Most chapters end with a set of problems. Most of these ask for computations and applications
based on the material in the text, some extend the theory and introduce further concepts, some
ask the reader to furnish or complete proofs omitted in the text, and some are essay questions 
which implicitly ask the reader to seek answers in other sources. Problems marked $*$
are more difficult or more peripheral to the main ideas. Most of these problems are 
included in the weekly problem sets that are an integral part of the course at Chicago. In fact, 
doing the problems is the heart of the course. (There are no exams and no grades; students are 
strongly encouraged to work together, and more work is assigned than a student can reasonably be 
expected to complete working alone.) {\em The reader is urged to try most of the problems: this is 
the way to learn the material}. The lectures focus on the ideas; their assimilation requires 
more calculational examples and applications than are included in the text.

I have ended with a brief and idiosyncratic guide to the literature for the 
reader interested in going further in algebraic topology.

These notes have evolved over many years, and I claim no originality for most of the material.
In particular, many of the problems, especially in the more classical chapters, are the same as, 
or are variants of, problems that appear in other texts. Perhaps this is unavoidable: interesting 
problems that are doable at an early stage of the development are few and far between. I am 
especially aware of my debts to earlier texts by Massey, Greenberg and Harper, Dold, and Gray.

I am very grateful to John Greenlees for his careful reading and suggestions, especially
of the last three chapters. I am also grateful to Igor Kriz for his suggestions
and for trying out the book at the University of Michigan. By far my greatest debt, 
a cumulative one, is to several generations of students, far too numerous to name. 
They have caught countless infelicities and outright blunders, and they have contributed 
quite a few of the details. You know who you are. Thank you.

\clearpage

\thispagestyle{empty}

\chapter{The fundamental group and some of its applications}

We introduce algebraic topology with a quick treatment of standard material about the
fundamental groups of spaces, embedded in a geodesic proof of the Brouwer fixed point 
theorem and the fundamental theorem of algebra.

\section{What is algebraic topology?}

A topological space $X$ is a set in which there is a notion of nearness of 
points. Precisely, there is given a collection of ``open'' subsets of $X$ 
which is closed under finite intersections and arbitrary unions. It suffices
to think of metric spaces. In that case, the open sets are the arbitrary unions 
of finite intersections of neighborhoods $U_{\epz}(x) = \{ y| d(x,y) < \epz \}$.

A function $p: X\rtarr Y$ is continuous if it takes nearby points to 
nearby points. Precisely, $p^{-1}(U)$ is open if $U$ is open. If $X$ and
$Y$ are metric spaces, this means that, for any  $x\in X$ and $\epz>0$, there
exists $\de>0$ such that $p(U_{\de}(x))\subset U_{\epz}(p(x))$.

Algebraic topology assigns discrete algebraic invariants to topological
spaces and continuous maps. More narrowly, one wants the algebra to be
invariant with respect to continuous deformations of the topology. 
Typically, one associates a group $A(X)$ to a space $X$ and a homomorphism 
$A(p): A(X)\rtarr A(Y)$ to a map $p: X\rtarr Y$; one usually writes $A(p) = p_*$.

A ``homotopy'' $h:p\htp q$\index{homotopy} between maps $p,q: X\rtarr Y$ is a continuous
map $h: X\times I \rtarr Y$ such that $h(x,0)=p(x)$ and $h(x,1)=q(x)$,
where $I$ is the unit interval $[0,1]$. We usually want $p_*=q_*$ if
$p\htp q$, or some invariance property close to this. 

In oversimplified outline, the way homotopy theory works is roughly this.
\begin{enumerate}
\item One defines some algebraic construction $A$ and proves that it is
suitably homotopy invariant.
\item One computes $A$ on suitable spaces and maps.
\item One takes the problem to be solved and deforms it to the point that
step 2 can be used to solve it.
\end{enumerate}

The further one goes in the subject, the more elaborate become the constructions
$A$ and the more horrendous become the relevant calculational techniques. This
chapter will give a totally self-contained paradigmatic illustration of the basic 
philosophy. Our construction $A$ will be the ``fundamental group.'' We will calculate 
$A$ on the circle $S^1$ and on some maps from $S^1$ to itself. We will then use the
computation to prove the ``Brouwer fixed point theorem'' and the ``fundamental
theorem of algebra.''

\section{The fundamental group}

Let $X$ be a space. Two paths $f,g: I\rtarr X$ from $x$ to $y$ are 
equivalent\index{equivalent!paths}
if they are homotopic through paths from $x$ to $y$. That is, there must exist
a homotopy $h:I\times I\rtarr X$ such that 
$$h(s,0)=f(s), \ \ h(s,1)=g(s), \ \ h(0,t)=x, \ \tand  \ h(1,t)=y$$
for all $s,t\in I$. Write $[f]$ for the equivalence class of $f$. We say that
$f$ is a loop\index{loop} if $f(0)=f(1)$. Define $\pi_1(X,x)$\index{fundamental group|(} to 
be the set of equivalence
classes of loops that start and end at $x$. 

For paths $f:x\to y$ and $g:y\to z$, define $g\cdot f$ to be the path
obtained by traversing first $f$ and then $g$, going twice as fast on each:
$$(g\cdot f)(s)=
\begin{cases}
f(2s) & \text{if $0\leq s\leq 1/2$} \\
g(2s-1) & \text{if $1/2\leq s\leq 1$}.
\end{cases}
$$
Define $f^{-1}$ to be $f$ traversed the other way around: $f^{-1}(s)=f(1-s)$. 
Define $c_x$ to be the constant loop at $x$: $c_x(s)=x$. Composition of paths
passes to equivalence classes via $[g][f]=[g\cdot f]$. It is easy to check that
this is well defined. Moreover, after passage to equivalence classes, this 
composition becomes associative and unital. It is easy enough to write down
explicit formulas for the relevant homotopies. It is more illuminating to draw a 
picture of the domain squares and to indicate schematically how the
homotopies are to behave on it. In the following, we assume given paths
$$f:x\to y,\ \ \ g:y\to z,\ \tand \ \  h: z\to w.$$

$$h\cdot (g\cdot f) \htp (h\cdot g)\cdot f$$
$$\diagram
\xline[0,4]^<(0.25)f ^<(0.65)g ^<(0.85)h \xline[4,0]_{c_x}
 & & \xline[4,-1] & \xline[4,-1] & \xline[4,0]^{c_w} \\
 & & & & \\
 & & & & \\
 & & & & \\
\xline[0,4]_<(0.15)f _<(0.35)g _<(0.75)h & & & & \\
\enddiagram$$

$$f\cdot c_x \htp f \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ c_y\cdot f\htp f$$
$$\diagram
\rrline^f \ddline_{c_x} \ddrline & & \ddline^{c_y} &
 & \rrline^f \ddline_{c_x}  & & \ddline^{c_y} \ddlline\\
& & & & & & \\
\rrline_<(0.25){c_x} _<(0.75)f & & & & \rrline_<(0.25){f} _<(0.75){c_y} & & \\
\enddiagram$$
Moreover, $[f^{-1}\cdot f] = [c_x]$ and $[f\cdot f^{-1}]=[c_y]$. For the first,
we have the following schematic picture and corresponding formula. In the
schematic picture, 
$$f_t=f|[0,t] \ \ \ \tand \ \ \ f^{-1}_t=f^{-1}|[1-t,1].$$
$$\diagram
\xline[0,4]^<(0.25){f} ^<(0.75){f^{-1}} \xline[4,0]_{c_x} 
& & \xline[4,-2] \xline[4,2] & & \xline[4,0]^{c_x} \\
& & & & \\
\xdotted[0,4]^<(0.15){f_t} ^<(0.5){c_{f(t)}} ^<(0.85){f^{-1}_t} & & & & \\
& & & & \\
\xline[0,4]_{c_x} & & & &\\
\enddiagram$$
$$h(s,t) =
\begin{cases}
f(2s) & \text{if $0\leq s\leq t/2$} \\
f(t) & \text{if $t/2\leq s\leq 1-t/2$} \\
f(2-2s) & \text{if $1-t/2\leq s\leq 1$.}
\end{cases}
$$
We conclude that $\pi_1(X,x)$ is a group with identity element $e=[c_x]$ and 
inverse elements $[f]^{-1}=[f^{-1}]$. It is called the fundamental 
group\index{fundamental group|)}  of
$X$, or the first homotopy group of $X$. There are higher homotopy groups $\pi_n(X,x)$ 
defined in terms of maps $S^n\rtarr X$. We will get to them later.

\section{Dependence on the basepoint}

For a path $a: x\to y$, define $\ga [a]: \pi_1(X,x)\rtarr \pi_1(X,y)$
by $\ga [a][f]=[a\cdot f\cdot a^{-1}]$. It is easy to check that $\ga [a]$ 
depends only on the equivalence class of $a$ and is a homomorphism of groups. 
For a path $b:y\to z$, we see that $\ga [b\cdot a]=\ga [b]\com \ga [a]$. It
follows that $\ga [a]$ is an isomorphism with inverse $\ga [a^{-1}]$. For
a path $b:y\to x$, we have $\ga[b\cdot a][f] = [b\cdot a][f][(b\cdot a)^{-1}]$.
If the group $\pi_1(X,x)$ happens to be Abelian, which may or may not be the
case, then this is just $[f]$. By taking $b=(a')^{-1}$ for another path
$a': x\to y$, we see that, when  $\pi_1(X,x)$ is Abelian, $\ga [a]$ is 
independent of the choice of the path class $[a]$. Thus, in this case, we have 
a canonical way to identify  $\pi_1(X,x)$ with  $\pi_1(X,y)$.

\section{Homotopy invariance}

For a map $p: X\rtarr Y$, define $p_*: \pi_1(X,x)\rtarr \pi_1(Y,p(x))$ by
$p_*[f] = [p\com f]$, where $p\com f$ is the composite of $p$ with the loop 
$f:I\rtarr X$. Clearly $p_*$ is a homomorphism. The identity map 
$\id: X\rtarr X$ induces the identity homomorphism. For a map
$q: Y\rtarr Z$, $q_*\com p_* = (q\com p)_*$.

Now suppose given two maps $p,q: X\rtarr Y$ and a homotopy $h: p\htp q$.
We would like to conclude that $p_*=q_*$, but this doesn't quite make sense
because homotopies needn't respect basepoints. However, the homotopy $h$
determines the path $a: p(x)\to q(x)$ specified by $a(t)=h(x,t)$, and the
next best thing happens.

\begin{prop}
The following diagram is commutative:
$$\diagram
& \pi_1(X,x) \dlto_{p_*} \drto^{q_*} & \\
\pi_1(Y,p(x)) \rrto_{\ga [a]} & & \pi_1(Y,q(x)).  \\
\enddiagram$$
\end{prop}
\begin{proof}
Let $f: I\rtarr X$ be a loop at $x$. We must show that $q\com f$ is 
equivalent to $a\cdot (p\com f)\cdot a^{-1}$. It is easy to check that this is 
equivalent to showing that $c_{p(x)}$ is equivalent to 
$a^{-1}\cdot (q\com f)^{-1}\cdot a\cdot (p\com f)$.
Define $j: I\times I\rtarr Y$ by $j(s,t)=h(f(s),t)$. Then
$$j(s,0)=(p\com f)(s), \ \ \ j(s,1)=(q\com f)(s),  \ \tand \  j(0,t)=a(t)=j(1,t).$$
Note that $j(0,0)=p(x)$. Schematically, on the boundary of the square, $j$ is
$$\diagram
\rrline^{q\com f}|\tip & &\\
& &\\
\uuline^{a}|\tip \rrline_{p\com f}|\tip & & \uuline_{a}|\tip \\
\enddiagram$$
Thus, going counterclockwise around the boundary starting at $(0,0)$, we
traverse $a^{-1}\cdot (q\com f)^{-1}\cdot a\cdot (p\com f)$. The map $j$ induces
a homotopy through loops between this composite and $c_{p(x)}$. Explicitly,
a homotopy $k$ is given by $k(s,t)=j(r_t(s))$, where $r_t: I\rtarr I\times I$
maps successive quarter intervals linearly onto the edges of the bottom left 
subsquare of $I\times I$ with edges of length $t$, starting at $(0,0)$:
$$\diagram
\rrline \ddline|<(0.75)\tip & & \ddline \\
& \lline|\tip & \\
\rrline|<(0.25)\tip & \uline|\tip & \\
\enddiagram$$
\end{proof} 

\section{Calculations: $\pi_1(\bR )=0$ and $\pi_1(S^1)=\bZ$}

Our first calculation is rather trivial. We take the origin $0$ as a convenient
basepoint for the real line $\bR$.

\begin{lem}
$\pi_1(\bR,0)=0$.
\end{lem}
\begin{proof}
Define $k: \bR\times I\rtarr \bR$ by $k(s,t)=(1-t)s$. Then $k$ is a homotopy
from the identity to the constant map at $0$. For a loop $f: I\rtarr \bR$ at
$0$, define $h(s,t)=k(f(s),t)$. The homotopy $h$ shows that $f$ is equivalent
to $c_0$.
\end{proof}

Consider the circle $S^1$ to be the set of complex numbers $x=y+iz$ of norm $1$,
$y^2+z^2=1$. Observe that $S^1$ is a group under multiplication of complex
numbers. It is a topological group: multiplication is a continuous function.\index{topological
group}
We take the identity element $1$ as a convenient basepoint for $S^1$.

\begin{thm}
$\pi_1(S^1,1)\iso \bZ$.
\end{thm}
\begin{proof}
For each integer $n$, define a loop $f_n$ in $S^1$ by $f_n(s)=e^{2\pi ins}$. This is
the composite of the map  $I\rtarr S^1$ that sends $s$ to $e^{2\pi is}$ and the $n$th
power map on $S^1$; if we identify the boundary points $0$ and $1$ of $I$, then the
first map induces the evident identification of $I/\pa I$ with $S^1$. It is easy
to check that $[f_m][f_n]=[f_{m+n}]$, and we define a homomorphism 
$i:\bZ\rtarr \pi_1(S^1,1)$ by $i(n)=[f_n]$. We claim that $i$ is 
an isomorphism. The idea of the proof is to use the
fact that, locally, $S^1$ looks just like $\bR$. 

Define $p: \bR\rtarr S^1$ by 
$p(s)=e^{2\pi i s}$. Observe that $p$ wraps each interval $[n,n+1]$ around 
the circle, starting at $1$ and going counterclockwise. Since the
exponential function converts addition to multiplication, we easily check that
$f_n=p\com \tilde{f}_n$, where $\tilde{f}_n$ is the path in $\bR$ defined by
$\tilde{f}_n(s)=sn$. 

This lifting of paths works generally. For any path
$f:I\rtarr S^1$ with $f(0)=1$, there is a unique path $\tilde{f}:I\rtarr\bR$
such that $\tilde{f}(0)=0$ and $p\com \tilde{f}=f$. To see this, observe that
the inverse image in $\bR$ of any small connected neighborhood in $S^1$ is a 
disjoint union of a copy of that neighborhood contained in each interval $(r+n,r+n+1)$ for 
some $r\in [0,1)$. Using the fact that $I$ is compact, we see that we can 
subdivide $I$ into finitely many closed subintervals such that $f$ carries each 
subinterval into one of these small connected neighborhoods. Now, proceeding 
subinterval by subinterval, we obtain the required unique lifting of $f$ by
observing that the lifting on each subinterval is uniquely determined by the 
lifting of its initial point. 

Define a function $j:\pi_1(S^1,1)\rtarr \bZ$ by 
$j[f]=\tilde{f}(1)$, the endpoint of the lifted path. This is an integer 
since $p(\tilde{f}(1)) =1$. We must show that this integer is independent 
of the choice of $f$ in its path class $[f]$. In fact, if we have a homotopy
$h: f\htp g$ through loops at $1$, then the homotopy lifts uniquely to a 
homotopy $\tilde{h}: I\times I\rtarr \bR$ such that $\tilde{h}(0,0)=0$ and
$p\com \tilde{h}=h$. The argument is just the same as for $\tilde{f}$: we use
the fact that $I\times I$ is compact to subdivide it into finitely many subsquares
such that $h$ carries each into a small connected neighborhood in $S^1$. We then
construct the unique lift $\tilde{h}$ by proceeding subsquare by subsquare, starting
at the lower left, say, and proceeding upward one row of squares at a time.  By
the uniqueness of lifts of paths, which works just as well for paths with any
starting point, $c(t)=\tilde{h}(0,t)$ and $d(t)=\tilde{h}(1,t)$ specify 
constant paths since $h(0,t)=1$ and $h(1,t)=1$ for all $t$. Clearly $c$ is
constant at $0$, so, again by the uniqueness of lifts of paths, we must have 
$$\tilde{f}(s)=\tilde{h}(s,0) \ \ \ \tand \ \ \ \tilde{g}(s)=\tilde{h}(s,1).$$
But then our second constant path $d$ starts at $\tilde{f}(1)$ and ends at 
$\tilde{g}(1)$. 

Since $j[f_n]=n$ by our explicit formula for $\tilde{f}_n$, 
the composite $j\com i:\bZ\rtarr \bZ$ is the identity. It suffices to 
check that the function $j$ is one-to-one, since then both $i$ and $j$ will be
one-to-one and onto. Thus suppose that $j[f]=j[g]$. This means that
$\tilde{f}(1)=\tilde{g}(1)$. Therefore $\tilde{g}^{-1}\cdot \tilde{f}$ is a 
loop at $0$ in $\bR$. By the lemma, $[\tilde{g}^{-1}\cdot \tilde{f}]=[c_0]$.
It follows upon application of $p_*$ that 
$$[g^{-1}][f]=[g^{-1}\cdot f]=[c_1].$$
Therefore $[f]=[g]$ and the proof is complete.
\end{proof}

\section{The Brouwer fixed point theorem}

Let $D^2$ be the unit disk $\{ y+iz| y^2+z^2\leq 1\}$. Its boundary is $S^1$,
and we let $i: S^1\rtarr D^2$ be the inclusion. Exactly as for $\bR$, we see that 
$\pi_1(D^2)=0$ for any choice of basepoint.

\begin{prop}
There is no continuous map $r: D^2\rtarr S^1$ such that $r\com i=\id$.
\end{prop}
\begin{proof} 
If there were such a map $r$, then the composite homomorphism
$$\diagram
\pi_1(S^1,1)\rto^<(0.2){i_*} & \pi_1(D^2,1) \rto^<(0.2){r_*} & \pi_1(S^1,1) \\
\enddiagram$$
would be the identity. Since the identity homomorphism of $\bZ$ does not factor 
through the zero group, this is impossible.
\end{proof}

\begin{thm}[Brouwer fixed point theorem]\index{Brouwer fixed point theorem}
Any continuous map 
$$f: D^2\rtarr D^2$$ 
has a fixed point.
\end{thm}
\begin{proof}
Suppose that $f(x)\neq x$ for all $x$. Define $r(x)\in S^1$ to be the 
intersection with $S^1$ of the ray that starts at $f(x)$ and passes through $x$.
Certainly $r(x)=x$ if $x\in S^1$. By writing an equation for $r$ in terms of $f$,
we see that $r$ is continuous. This contradicts the proposition.
\end{proof}

\section{The fundamental theorem of algebra}

Let $\io\in\pi_1(S^1,1)$ be a generator. For a map $f: S^1\rtarr S^1$, define
an integer $\text{deg}(f)$\index{degree of a map} by letting the composite
$$\diagram
\pi_1(S^1,1)\rto^<(0.2){f_*} & \pi_1(S^1,f(1)) \rto^{\ga [a]} & \pi_1(S^1,1) \\
\enddiagram$$
send $\io$ to $\text{deg}(f)\io$. Here $a$ is any path $f(1)\to 1$; $\ga [a]$ is 
independent of the choice of $[a]$ since $\pi_1(S^1,1)$ is Abelian. If $f\htp g$,
then $\text{deg}(f)=\text{deg}(g)$ by our homotopy invariance diagram and this independence of
the choice of path. Conversely, our calculation of $\pi_1(S^1,1)$ implies that if 
$\text{deg}(f)=\text{deg}(g)$, then $f\htp g$, but we will not need that for the moment. It is 
clear that $\text{deg}(f)=0$ if $f$ is the constant map at some point. It is also clear that if
$f_n(x)=x^n$, then $\text{deg}(f_n)=n$: we built that fact into our proof that $\pi_1(S^1,1)=\bZ$.

\begin{thm}[Fundamental theorem of algebra]\index{fundamental theorem!of algebra} Let 
$$f(x) = x^n+c_1x^{n-1}+\cdots + c_{n-1}x + c_n$$
be a polynomial with complex coefficients $c_i$, where $n>0$. Then there is a
complex number $x$ such that $f(x)=0$. Therefore there are $n$ such complex
numbers (counted with multiplicities).
\end{thm}
\begin{proof}
Using $f(x)/(x-c)$ for a root $c$, we see that the last statement will follow 
by induction from the first. We may as well assume that $f(x)\neq 0$ for 
$x\in S^1$. This allows us to define $\hat{f}: S^1\rtarr S^1$ by 
$\hat{f}(x)=f(x)/|f(x)|$.  We proceed to calculate $\text{deg}(\hat{f})$. 
Suppose first that $f(x)\neq 0$ for all $x$ such that $|x|\leq 1$.
This allows us to define $h: S^1\times I\rtarr S^1$ by $h(x,t)=f(tx)/|f(tx)|$. 
Then $h$ is a homotopy from the constant map at $f(0)/|f(0)|$ to $\hat{f}$,
and we conclude that $\deg(\hat{f})=0$. Suppose next that $f(x)\neq 0$ for all 
$x$ such that $|x|\geq 1$. This allows us to define $j:S^1\times I\rtarr S^1$ by
$j(x,t)=k(x,t)/|k(x,t)|$, where
$$k(x,t)=t^nf(x/t)=x^n+t(c_1x^{n-1}+tc_2x^{n-2}+\cdots +t^{n-1}c_n).$$
Then $j$ is a homotopy from $f_n$ to $\hat{f}$, and we conclude that 
$\deg(\hat{f})=n$. One of our suppositions had better be false!
\end{proof}

It is to be emphasized how technically simple this is, requiring nothing remotely
as deep as complex analysis. Nevertheless, homotopical proofs like this are 
relatively recent. Adequate language, elementary as it is, was not developed
until the 1930s.

\vspace{.1in}

\begin{center}
PROBLEMS
\end{center}
\begin{enumerate}
\item  Let  $p$  be a polynomial function on $\bC$ which has no root on  $S^1$.
Show that the number of roots of $p(z) = 0$  with $|z| < 1$  is the degree 
of the map  $\hat{p}: S^1\rtarr S^1$  specified by $\hat{p}(z) = p(z)/|p(z)|$.
\item Show that any map  $f: S^1 \rtarr S^1$  such that $\text{deg}(f)\neq 1$  has 
a fixed point.
\item Let $G$  be a topological group and take its identity element $e$ as its basepoint.  
Define the pointwise product of loops $\al$ and $\be$ by $(\al\be)(t) = \al (t)\be (t)$.  
Prove that $\al\be$ is equivalent to the composition of paths $\be\cdot\al$.  
Deduce that $\pi_1(G,e)$ is Abelian. 
\end{enumerate}

\clearpage

\thispagestyle{empty}

\chapter{Categorical language and the van Kampen theorem}

We introduce categorical language and ideas and use them to prove the van Kampen
theorem. This method of computing fundamental groups illustrates the general principle 
that calculations in algebraic topology usually work by piecing together a few pivotal 
examples by means of general constructions or procedures.

\section{Categories}

Algebraic topology concerns mappings from topology to algebra. Category theory gives
us a language to express this. We just record the basic terminology, without being
overly pedantic about it. 

A category\index{category} $\sC$ consists of a collection of objects, a set $\sC(A,B)$ of 
morphisms (also called maps) between
any two objects, an identity morphism $\id_A\in\sC(A,A)$ for each object $A$ (usually
abbreviated $\id$), and a composition law 
$$\com : \sC (B,C)\times\sC (A,B)\rtarr \sC (A,C)$$
for each triple of  objects $A$, $B$, $C$. Composition must be associative, and identity
morphisms must behave as their names dictate:
$$h\com (g\com f)=(h\com g)\com f,\ \ \ \id\com f =f,  \ \tand \  f\com\id=f$$
whenever the specified composites are defined. A category is ``small''\index{category!small} if 
it has a set of objects.

We have the category $\sS$ of sets and functions, the category $\sU$\index{U@$\sU$} of 
topological spaces and continuous functions, the category $\sG$ of groups and homomorphisms, the 
category $\sA b$ of Abelian groups and homomorphisms, and so on.

\section{Functors}

A functor\index{functor} $F:\sC \rtarr \sD$ is a map of categories. It assigns an object 
$F(A)$ of $\sD$
to each object $A$ of $\sC$ and a morphism $F(f):F(A)\rtarr F(B)$ of $\sD$ to each morphism
$f:A\rtarr B$ of $\sC$ in such a way that
$$F(\id_A)=\id_{F(A)}  \ \ \tand \ \ F(g\com f)=F(g)\com F(f).$$
More precisely, this is a covariant functor\index{functor!covariant}. A contravariant 
functor\index{functor!contravariant} $F$ reverses the
direction of arrows, so that $F$ sends $f:A\rtarr B$ to $F(f):F(B)\rtarr F(A)$ and
satisfies $F(g\com f)=F(f)\com F(g)$. A category $\sC$ has an 
opposite category\index{category!opposite}
$\sC^{op}$ with the same objects and with $\sC^{op}(A,B)=\sC(B,A)$. A contravariant
functor $F:\sC\rtarr \sD$ is just a covariant functor $\sC^{op}\rtarr \sD$. 

For example, we have forgetful functors from spaces to sets and from Abelian groups
to sets, and we have the free Abelian group functor from sets to Abelian groups.

\section{Natural transformations}

A natural transformation\index{natural transformation}
$\al: F\rtarr G$ between functors $\sC\rtarr \sD$ is a map 
of functors. It consists of a morphism $\al_A: F(A)\rtarr G(A)$ for each object $A$ of $\sC$
such that the following diagram commutes for each morphism $f:A\rtarr B$ of $\sC$:
$$\diagram
F(A) \rto^{F(f)} \dto_{\al_A} & F(B) \dto^{\al_B}\\
G(A) \rto_{G(f)} & G(B).\\
\enddiagram$$
Intuitively, the maps $\al_A$ are defined in the same way for every $A$.

For example, if $F:\sS\rtarr \sA b$ is the functor that sends a set to the free
Abelian group that it generates and $U:\sA b\rtarr \sS$ is the forgetful functor 
that sends an Abelian group to its underlying set, then we have a natural inclusion of 
sets $S\rtarr UF(S)$. The functors $F$ and $U$ are left adjoint and right 
adjoint\index{functor!adjoint}\index{adjoint functors} to each
other, in the sense that we have a natural isomorphism
$$\sA b(F(S),A)\iso \sS(S,U(A))$$
for a set $S$ and an Abelian group $A$. This just expresses 
the ``universal property''\index{universal property} of 
free objects: a map of sets $S\rtarr U(A)$ extends uniquely to a homomorphism of groups
$F(S)\rtarr A$. Although we won't bother with a formal definition, the notion of an
adjoint pair of functors will play an important role later on.

Two categories $\sC$ and $\sD$ are equivalent\index{equivalent!categories} if there are 
functors $F:\sC\rtarr \sD$
and $G:\sD\rtarr \sC$ and natural isomorphisms $FG\rtarr \Id$ and $GF\rtarr \Id$,
where the $\Id$ are the respective identity functors.

\section{Homotopy categories and homotopy equivalences}

Let $\sT$\index{T@$\sT$} be the category of spaces $X$ with a chosen basepoint $x\in X$; its 
morphisms are continuous maps $X\rtarr Y$ that carry the basepoint of $X$ to the basepoint of $Y$.
The fundamental group specifies a functor $\sT\rtarr\sG$, where $\sG$ is the category of
groups and homomorphisms. 

When we have a (suitable) relation of homotopy between maps in a category $\sC$, we define
the homotopy category\index{category!homotopy}\index{homotopy category} $h\sC$ to be the 
category with the same objects as $\sC$ but with 
morphisms the homotopy classes of maps. We have the homotopy category $h\sU$ of unbased
spaces. On $\sT$, we require homotopies to map basepoint to basepoint at all times $t$,
and we obtain the homotopy category $h\sT$ of based spaces. The fundamental group is a 
homotopy invariant functor\index{functor!homotopy invariant} on $\sT$, in the sense that it 
factors through a functor $h\sT\rtarr \sG$. 

A homotopy equivalence\index{homotopy equivalence} in $\sU$ is an isomorphism in $h\sU$. Less 
mysteriously, a map
$f:X\rtarr Y$ is a homotopy equivalence if there is a map $g: Y\rtarr X$ such that
both $g\com f\htp \id$ and $f\com g\htp \id$. Working in $\sT$, we obtain the
analogous notion of a based homotopy equivalence. Functors carry isomorphisms to 
isomorphisms, so we see that a based homotopy equivalence induces an isomorphism
of fundamental groups. The same is true, less obviously, for unbased homotopy equivalences.

\begin{prop}
If $f: X\rtarr Y$ is a homotopy equivalence, then 
$$f_*:\pi_1(X,x)\rtarr \pi_1(Y,f(x))$$
is an isomorphism for all $x\in X$.
\end{prop}
\begin{proof}
Let $g:Y\rtarr X$ be a homotopy inverse of $f$. By our homotopy invariance diagram, we
see that the composites
$$\pi_1(X,x)\overto{f_*} \pi_1(Y,f(x)) \overto{g_*} \pi_1(X,(g\com f)(x))$$
and
$$\pi_1(Y,y)\overto{g_*} \pi_1(X,g(y)) \overto{f_*} \pi_1(Y,(f\com g)(y))$$
are isomorphisms determined by paths between basepoints given by chosen homotopies
$g\com f\htp\id$ and $f\com g\htp\id$. Therefore, in each displayed composite, the
first map is a monomorphism and the second is an epimorphism. Taking $y=f(x)$ in
the second composite, we see that the second map in the first composite is an
isomorphism. Therefore so is the first map.
\end{proof}

A space $X$ is said to be contractible\index{contractible space} if it is homotopy 
equivalent to a point.

\begin{cor}
The fundamental group of a contractible space is zero.
\end{cor}

\section{The fundamental groupoid}

While algebraic topologists often concentrate on connected spaces with chosen basepoints,
it is valuable to have a way of studying fundamental groups that does not require such
choices. For this purpose, we define the ``fundamental groupoid''\index{fundamental groupoid} 
$\PI(X)$ of a space $X$ to be the category whose objects are the points of $X$ and whose 
morphisms $x\rtarr y$ 
are the equivalence classes of paths from $x$ to $y$. Thus the set of endomorphisms of 
the object $x$ is exactly the fundamental group $\pi_1(X,x)$. 

The term ``groupoid''\index{groupoid} is used for a category all morphisms of which are 
isomorphisms. 
The idea is that a group may be viewed as a groupoid with a single object. Taking morphisms 
to be functors, we obtain the category $\sG\sP$ of groupoids. Then we may view $\PI$ as a 
functor $\sU\rtarr \sG\sP$.

There is a useful notion of a skeleton\index{skeleton!of a category} $sk\sC$ of a category $\sC$. 
This is a ``full'' subcategory \index{full subcategory} with one object from each isomorphism 
class of objects of $\sC$, ``full'' meaning 
that the morphisms between two objects of $sk\sC$ are all of the morphisms between these 
objects in $\sC$. The inclusion functor $J: sk\sC\rtarr \sC$ is an 
equivalence of categories. An inverse functor $F:\sC\rtarr sk\sC$ is obtained by letting 
$F(A)$ be the unique object in $sk\sC$ that is isomorphic to $A$, choosing an isomorphism 
$\al_A: A\rtarr F(A)$, and defining $F(f)=\al_B\com f\com\al_A^{-1}:F(A)\rtarr F(B)$ for
a morphism $f:A\rtarr B$ in $\sC$. We choose $\al$ to be the identity morphism if $A$ is
in $sk\sC$, and then $FJ=\Id$; the $\al_A$ specify a natural isomorphism $\al:\Id\rtarr JF$.

A category $\sC$ is said to be connected\index{category!connected} if any two of its objects 
can be connected by a sequence of morphisms. For example, a sequence $A \longleftarrow B \rtarr C$ 
connects $A$ to $C$, although there need be no morphism $A\rtarr C$. However, a groupoid $\sC$ is  
connected if and only if any two of its objects are isomorphic. The group of endomorphisms of
any object $C$ is then a skeleton of $\sC$. Therefore the previous paragraph specializes to give
the following relationship between the fundamental group and the fundamental groupoid of a 
path connected space $X$.

\begin{prop} Let $X$ be a path connected space. For each point $x\in X$, the inclusion
$\pi_1(X,x)\rtarr \PI(X)$ is an equivalence of categories.
\end{prop}
\begin{proof}
We are regarding $\pi_1(X,x)$ as a category with a single object $x$, and it is a 
skeleton of $\PI(X)$.
\end{proof}

\section{Limits and colimits}

Let $\sD$ be a small category and let $\sC$ be any category. A $\sD$-shaped diagram
in $\sC$ is a functor $F:\sD\rtarr \sC$. A morphism $F\rtarr F'$ of $\sD$-shaped
diagrams\index{diagram!$\sD$-shaped} is a natural transformation, and we have the 
category $\sD[\sC]$ of $\sD$-shaped
diagrams in $\sC$. Any object $C$ of $\sC$ determines the constant diagram $\ul{C}$ that 
sends each object of $\sD$ to $C$ and sends each morphism of $\sD$ to the identity 
morphism of $C$. 

The colimit, $\colim F$,\index{colimit} of a $\sD$-shaped diagram $F$ is an object of $\sC$ 
together with a morphism of diagrams $\io: F\rtarr \ul{\colim F}$ that is 
initial among all such morphisms. This means that if $\et: F\rtarr \ul A$ is 
a morphism of diagrams, then there is a unique map $\tilde{\et}: \colim F\rtarr A$
in $\sC$ such that $\tilde{\eta}\com \io=\et$. Diagrammatically, this property
is expressed by the assertion that, for each map $d:D\rtarr D'$ in $\sD$, we have a 
commutative diagram
$$\diagram
F(D)\rrto^{F(d)} \drto_{\io} \ddrto_{\et} && F(D') \dlto^{\io} \ddlto^{\et}\\
 & \colim F \dto^{\tilde{\et}} & \\
 & A. &\\
\enddiagram$$

The limit\index{limit} of $F$ is defined by reversing arrows: it is an object $\lim F$ of $\sC$ 
together with a morphism of diagrams $\pi: \ul{\lim F}\rtarr F $ that is 
terminal among all such morphisms. This means that if $\epz: \ul A\rtarr F$ is 
a morphism of diagrams, then there is a unique map $\tilde{\epz}: A\rtarr\lim F$
in $\sC$ such that $\pi\com\tilde{\epz}=\epz$. Diagrammatically, this property
is expressed by the assertion that, for each map $d:D\rtarr D'$ in $\sD$, we have a 
commutative diagram
$$\diagram
F(D)\rrto^{F(d)} && F(D') \\
 & \lim F \ulto^{\pi}    \urto_{\pi} & \\
 & A. \uto_{\tilde{\epz}}  \uulto^{\epz} \uurto_{\epz} &\\
\enddiagram$$

If $\sD$ is a set regarded as a discrete category\index{category!discrete} (only identity 
morphisms), then
colimits and limits indexed on $\sD$ are coproducts\index{coproduct} and products\index{product} 
indexed on the set $\sD$.
Coproducts are disjoint unions in $\sS$ or $\sU$, wedges (or one-point unions)\index{wedge} in 
$\sT$, free products in $\sG$, and direct sums in $\sA b$. Products are Cartesian products in
all of these categories; more precisely, they are Cartesian products of underlying sets,
with additional structure. If $\sD$ is the category displayed schematically as
$$\diagram
e & d \lto \rto & f  & \text{or} & d \rto<.5ex> \rto<-.5ex> & d',\\
\enddiagram$$
where we have displayed all objects and all non-identity morphisms, then the co\-limits
indexed on $\sD$ are called pushouts\index{pushout} or coequalizers\index{coequalizer}, 
respectively. Similarly, if $\sD$ is 
displayed schematically as 
$$\diagram
e \rto & d  & f \lto & \text{or} & d \rto<.5ex> \rto<-.5ex> & d',\\
\enddiagram$$
then the limits indexed on $\sD$ are called pullbacks\index{pullback} or 
equalizers,\index{equalizer} respectively.

A given category may or may not have all colimits, and it may have some but not others.
A category is said to be cocomplete\index{category!cocomplete} if it has all colimits, 
complete\index{category!complete} if it has all
limits. The categories $\sS$, $\sU$, $\sT$, $\sG$, and $\sA b$ are complete and cocomplete.
If a category has coproducts and coequalizers, then it is cocomplete, and similarly for
completeness. The proof is a worthwhile exercise.

\section{The van Kampen theorem}

The following is a modern dress treatment of the van Kampen theorem. I should admit that,
in lecture, it may make more sense not to introduce the fundamental groupoid and to
go directly to the fundamental group statement. The direct proof is shorter, but
not as conceptual. However, as far as I know, the deduction of the fundamental group version 
of the van Kampen theorem from the fundamental groupoid version does not appear in the literature
in full generality. The proof well illustrates how to manipulate colimits formally. We have used 
the van Kampen theorem as an excuse to introduce some basic categorical language, and we shall use 
that language heavily in our treatment of covering spaces in the next chapter.

\begin{thm}[van Kampen]\index{van Kampen theorem|(}
Let $\sO=\sset{U}$ be a cover of a space $X$ by path connected open subsets such that 
the intersection of finitely many subsets in $\sO$ is again in $\sO$. Regard $\sO$ as 
a category whose morphisms are the inclusions of subsets and observe that the functor
$\PI$, restricted to the spaces and maps in $\sO$, gives a diagram 
$$\PI|\sO: \sO\rtarr \sG\sP$$ 
of groupoids. The groupoid $\PI(X)$ is the colimit of this diagram. In symbols,
$$\PI (X) \iso \colim_{U\in\sO} \PI(U).$$
\end{thm}
\begin{proof}
We must verify the universal property. For a groupoid $\sC$ and a map 
$\et: \PI|\sO\rtarr \ul{\sC}$ of $\sO$-shaped diagrams of groupoids, we must 
construct a map $\tilde{\et}: \PI(X)\rtarr  \sC$ of groupoids that restricts to $\et_U$ on 
$\PI(U)$ for each $U\in\sO$. On objects, that is on points of $X$, we must define 
$\tilde{\et}(x)=\et_U(x)$ for $x\in U$. This is independent of the choice of $U$
since $\sO$ is closed under finite intersections. If a path $f:x\to y$ lies entirely 
in a particular $U$, then we must define $\tilde{\et}[f]=\et([f])$. Again, since $\sO$
is closed under finite intersections, this specification is independent of the choice 
of $U$ if $f$ lies entirely in more than one $U$. Any path $f$ is the composite of 
finitely many paths $f_i$, each of which does lie in a single $U$, and we must 
define $\tilde{\et}[f]$ to be the composite of the $\tilde{\et}[f_i]$. Clearly this
specification will give the required unique map $\tilde{\et}$, provided that $\tilde{\et}$
so specified is in fact well defined. Thus suppose that $f$ is equivalent to $g$. The
equivalence is given by a homotopy $h:f\htp g$ through paths $x\to y$. We may subdivide 
the square $I\times I$ into subsquares, each of which is mapped into one of the $U$.
We may choose the subdivision so that the resulting subdivision of $I\times\sset{0}$ refines the 
subdivision used to decompose $f$ as the composite of paths $f_i$, and similarly for $g$ and
the resulting subdivision of $I\times\sset{1}$. We see that the relation $[f]=[g]$ in $\PI(X)$ is 
a consequence of a finite number of relations, each of which holds in one of the $\PI(U)$. 
Therefore $\tilde\et([f])=\tilde\eta([g])$. This verifies the universal property and proves 
the theorem.
\end{proof}

The fundamental group version of the van Kampen theorem ``follows formally.'' That 
is, it is an essentially categorical consequence of the version just proved. Arguments 
like this are sometimes called proof by categorical nonsense.

\begin{thm}[van Kampen]\index{van Kampen theorem|)}
Let $X$ be path connected and choose a basepoint $x\in X$. Let $\sO$ 
be a cover of $X$ by path connected open subsets such that the intersection of finitely 
many subsets in $\sO$ is again in $\sO$ and $x$ is in each $U\in\sO$. Regard $\sO$ as 
a category whose morphisms are the inclusions of subsets and observe that the functor
$\pi_1(-,x)$, restricted to the spaces and maps in $\sO$, gives a diagram 
$$\pi_1|\sO: \sO\rtarr \sG$$ 
of groups. The group $\pi_1(X,x)$ is the colimit of this diagram. In symbols,
$$\pi_1(X,x) \iso \colim_{U\in\sO} \pi_1(U,x).$$
\end{thm}

We proceed in two steps.

\begin{lem} The van Kampen theorem holds when the cover $\sO$ is finite.
\end{lem}
\begin{proof} This step is based on the nonsense above about skeleta of categories.
We must verify the universal property, this time in the category of groups. For a group $G$ 
and a map $\et: \pi_1|\sO\rtarr \ul{G}$ of $\sO$-shaped diagrams of groups, we must show 
that there is a unique homomorphism $\tilde{\et}: \pi_1(X,x)\rtarr G$ that restricts to 
$\et_U$ on $\pi_1(U,x)$. Remember that we think of a group as a groupoid with a single 
object and with the elements of the group as the morphisms. The inclusion of categories
$J: \pi_1(X,x)\rtarr \PI(X)$ is an equivalence. An inverse equivalence $F: \PI(X)\rtarr \pi_1(X,x)$
is determined by a choice of  path classes $x\rtarr y$ for $y\in X$; we choose $c_x$ when
$y=x$ and so ensure that $F\com J = \Id$. Because the cover $\sO$ is finite and closed 
under finite intersections, we can choose our paths inductively so that the path $x\rtarr y$ 
lies entirely in $U$ whenever $y$ is in $U$. This ensures that the chosen paths determine 
compatible inverse equivalences 
$F_U: \PI(U)\rtarr \pi_1(U,x)$ to the inclusions $J_U: \pi_1(U,x)\rtarr \PI(U)$. Thus the functors 
$$\diagram
\PI(U)\rto^{F_U} & \pi_1(U,x) \rto^(0.6){\et_U} & G
\enddiagram$$
specify an $\sO$-shaped diagram of groupoids $\PI|\sO\rtarr \ul{G}$. By the fundamental groupoid 
version of the van Kampen theorem, there is a unique map of groupoids
$$\xi: \PI(X)\rtarr G$$
that restricts to $\et_U\com F_U$ on $\PI(U)$ for each $U$. The composite
$$\diagram
\pi_1(X,x)\rto^{J} & \PI(X)\rto^(0.6){\xi} & G
\enddiagram$$
is the required homomorphism $\tilde{\et}$. It restricts to $\et_U$ on $\pi_1(U,x)$
by a little ``diagram chase'' and the fact that $F_U\com J_U=\Id$. It is unique because
$\xi$ is unique. In fact, if we are given $\tilde{\et}: \pi_1(X,x)\rtarr G$ that restricts 
to $\et_U$ on each $\pi_1(U,x)$, then $\tilde{\et}\com F:\PI(X)\rtarr G$ restricts to
$\et_U\com F_U$ on each $\PI(U)$; therefore $\xi = \tilde{\et}\com F$ and thus 
$\xi\com J = \tilde\et$. 
\end{proof}

\begin{proof}[Proof of the van Kampen theorem]
We deduce the general case from the case just proved. Let $\sF$ be the set of those finite
subsets of the cover $\sO$ that are closed under finite intersection. For $\sS\in \sF$, 
let $U_{\sS}$ be the union of the $U$ in $\sS$. Then $\sS$ is a cover of $U_{\sS}$ to
which the lemma applies. Thus 
$$ \colim_{U\in\sS}\pi_1(U,x)\iso \pi_1(U_{\sS},x).$$
Regard $\sF$ as a category with a morphism $\sS\rtarr \sT$ whenever $U_{\sS}\subset U_{\sT}$.
We claim first that
$$\colim_{\sS\in\sF}\pi_1(U_{\sS},x)\iso \pi_1(X,x).$$
In fact, by the usual subdivision argument, any loop $I\rtarr X$ and any 
equivalence $h: I\times I\rtarr X$ between loops has image in some $U_{\sS}$. 
This implies directly that $\pi_1(X,x)$, together with the homomorphisms 
$\pi_1(U_{\sS},x)\rtarr \pi_1(X,x)$, has the universal property that characterizes
the claimed colimit. We claim next that
$$\colim_{U\in\sO}\pi_1(U,x)\iso \colim_{\sS\in\sF}\pi_1(U_{\sS},x),$$
and this will complete the proof. Substituting in the colimit on the right, we have 
$$\colim_{\sS\in\sF}\pi_1(U_{\sS},x)\iso \colim_{\sS\in\sF}\colim_{U\in\sS}\pi_1(U,x).$$
By a comparison of universal properties, this iterated colimit is isomorphic to the single 
colimit 
$$\colim_{(U,\sS)\in (\sO,\sF)}\pi_1(U,x).$$ 
Here the indexing category $(\sO,\sF)$ has objects the pairs $(U,\sS)$ with $U\in \sS$; there 
is a morphism $(U,\sS)\rtarr (V,\sT)$ whenever both $U\subset V$ and $U_{\sS}\subset U_{\sT}$. 
A moment's reflection on the relevant universal properties should convince the reader of the 
claimed identification of colimits: the system on the right differs from the system on the left
only in that the homomorphisms $\pi_1(U,x)\rtarr \pi_1(V,x)$ occur many times in the system
on the right, each appearance making the same contribution to the colimit. If we assume known
a priori that colimits of groups exist, we can formalize this as follows. We have a functor 
$\sO\rtarr \sF$ that sends $U$ to the singleton set $\sset{U}$ and thus a functor 
$\sO\rtarr (\sO,\sF)$ that sends $U$ to $(U,\sset{U})$. The functor $\pi_1(-,x):\sO\rtarr \sG$ 
factors through $(\sO,\sF)$, hence we have an induced map of colimits
$$\colim_{U\in\sO}\pi_1(U,x)\rtarr \colim_{(U,\sS)\in (\sO,\sF)}\pi_1(U,x).$$ 
Projection to the first coordinate gives a functor $(\sO,\sF)\rtarr \sO$. Its composite with
$\pi_1(-,x):\sO\rtarr \sG$ defines the colimit on the right, hence we have an induced map of
colimits
$$\colim_{(U,\sS)\in (\sO,\sF)}\pi_1(U,x)\rtarr \colim_{U\in\sO}\pi_1(U,x).$$
These maps are inverse isomorphisms.
\end{proof}

\section{Examples of the van Kampen theorem}

So far, we have only computed the fundamental groups of the circle and of contractible
spaces. The van Kampen theorem lets us extend these calculations. We now drop notation 
for the basepoint, writing $\pi_1(X)$ instead of $\pi_1(X,x)$.

\begin{prop}
Let $X$ be the wedge of a set of path connected based spaces $X_i$, each of which
contains a contractible neighborhood $V_i$ of its basepoint. Then $\pi_1(X)$ is the
coproduct (= free product)\index{free product} of the groups $\pi_1(X_i)$.
\end{prop}
\begin{proof}
Let $U_i$ be the union of $X_i$ and the $V_j$ for $j\neq i$. We apply the van Kampen
theorem with $\sO$ taken to be the $U_i$ and their finite intersections. Since any 
intersection of two or more of the $U_i$ is contractible, the intersections make no 
contribution to the colimit and the conclusion follows.
\end{proof}

\begin{cor}
The fundamental group of a wedge of circles is a free group with one generator for
each circle.
\end{cor}

Any compact surface is homeomorphic to a sphere, or to a connected sum of tori\index{torus}
$T^2=S^1\times S^1$, or to a connected sum of projective planes\index{projective plane} 
$\bR P^2=S^2/\bZ_2$ (where we write $\bZ_2 = \bZ/2\bZ$).  
We shall see shortly that $\pi_1(\bR P^2) =\bZ_2$. We also have the following observation,
which is immediate from the universal property of products. Using this information,
it is an exercise to compute the fundamental group of any compact surface from the
van Kampen theorem.

\begin{lem}
For based spaces $X$ and $Y$, $\pi_1(X\times Y)\iso \pi_1(X)\times \pi_1(Y)$.
\end{lem}

We shall later use the following application of the van Kampen theorem to prove 
that any group is the fundamental group of some space. We need a definition.

\begin{defn}
A space $X$ is said to be simply connected if it is path connected and satisfies
$\pi_1(X)=0$. 
\end{defn}

\begin{prop}
Let $X=U\cup V$, where $U$, $V$, and $U\cap V$ are path connected open neighborhoods of 
the basepoint of $X$ and $V$ is simply connected. Then $\pi_1(U)\rtarr \pi_1(X)$ is 
an epimorphism whose kernel is the smallest normal subgroup of $\pi_1(U)$ that contains
the image of $\pi_1(U\cap V)$.
\end{prop}
\begin{proof}
Let $N$ be the cited kernel and consider the diagram
$$\diagram
 & \pi_1(U) \drto \drrto & & \\
\pi_1(U\cap V) \drto \urto & &\pi_1(X) \rdashed^(0.4){\xi}|>\tip & \pi_1(U)/N\\
& \pi_1(V)=0 \urto \urrto & & \\
\enddiagram$$
The universal property gives rise to the map $\xi$, and 
$\xi$ is an isomorphism since, by an easy algebraic inspection, $\pi_1(U)/N$ is the 
pushout in the category of groups of the homomorphisms $\pi_1(U\cap V)\rtarr \pi_1(U)$ 
and $\pi_1(U\cap V)\rtarr 0$.
\end{proof}

\vspace{.1in}

\begin{center}
PROBLEMS
\end{center}
\begin{enumerate}
\item Compute the fundamental group of the two-holed torus (the compact surface of genus $2$
obtained by sewing together two tori along the boundaries of an open disk removed from each).
\item The Klein bottle\index{Klein bottle}
$K$ is the quotient space of $S^1\times I$  obtained by identifying 
$(z,0)$  with  $(z^{-1},1)$ for $z\in S^1$.  Compute $\pi_1(K)$.  
\item$*$ Let  $X = \{ (p,q)| p \neq -q\}\subset S^n\times S^n$.  Define a map $f: S^n\rtarr X$ by  
$f(p) = (p,p)$.  Prove that $f$ is a homotopy equivalence.
\item Let $\sC$ be a category that has all coproducts and coequalizers. Prove that $\sC$ is
cocomplete (has all colimits). Deduce formally, by use of opposite categories, that a category 
that has all products and equalizers is complete.
\end{enumerate}

\chapter{Covering spaces}

We run through the theory of covering spaces and their relationship to
fundamental groups and fundamental groupoids. This is standard material, 
some of the oldest in algebraic topology. However, I know of no published 
source for the use that we shall make of the orbit category $\sO(\pi_1(B,b))$ 
in the classification of coverings of a space $B$. This point of view gives us
the opportunity to introduce some ideas that are central to equivariant 
algebraic topology, the study of spaces with group actions. In any case, this 
material is far too important to all branches of mathematics to omit.

\section{The definition of covering spaces}

While the reader is free to think about locally contractible spaces,
weaker conditions are appropriate for the full generality of the theory
of covering spaces. A space $X$ is
said to be locally path connected\index{locally path connected space} if for 
any $x\in X$ and any neighborhood 
$U$ of $x$, there is a smaller neighborhood V of $x$ each of whose points 
can be connected to $x$ by a path in $U$. This is equivalent to the 
seemingly more stringent requirement that the topology of $X$ have a basis 
consisting of path connected open sets. In fact, if $X$ is locally path
connected and $U$ is an open neighborhood of a point $x$, then the set
$$V=\{y\,|\,y \ \text{can be connected to}\ x \ \text{by a path in}\ U\}$$
is a path connected open neighborhood of $x$ that is contained in $U$. 
Observe that if $X$ is connected and locally path connected, then it is
path connected. Throughout this chapter, we assume that all given spaces
are connected and locally path connected. 

\begin{defn}
A map $p: E\rtarr B$ is a covering (or cover, or covering 
space)\index{cover}\index{covering}\index{covering space} if it is 
surjective and if each 
point $b\in B$ has an open neighborhood $V$ such that each component of
$p^{-1}(V)$ is open in $E$ and is mapped homeomorphically onto $V$ by $p$.
We say that a path connected open subset $V$ with this property is a 
fundamental neighborhood\index{fundamental neighborhood} of $B$. We call 
$E$ the total space,\index{total space} $B$ the base
space,\index{base space} and $F_b=p^{-1}(b)$ a fiber\index{fiber} of the 
covering $p$.
\end{defn}

Any homeomorphism is a cover. A product of covers is a cover. The projection
$\bR\rtarr S^1$ is a cover. Each $f_n: S^1\rtarr S^1$ is a cover. The
projection $S^n\rtarr \bR P^n$ is a cover, where the real projective 
space $\bR P^n$ is obtained from $S^n$ by identifying antipodal points. If
$f: A\rtarr B$ is a map (where $A$ is connected and locally path connected)
and $D$ is a component of the pullback of $f$ along $p$, then $p: D\rtarr A$ 
is a cover.

\section{The unique path lifting property}

The following result is abstracted from what we saw in the case of the 
particular cover $\bR\rtarr S^1$. It describes the behavior of $p$ with
respect to path classes and fundamental groups.

\begin{thm}[Unique path lifting]\index{unique path lifting theorem} Let 
$p: E\rtarr B$ be a covering, let $b\in B$, 
and let $e,e'\in F_b$.
\begin{enumerate}
\item[(i)] A path $f:I\rtarr B$ with $f(0)=b$ lifts uniquely to a path
$g:I\rtarr E$ such that $g(0)=e$ and $p\com g=f$.
\item[(ii)] Equivalent paths $f\htp f':I\rtarr B$ that start at $b$ lift
to equivalent paths $g\htp g': I\rtarr E$ that start at $e$, hence 
$g(1)=g'(1)$.
\item[(iii)] $p_*: \pi_1(E,e)\rtarr \pi_1(B,b)$ is a monomorphism.
\item[(iv)] $p_*(\pi_1(E,e'))$ is conjugate to $p_*(\pi_1(E,e))$.
\item[(v)] As $e'$ runs through $F_b$, the groups $p_*(\pi_1(E,e'))$
run through all conjugates of $p_*(\pi_1(E,e))$ in $\pi_1(B,b)$.
\end{enumerate}
\end{thm}
\begin{proof} For (i), subdivide $I$ into subintervals each of which maps to a 
fundamental neighborhood under $f$, and lift $f$ to $g$ inductively by use of 
the prescribed homeomorphism property of fundamental neighborhoods. For (ii),
let $h:I\times I\rtarr B$ be a homotopy $f\htp f'$ through paths
$b\rtarr b'$. Subdivide the square into subsquares each of which maps to a 
fundamental neighborhood under $f$. Proceeding inductively, we see that $h$ lifts
uniquely to a homotopy $H:I\times I\rtarr E$ such that $H(0,0)=e$ and $p\com H=h$.
By uniqueness, $H$ is a homotopy $g\htp g'$ through paths $e\rtarr e'$, where
$g(1)=e'=g'(1)$. Parts (iii)--(v) are formal consequences of (i) and (ii), as we 
shall see in the next section.
\end{proof}

\begin{defn}
A covering $p: E\rtarr B$ is regular\index{covering!regular} if $p_*(\pi_1(E,e))$ is a 
normal subgroup of $\pi_1(B,b)$. It is universal\index{covering!universal} if $E$ is 
simply connected.
\end{defn}

As we shall explain in \S4, for a universal cover $p: E\rtarr B$, the
elements of $F_b$ are in bijective correspondence with the elements of  $\pi_1(B,b)$.
We illustrate the force of this statement.

\begin{exmp}
For $n\geq 2$, $S^n$ is a universal cover of $\bR P^n$. Therefore $\pi_1(\bR P^n)$
has only two elements. There is a unique group with two elements, and this proves
our earlier claim that $\pi_1(\bR P^n)=\bZ_2$. 
\end{exmp}

\section{Coverings of groupoids}

Much of the theory of covering spaces can be recast conceptually in terms 
of fundamental groupoids. This point of view separates the essentials
of the topology from the formalities and gives a convenient language in 
which to describe the algebraic classification of coverings.

\begin{defn} (i) Let $\sC$ be a category and $x$ be an object of $\sC$. The
category $x\backslash \sC$ of objects under $x$ has objects the maps
$f: x\rtarr y$ in $\sC$; for objects $f:x\rtarr y$ and $g: x\rtarr z$, the
morphisms $\ga: f\rtarr g$ in $x\backslash\sC$ are the morphisms
$\ga: y\rtarr z$ in $\sC$ such that $\ga\com f = g:x\rtarr z$. Composition and 
identity maps are given by composition and identity maps in $\sC$. When
$\sC$ is a groupoid, $\ga = g\com f^{-1}$, and the objects of 
$x\backslash \sC$ therefore determine the category. 

(ii) Let $\sC$ be a small groupoid. Define the star\index{star} of $x$, denoted $St(x)$ 
or $St_{\sC}(x)$, to be the set of objects of $x\backslash \sC$, that is, the
set of morphisms of $\sC$ with source $x$. Write $\sC(x,x)=\pi(\sC,x)$ for the 
group of automorphisms of the object $x$.

(iii) Let $\sE$ and $\sB$ be small connected groupoids. A 
covering\index{covering!of a groupoid}
$p: \sE \rtarr \sB$ is a functor that is surjective on objects and
restricts to a bijection 
$$p: St(e)\rtarr St(p(e))$$ 
for each object $e$ of $\sE$. For an object $b$ of $\sB$, let $F_b$ 
denote the set of objects of $\sE$ such that $p(e) = b$. Then
$p^{-1}(St(b))$ is the disjoint union over $e\in F_b$ of $St(e)$.
\end{defn}

Parts (i) and (ii) of the unique path lifting theorem can be restated 
as follows.

\begin{prop} If $p: E\rtarr B$ is a covering of spaces, then the induced
functor $\PI(p): \PI(E)\rtarr \PI(B)$ is a covering of groupoids.
\end{prop}

Parts (iii), (iv), and (v) of the unique path lifting theorem are categorical 
consequences that apply to any covering of groupoids, where they read as follows. 

\begin{prop} Let $p: \sE\rtarr \sB$ be a covering of groupoids, let $b$
be an object of $\sB$, and let $e$ and $e'$ be objects of $F_b$.
\begin{enumerate}
\item[(i)] $p: \pi(\sE ,e)\rtarr \pi(\sB ,b)$ is a monomorphism.
\item[(ii)] $p(\pi(\sE ,e'))$ is conjugate to $p(\pi(\sE,e))$.
\item[(iii)] As $e'$ runs through $F_b$, the groups $p(\pi(E,e'))$
run through all conjugates of $p(\pi(\sE ,e))$ in $\pi(\sB ,b)$.
\end{enumerate}
\end{prop}
\begin{proof} For (i), if $g,g'\in \pi(\sE ,e)$ and $p(g) = p(g')$, then
$g = g'$ by the injectivity of $p$ on $St(e)$. For (ii), there is a map
$g: e\rtarr e'$ since $\sE$ is connected. Conjugation by $g$ gives a 
homomorphism $\pi(\sE ,e)\rtarr \pi(\sE, e')$ that maps under $p$ to 
conjugation of $\pi(\sB,b)$ by its element $p(g)$.
For (iii), the surjectivity of $p$ on $St(e)$ gives that any $f\in\pi(\sB ,b)$
is of the form $p(g)$ for some $g\in St(e)$. If $e'$ is the target of $g$, 
then $p(\pi(\sE,e'))$ is the conjugate of $p(\pi(\sE,e))$ by $f$. 
\end{proof}

The fibers $F_b$ of a covering of groupoids are related by translation functions.

\begin{defn}  Let $p: \sE\rtarr \sB$ be a covering of groupoids. Define the
fiber translation functor\index{fiber translation functor} $T=T(p): \sB\rtarr \sS$ 
as follows. For an object $b$ of 
$\sB$, $T(b) = F_b$. For a morphism $f: b\rtarr b'$ of $\sB$, $T(f): F_b\rtarr F_{b'}$
is specified by $T(f)(e) = e'$, where $e'$ is the target of the unique $g$ in
$St(e)$ such that $p(g)=f$.
\end{defn}

It is an exercise from the definition of a covering of a groupoid to verify that 
$T$ is a well defined functor. For a covering space $p: E\rtarr B$ and a path  
$f: b\rtarr b'$, $T(f): F_b\rtarr F_{b'}$ is given by $T(f)(e)=g(1)$ where $g$ 
is the path in $E$ that starts at $e$ and covers $f$.

\begin{prop} Any two fibers $F_b$ and $F_{b'}$ of a covering of groupoids have the
same cardinality. Therefore any two fibers of a covering of spaces have the same
cardinality.
\end{prop}
\begin{proof}
For $f: b\rtarr b'$, $T(f): F_b\rtarr F_{b'}$ is a bijection with inverse $T(f^{-1})$.
\end{proof}

\section{Group actions and orbit categories}

The classification of coverings is best expressed in categorical language
that involves actions of groups and groupoids on sets.

A (left) action of a group $G$\index{group action} on a set $S$ is a function 
$G\times S\rtarr S$ such that $es=s$ (where $e$ is the identity element) and 
$(g'g)s=g'(gs)$ for all $s\in S$. The {\em isotropy group}\index{isotropy group} 
$G_s$ of a point $s$ is the subgroup $\{ g| gs=s\}$ of $G$. 
An action is {\em free}\index{group action!free} if $gs=s$ implies $g=e$, that is, 
if $G_s=e$ for every $s\in S$. 

The orbit generated by a point $s$ is $\{ gs|g\in G\}$. An action is 
{\em transitive}\index{group action!transitive} 
if for every pair $s,s'$ of elements of $S$, there is an element $g$ of $G$ such 
that $gs=s'$. Equivalently, $S$ consists of a single orbit. If $H$ is a subgroup 
of $G$, the set $G/H$ of cosets $gH$ is a transitive $G$-set. When $G$ acts 
transitively on a set $S$, we obtain an isomorphism of $G$-sets between $S$ and 
the $G$-set $G/G_s$ for any fixed $s\in S$ by sending $gs$ to the coset $gG_s$. 

The following lemma describes the group of automorphisms of a transitive \linebreak
$G$-set $S$.
For a subgroup $H$ of $G$, let $NH$ denote the normalizer of $H$ in $G$ and define 
$WH=NH/H$. Such quotient groups $WH$ are sometimes called Weyl groups.\index{Weyl group}

\begin{lem} Let $G$ act transitively on a set $S$, choose $s\in S$, and let $H=G_s$.
Then $WH$ is isomorphic to the group {\em Aut}$_G(S)$\index{AutGS@Aut$_G(S)$} of automorphisms 
of the $G$-set $S$.
\end{lem}
\begin{proof}
For $n\in NH$ with image $\bar{n}\in WH$, define an automorphism  $\ph(\bar{n})$ of $S$
by $\ph(\bar{n})(gs)=gns$. For an automorphism $\ph$ of $S$, we have $\ph(s)=ns$ for some
$n\in G$. For $h\in H$, $hns=\ph(hs)=\ph(s)=ns$, hence $n^{-1}hn\in G_s=H$ and $n\in NH$. 
Clearly $\ph=\ph(\bar{n})$, and it is easy to check that this bijection between $WH$ and 
Aut$_G(S)$ is an isomorphism of groups.
\end{proof}

We shall also need to consider $G$-maps between different $G$-sets $G/H$.

\begin{lem} A $G$-map $\al: G/H\rtarr G/K$ has the form $\al(gH) = g\ga K$, where
the element $\ga\in G$ satisfies $\ga^{-1} h \ga\in K$ for all $h\in H$. 
\end{lem}
\begin{proof}
If $\al(eH) = \ga K$, then the relation
$$\ga K = \al (eH) = \al(hH) = h\al(eH) = h\ga K$$
implies that $\ga^{-1}h\ga\in K$ for $h\in H$.
\end{proof}

\begin{defn} The category $\sO(G)$ of canonical 
orbits\index{category!of canonical orbits}\index{orbit category} has 
objects the $G$-sets $G/H$ and morphisms the $G$-maps of $G$-sets.
\end{defn}

The previous lemmas give some feeling for the structure of $\sO(G)$ and lead
to the following alternative description.

\begin{lem} The category $\sO(G)$ is isomorphic to the category $\sG$ whose
objects are the subgroups of $G$ and whose morphisms are the distinct 
subconjugacy relations $\ga^{-1}H\ga \subset K$ for $\ga\in G$.
\end{lem}

If we regard $G$ as a category with a single object, then a (left) action of 
$G$ on a set $S$ is the same thing as a covariant functor $G\rtarr \sS$.  
(A right action is the same thing as a contravariant functor.)  If $\sB$ is
a small groupoid, it is therefore natural to think of a covariant functor 
$T:\sB\rtarr \sS$ as a generalization of a group action.\index{groupoid action} For 
each object $b$ of $\sB$, $T$ restricts to an action of $\pi(\sB ,b)$ on $T(b)$. We 
say that the functor $T$ is {\em transitive}\index{groupoid action!transitive} if 
this group action is transitive for each object $b$. If $\sB$ is connected, this
holds for all objects $b$ if it holds for any one object $b$.

For example, for a covering of groupoids $p: \sE\rtarr \sB$, the fiber translation 
functor $T$ restricts to give an action of $\pi(\sB,b)$ on the set $F_b$. For $e\in F_b$, 
the isotropy group of $e$ is precisely $p(\pi(\sE,e))$. That is, $T(f)(e)=e$ if and only 
if the lift of $f$ to an element of $St(e)$ is an automorphism of $e$. Moreover, the 
action is transitive since there is an isomorphism in $\sE$ connecting any two points of 
$F_b$. Therefore, as a $\pi(\sB,b)$-set,
$$F_b\iso \pi(\sB,b)/p(\pi(\sE,e)).$$

\begin{defn} A covering $p: \sE\rtarr \sB$ of groupoids is 
regular\index{covering!regular} if 
$p(\pi(\sE,e))$ is a normal subgroup of $\pi(\sB,b)$. It is 
universal\index{covering!universal} if $p(\pi(\sE,e))=\sset{e}$.
Clearly a covering space is regular or universal if and only if its 
associated covering of fundamental groupoids is regular or universal.
\end{defn}

A covering of groupoids is universal if and only if $\pi(\sB,b)$ acts freely on $F_b$, 
and then $F_b$ is isomorphic to $\pi(\sB,b)$ as a $\pi(\sB,b)$-set. Specializing
to covering spaces, this sharpens our earlier claim that the elements of $F_b$ and 
$\pi_1(B,b)$ are in bijective correspondence.

\section{The classification of coverings of groupoids}

Fix a small connected groupoid $\sB$ throughout this section and the next. We explain 
the classification of coverings of $\sB$. This gives an algebraic prototype for
the classification of coverings of spaces. We begin with a result that should be 
called the fundamental theorem\index{fundamental theorem!of covering groupoid theory}
of covering groupoid theory. We assume once and for 
all that all given groupoids are small and connected. 

\begin{thm}
Let $p: \sE\rtarr \sB$ be a covering of groupoids, let $\sX$ be a groupoid, and let 
$f: \sX\rtarr \sB$ be a functor. Choose a base object $x_0\in \sX$, let $b_0=f(x_0)$, and 
choose $e_0\in F_{b_0}$. Then there exists a functor $g: \sX\rtarr \sE$ such that 
$g(x_0)=e_0$ and $p\com g= f$ if and only if
$$f(\pi(\sX,x_0))\subset p(\pi(\sE,e_0))$$
in $\pi(\sB,b_0)$. When this condition holds, there is a unique such functor $g$.
\end{thm}
\begin{proof}
If $g$ exists, its properties directly imply that $\im(f)\subset\im(p)$. For an
object $x$ of $\sX$ and a map $\al: x_0\rtarr x$ in $\sX$, let $\tilde \al$
be the unique element of $St(e_0)$ such that $p(\tilde\al)=f(\al)$. If $g$ 
exists, $g(\al)$ must be $\tilde\al$ and therefore $g(x)$ must be the target
$T(f(\al))(e_0)$ of $\tilde\al$. The inclusion $f(\pi(\sX,x_0))\subset p(\pi(\sE,e_0))$
ensures that $T(f(\al))(e_0)$ is independent of the choice of $\al$, so that $g$ so 
specified is a well defined functor. In fact, given another map $\al': x_0\rtarr x$,
$\al^{-1}\com\al'$ is an element of $\pi(\sX,x_0)$. Therefore 
$$f(\al)^{-1}\com f(\al') = f(\al^{-1}\com\al') = p(\be)$$
for some $\be\in \pi(\sE,e_0)$. Thus
$$p(\tilde\al\com \be) = f(\al)\com p(\be) = f(\al)\com f(\al)^{-1}\com f(\al') = f(\al').$$
This means that $\tilde\al\com\be$ is the unique element $\tilde\al'$ of $St(e_0)$ such that
$p(\tilde\al')=f(\al')$, and its target is the target of $\tilde\al$, as required.
\end{proof}

\begin{defn}
A map $g: \sE\rtarr \sE'$ of coverings of $\sB$ is a functor $g$ such that the 
following diagram of functors is commutative:
$$\diagram
\sE \drto_p \rrto^g & & \sE' \dlto^{p'}\\
& \sB. & \\
\enddiagram$$
Let Cov$(\sB)$\index{CovBb@Cov$(\sB)$} denote the category of coverings of $\sB$; when $\sB$ 
is understood, we write Cov$(\sE,\sE')$ for the set of maps $\sE \rtarr \sE'$ of coverings 
of $\sB$.
\end{defn}

\begin{lem}
A map $g:\sE\rtarr \sE'$ of coverings is itself a covering.
\end{lem}
\begin{proof}
The functor $g$ is surjective on objects since, if $e'\in \sE'$ and we choose an object 
$e\in \sE$ and a map $f: g(e)\rtarr e'$ in $\sE'$, then $e'= g(T(p'(f))(e))$. 
The map $g: St_{\sE}(e)\rtarr St_{\sE'}(g(e))$ is a bijection since its 
composite with the bijection $p': St_{\sE'}(g(e))\rtarr St_{\sB}(p'(g(e)))$ 
is the bijection $p: St_{\sE}(e)\rtarr St_{\sB}(p(e))$. 
\end{proof}

The fundamental theorem immediately determines all maps of coverings of $\sB$ in
terms of group level data.

\begin{thm}
Let $p:\sE\rtarr \sB$ and $p': \sE'\rtarr \sB$ be coverings and choose base objects
$b\in \sB$, $e\in \sE$, and $e'\in \sE'$ such that $p(e)=b=p'(e')$. There exists a map 
$g:\sE\rtarr \sE'$ of coverings with $g(e)=e'$ if and only if
$$p(\pi(\sE,e))\subset p'(\pi(\sE',e')),$$
and there is then only one such $g$. In particular, two maps of covers $g,g': \sE\rtarr \sE'$ 
coincide if $g(e)=g'(e)$ for any one object $e\in \sE$. Moreover, $g$ is an isomorphism if and 
only if the displayed inclusion of subgroups of $\pi(\sB,b)$ is an equality. Therefore 
$\sE$ and $\sE'$ are isomorphic if and only if $p(\pi(\sE,e))$ and $p'(\pi(\sE',e'))$ are 
conjugate whenever $p(e)=p'(e')$. 
\end{thm}

\begin{cor} If it exists, the universal cover of $\sB$ is unique up to isomorphism and 
covers any other cover. 
\end{cor}

That the universal cover does exist will be proved in the next section.
It is useful to recast the previous theorem in terms of actions on fibers.

\begin{thm} Let $p:\sE\rtarr \sB$ and $p': \sE'\rtarr \sB$ be coverings, choose a base 
object $b\in \sB$, and let $G=\pi(\sB,b)$. If $g: \sE\rtarr \sE'$ is a map of coverings, 
then $g$ restricts to a map $F_b\rtarr F'_b$ of $G$-sets, and restriction to fibers 
specifies a bijection between {\em Cov}$(\sE,\sE')$ and the set of $G$-maps $F_b\rtarr F'_{b}$.
\end{thm}
\begin{proof}
Let $e\in F_b$ and $f\in \pi(\sB,b)$. By definition, $fe$ is the target of the map 
$\tilde f\in St_{\sE}(e)$ such that $p(\tilde f)=f$. Clearly $g(fe)$ is the target
of $g(\tilde f)\in St_{\sE'}(g(e))$ and $p'(g(\tilde f))= p(\tilde f) = f$. Again
by definition, this gives $g(fe) = fg(e)$. The previous theorem shows that restriction
to fibers is an injection on Cov$(\sE,\sE')$. To show surjectivity, let $\al: F_b\rtarr F'_{b}$ 
be a $G$-map. Choose $e\in F_b$ and let $e'=\al(e)$. Since $\al$ is a $G$-map, the isotropy 
group $p(\pi(\sE,e))$ of $e$ is contained in the isotropy group $p'(\pi(\sE',e'))$ of $e'$. 
Therefore the previous theorem ensures the existence of a covering map $g$ that restricts to 
$\al$ on fibers.
\end{proof}

\begin{defn} Let Aut$(\sE)\subset$\, Cov$(\sE,\sE)$\index{AutEb@Aut($\sE$)} denote the group of 
automorphisms of a cover $\sE$. Note that, since it is possible to have conjugate subgroups 
$H$ and $H'$ of a group $G$ such that $H$ is a proper subgroup of $H'$, it is possible to have 
a map of covers $g: \sE \rtarr \sE$ such that $g$ is not an isomorphism.
\end{defn}

\begin{cor}
Let $p:\sE\rtarr \sB$ be a covering and choose objects $b\in \sB$ and $e\in F_b$. Write 
$G=\pi(\sB,b)$ and $H=p(\pi(\sE,e))$. Then {\em Aut}$(\sE)$ is isomorphic to the group of 
automorphisms of the $G$-set $F_b$ and therefore to the group $WH$. If $p$ is 
regular,\index{covering!regular} then {\em Aut}$(\sE)\iso G/H$. If $p$ is 
universal,\index{covering!universal} then {\em Aut}$(\sE)\iso G$.
\end{cor}

\section{The construction of coverings of groupoids}

We have given an algebraic classification of all possible covers of $\sB$: there
is at most one isomorphism class of covers corresponding to each conjugacy class of
subgroups of $\pi(\sB,b)$. We show that all of these possibilities are actually 
realized. Since this algebraic result is not needed in the proof of its topological 
analogue, we shall not give complete details. 

\begin{thm} Choose a base object $b$ of $\sB$ and let $G=\pi(\sB,b)$. There is a functor 
$$\sE(-): \sO(G)\rtarr \text{\em{Cov}}(\sB)$$
that is an equivalence of categories. For each subgroup $H$ of $G$, the covering 
$p: \sE(G/H)\rtarr \sB$ has a canonical base object $e$ in its fiber over $b$ such that 
$$p(\pi(\sE(G/H),e)) = H.$$
Moreover, $F_b=G/H$ as a $G$-set and, for a $G$-map $\al: G/H\rtarr G/K$ in $\sO(G)$, 
the restriction of  $\sE(\al): \sE(G/H)\rtarr \sE(G/K)$ to fibers over $b$ coincides 
with $\al$. 
\end{thm}
\begin{proof}
The idea is that, up to bijection, $St_{\sE(G/H)}(e)$ must be the same set for each $H$, 
but the nature of its points can differ with $H$. At one extreme, $\sE(G/G)=\sB$, $p=\id$, 
$e=b$, and the set of morphisms from $b$ to any other object $b'$ is a copy of $\pi(\sB,b)$. 
At the other extreme, $\sE(G/e)$ is a universal cover of $\sB$ and there is just one 
morphism from $e$ to any other object $e'$. In general, the set of objects of $\sE(G/H)$
is defined to be $St_{\sB}(b)/H$, the coset of the identity morphism being $e$. Here $G$ 
and hence its subgroup $H$ act from the right on $St_{\sB}(b)$ by composition in $\sB$.  
We define $p: \sE(G/H)\rtarr \sB$ on objects by letting $p(fH)$ be the target of $f$, which 
is independent of the coset representative $f$. We define morphism sets by
$$ \sE(G/H)(fH,f'H) = \sset{f'\com h\com f^{-1} | h\in H} \subset \sB(p(fH),p(f'H)).$$
Again, this is independent of the choices of coset representatives $f$ and $f'$. Composition 
and identities are inherited from those of $\sB$, and $p$ is given on morphisms by the 
displayed inclusions. It is easy to check that $p: \sE(G/H)\rtarr \sB$ is a covering, 
and it is clear that $p(\pi(\sE(G/H),e)) = H$.

This defines the object function of the functor $\sE: \sO(G)\rtarr \text{Cov}(\sB)$.
To define $\sE$ on morphisms, consider $\al: G/H\rtarr G/K$. If $\al(eH) =gK$, then
$g^{-1}Hg\subset K$ and $\al(fH)=fg K$.  The functor $\sE(\al):\sE(G/H)\rtarr \sE(G/K)$ 
sends the object $fH$ to the object $\al(fH)=fgK$ and sends the morphism $f'\com h\com f^{-1}$
to the same morphism of $\sB$ regarded as $f'g\com g^{-1}hg\com g^{-1}f^{-1}$. It is easily
checked that each $\sE(\al)$ is a well defined functor, and that $\sE$ is functorial in $\al$.

To show that the functor $\sE(-)$ is an equivalence of categories, it suffices to show that 
it maps the morphism set $\sO(G)(G/H,G/K)$ bijectively onto the morphism set 
Cov$(\sE(G/H),\sE(G/K))$ and that every covering of $\sB$ is isomorphic to one of the 
coverings $\sE(G/H)$. These statements are immediate from the results of the previous section.
\end{proof}

The following remarks place the orbit category $\sO(\pi(\sB,b))$ in perspective by relating 
it to several other equivalent categories.

\begin{rem} Consider the category $\sS^{\sB}$ of functors $T:\sB\rtarr \sS$ and natural 
transformations. Let $G=\pi(\sB,b)$. Regarding $G$ as a category with one object $b$, it is
a skeleton of $\sB$, hence the inclusion $G \subset \sB$ is an equivalence of categories.
Therefore, restriction of functors $T$ to $G$-sets $T(b)$ gives an equivalence of categories 
from $\sS^{\sB}$ to the category of $G$-sets. This restricts to an equivalence between the
respective subcategories of transitive objects. We have chosen to focus on transitive objects
since we prefer to insist that coverings be connected. The inclusion of the orbit category
$\sO(G)$ in the category of transitive $G$-sets is an equivalence of categories because 
$\sO(G)$ is a full subcategory that contains a skeleton. We could shrink $\sO(G)$ to a 
skeleton by choosing one $H$ in each conjugacy class of subgroups of $G$, but the resulting 
equivalent subcategory is a less natural mathematical object. 
\end{rem}

\section{The classification of coverings of spaces}

In this section and the next, we shall classify covering spaces and their maps by arguments 
precisely parallel to those for covering groupoids in the previous sections. In fact, applied 
to the associated coverings of fundamental groupoids, some of the algebraic results directly 
imply their topological analogues. We begin with the following result, which deserves
to be called the fundamental theorem of covering space theory and has many other
applications. It asserts that the fundamental group gives the only ``obstruction''
to solving a certain lifting problem. Recall our standing assumption that all
given spaces are connected and locally path connected.\index{fundamental 
theorem!of covering space theory}

\begin{thm}
Let $p: E\rtarr B$ be a covering and let $f: X\rtarr B$ be a continuous map.
Choose $x\in X$, let $b=f(x)$, and choose $e\in F_{b}$. There exists
a map $g: X\rtarr E$ such that $g(x)=e$ and $p\com g= f$ if and only if
$$f_*(\pi_1(X,x))\subset p_*(\pi_1(E,e))$$
in $\pi_1(B,b)$. When this condition holds, there is a unique such map $g$.
\end{thm}
\begin{proof}
If $g$ exists, its properties directly imply that $\im(f_*)\subset\im(p_*)$. 
Thus assume that  $\im(f_*)\subset\im(p_*)$. Applied to the covering 
$\PI(p): \PI(E)\rtarr \PI(B)$, the analogue for groupoids gives a functor 
$\PI(X)\rtarr \PI(E)$ that restricts on objects to the unique map $g: X\rtarr E$ 
of sets such that $g(x)=e$ and $p\com g= f$. We need only check that $g$ is 
continuous, and this holds because $p$ is a local homeomorphism. In detail, if 
$y\in X$ and $g(y)\in U$, where $U$ is an open subset of $E$, then there is a smaller 
open neighborhood $U'$ of $g(y)$ that $p$ maps homeomorphically onto an open subset $V$ 
of $B$. If $W$ is any path connected neighborhood of $y$ such that $f(W)\subset V$, 
then $g(W)\subset U'$ by inspection of the definition of $g$.
\end{proof}

\begin{defn}
A map $g: E\rtarr E'$ of coverings over $B$ is a map $g$ such that the following
diagram is commutative:
$$\diagram
E \drto_p \rrto^g & & E' \dlto^{p'}\\
& B. & \\
\enddiagram$$
Let Cov$(B)$\index{CovBa@Cov$(B)$} denote the category of coverings of the space $B$; when $B$ 
is understood,
we write Cov$(E,E')$ for the set of maps $E \rtarr E'$ of coverings of $B$.
\end{defn}

\begin{lem}
A map $g:E\rtarr E'$ of coverings is itself a covering.
\end{lem}
\begin{proof}
The map $g$ is surjective by the algebraic analogue. The fundamental neighborhoods 
for $g$ are the components of the inverse images in $E'$ of the neighborhoods of 
$B$ which are fundamental for both $p$ and $p'$. 
\end{proof}

The following remarkable theorem is an immediate consequence of the
fundamental theorem of covering space theory.

\begin{thm}
Let $p:E\rtarr B$ and $p': E'\rtarr B$ be coverings and choose $b\in B$, $e\in E$, and 
$e'\in E'$ such that $p(e)=b=p'(e')$.
There exists a map $g:E\rtarr E'$ of coverings with $g(e)=e'$ if and only if
$$p_*(\pi_1(E,e))\subset p'_*(\pi_1(E',e')),$$
and there is then only one such $g$. In particular, two maps of covers $g,g': E\rtarr E'$
coincide if $g(e)=g'(e)$ for any one $e\in E$. Moreover, $g$ is a homeomorphism if and only
if the displayed inclusion of subgroups of $\pi_1(B,b)$ is an equality. Therefore $E$ and
$E'$ are homeomorphic if and only if $p_*(\pi_1(E,e))$ and $p'_*(\pi_1(E',e'))$ are
conjugate whenever $p(e)=p'(e')$.
\end{thm}

\begin{cor} If it exists, the universal cover of $B$ is unique up to isomorphism and 
covers any other cover. 
\end{cor}

Under a necessary additional hypothesis on $B$, we shall prove in the next section that 
the universal cover does exist. 

We hasten to add that the theorem above is atypical of algebraic topology. It is not 
usually the case that algebraic invariants like the fundamental group totally determine 
the existence and uniqueness of maps of topological spaces with prescribed properties. 
The following immediate implication of the theorem gives one explanation.

\begin{cor} The fundamental groupoid functor induces a bijection
$$\text{\em Cov}(E,E') \rtarr \text{\em Cov}(\PI(E),\PI(E')).$$
\end{cor}

Just as for groupoids, we can recast the theorem in terms of fibers. In fact, via the
previous corollary, the following result is immediate from its analogue for groupoids.

\begin{thm} Let $p:E\rtarr B$ and $p': E'\rtarr B$ be coverings, choose a basepoint 
$b\in B$, and let $G=\pi_1(B,b)$. If $g: E\rtarr E'$ is a map of coverings, 
then $g$ restricts to a map $F_b\rtarr F'_b$ of $G$-sets, and restriction to fibers 
specifies a bijection between {\em Cov}$(E,E')$ and the set of $G$-maps $F_b\rtarr F'_{b}$.
\end{thm}

\begin{defn} Let Aut$(E)\subset$\,Cov$(E,E)$\index{AutEb@Aut($\sE$)} denote the group of 
automorphisms of a cover $E$. Again, just as for groupoids, it is possible to have 
a map of covers $g: E \rtarr E$ such that $g$ is not an isomorphism.
\end{defn}

\begin{cor}
Let $p:E\rtarr B$ be a covering and choose $b\in B$ and $e\in F_b$. Write $G=\pi_1(B,b)$
and $H=p_*(\pi_1(E,e))$. Then {\em Aut}$(E)$ is isomorphic to the group of automorphisms of the 
$G$-set $F_b$ and therefore to the group $WH$. If $p$ is regular,\index{covering!regular}
then {\em Aut}$(E)\iso G/H$. If $p$ is universal,\index{covering!universal} then 
{\em Aut}$(E)\iso G$.
\end{cor}

\section{The construction of coverings of spaces}

We have now given an algebraic classification of all possible covers of $B$: there
is at most one isomorphism class of covers corresponding to each conjugacy class of
subgroups of $\pi_1(B,b)$. We show here that all of these possibilities are actually 
realized.  We shall first construct universal covers and then show that the existence 
of universal covers implies the existence of all other possible covers. Again, while 
it suffices to think in terms of locally contractible spaces, appropriate generality
demands a weaker hypothesis. We say that a space $B$ is semi-locally simply
connected\index{semi-locally simply connected space} if every point $b\in B$ has a 
neighborhood $U$ such that $\pi_1(U,b)\rtarr \pi_1(B,b)$ is the trivial homomorphism.

\begin{thm}
If $B$ is connected, locally path connected, and semi-locally simply connected,
then $B$ has a universal cover.\index{covering!universal}
\end{thm}
\begin{proof}
Fix a basepoint $b\in B$. We turn the properties of paths that must hold in
a universal cover into a construction. Define $E$ to be the set of equivalence
classes of paths $f$ in $B$ that start at $b$ and define $p:E\rtarr B$ by
$p[f]=f(1)$. Of course, the equivalence relation is homotopy through paths from
$b$ to a given endpoint, so that $p$ is well defined. Thus, as a set, $E$ is 
just $St_{\PI(B)}(b)$, exactly as in the construction of the universal cover of $\PI(B)$.
The topology of $B$ has a basis consisting of path connected open subsets $U$ such that 
$\pi_1(U,u)\rtarr \pi_1(B,u)$ is trivial for all $u\in U$. Since every loop in $U$ is 
equivalent in $B$ to the trivial loop, any two paths $u\rtarr u'$ in such a $U$ are 
equivalent in $B$. We shall topologize $E$ so that $p$ is a cover with these $U$ as 
fundamental neighborhoods. For a path $f$ in $B$ that starts at $b$ and ends in $U$, 
define a subset $U[f]$ of $E$ by
$$U[f]=\{[g]\,|\,[g]=[c\cdot f]\ \text{for some}\ c:I\rtarr U\}.$$
The set of all such $U[f]$ is a basis for a topology on $E$ since if $U[f]$ and 
$U'[f']$ are two such sets and $[g]$ is in their intersection, then
$$W[g]\subset U[f]\cap U'[f']$$
for any open set $W$ of $B$ such that $p[g]\in W\subset U\cap U'$. For $u\in U$, 
there is a unique $[g]$ in each $U[f]$ such that $p[g]=u$. Thus $p$ maps $U[f]$
homeomorphically onto $U$ and, if we choose a basepoint $u$ in $U$, then $p^{-1}(U)$ 
is the disjoint union of those $U[f]$ such that $f$ ends at $u$. It only remains to show 
that $E$ is connected, locally path connected, and simply connected, and the second of
these is clear. Give $E$ the basepoint $e=[c_b]$. For $[f]\in E$, define a path 
$\tilde{f}:I\rtarr E$ by $\tilde{f}(s)=[f_s]$, where $f_s(t)=f(st)$; $\tilde{f}$ is 
continuous since each $\tilde{f}^{-1}(U[g])$ is open by the definition of $U[g]$ and 
the continuity of $f$. Since $\tilde{f}$ starts at $e$ and ends at $[f]$, $E$ is path 
connected. Since $f_s(1)=f(s)$, $p\com\tilde{f}=f$. Thus, by definition,
$$T[f](e)=[\tilde{f}(1)]=[f].$$
Restricting attention to loops $f$, we see that $T[f](e)=e$ if and only if $[f]=e$
as an element of $\pi_1(B,b)$. Thus the action of $\pi_1(B,b)$ on $F_b$ is free and
the isotropy group $p_*(\pi_1(E,e))$ is trivial. 
\end{proof}

We shall construct general covers by passage to orbit spaces from the universal cover,
and we need some preliminaries.

\begin{defn} A $G$-space $X$ is a space $X$ that is a $G$-set with continuous
action map $G\times X\rtarr X$. Define the orbit space\index{orbit space} $X/G$ to 
be the set of orbits $\sset{Gx|x\in X}$ with its topology as a quotient space of $X$. 
\end{defn}

The definition makes sense for general topological groups $G$. However, our interest
here is in discrete groups $G$, for which the continuity condition just means that 
action by each element of $G$ is a homeomorphism. The functoriality on $\sO(G)$ of
our construction of general covers will be immediate from the following observation.

\begin{lem} Let $X$ be a $G$-space. Then passage to orbit spaces
defines a functor $X/(-): \sO(G)\rtarr \sU$. 
\end{lem}
\begin{proof}
The functor sends $G/H$ to $X/H$ and sends a map $\al: G/H\rtarr G/K$
to the map $X/H\rtarr X/K$ that sends the coset $Hx$ to the coset $K\ga^{-1}x$, 
where $\al$ is given by the subconjugacy relation $\ga^{-1}H\ga\subset K$.
\end{proof}

The starting point of the construction of general covers is the following 
description of regular covers and in particular of the universal cover.

\begin{prop}
Let $p: E\rtarr B$ be a cover such that {\em Aut}$(E)$ acts transitively on $F_b$. Then the 
cover $p$ is regular and $E/$\,{\em Aut}$(E)$ is homeomorphic to $B$. 
\end{prop}
\begin{proof} For any points $e,e'\in F_b$, there exists $g\in \text{Aut}(E)$ 
such that $g(e)=e'$ and thus $p_*(\pi_1(E,e))=p_*(\pi_1(E,e'))$.  Therefore all conjugates of 
$p_*(\pi_1(E,e))$ are equal to $p_*(\pi_1(E,e))$ and $p_*(\pi_1(E,e))$ is a normal subgroup 
of $\pi_1(B,b)$. The homeomorphism is clear since, locally, both $p$ and passage 
to orbits identify the different components of the inverse images of fundamental 
neighborhoods.
\end{proof}

\begin{thm} Choose a basepoint $b\in B$ and let $G=\pi_1(B,b)$. There is a functor 
$$E(-):\sO(G)\rtarr \text{\em Cov}(B)$$
that is an equivalence of categories. For each subgroup $H$ of $G$, the covering 
$p: E(G/H)\rtarr B$ has a canonical basepoint $e$ in its fiber over $b$ such that 
$$p_*(\pi_1(E(G/H),e)) = H.$$
Moreover, $F_b\iso G/H$ as a $G$-set and, for a $G$-map $\al: G/H\rtarr G/K$ in $\sO(G)$, 
the restriction of  $E(\al): E(G/H)\rtarr E(G/K)$ to fibers over $b$ coincides with $\al$. 
\end{thm}
\begin{proof} Let $p: E\rtarr B$ be the universal cover of $B$ and fix $e\in E$ such that $p(e)=b$. 
We have the isomorphism Aut$(E)\iso\pi_1(B,b)$ given by mapping $g: E\rtarr E$ to the path 
class $[f]\in G$ such that $g(e)=T(f)(e)$, where $T(f)(e)$ is the endpoint of the path 
$\tilde f$ that starts at $e$ and lifts $f$. We identify subgroups of $G$ with subgroups of 
Aut$(E)$ via this isomorphism. We define $E(G/H)$ to be the orbit space $E/H$ and we let 
$q: E\rtarr E/H $ be the quotient map. We may identify $B$ with $E/\text{Aut}(E)$, and 
inclusion of orbits specifies a map $p': E/H\rtarr B$ such that $p'\com q=p: E\rtarr B$. 
If $U\subset B$ is a fundamental neighborhood for $p$ and $V$ is a component of 
$p^{-1}(U)\subset E$, then
$$p^{-1}(U)=\textstyle{\coprod}_{g\in \text{Aut}(E)}\, gV.$$
Passage to orbits over $H$ simply identifies some of these components, and we see
immediately that both $p'$ and $q$ are covers. If $e'=q(e)$, then $p_*'$ maps
$\pi_1(E/H,e')$ isomorphically onto $H$ since, by construction, the isotropy group
of $e'$ under the action of $\pi_1(B,b)$ is precisely $H$. Rewriting $p'=p$ and $e'=e$
generically, this gives the stated properties of the coverings $E(G/H)$. 
The functoriality on $\sO(G)$ follows directly from the previous lemma. 

The functor $E(-)$ is an equivalence of categories since the results of the previous
section imply that it maps the morphism set $\sO(G)(G/H,G/K)$ bijectively onto the 
morphism set Cov$(E(G/H),E(G/K))$ and that every covering of $B$ is isomorphic 
to one of the coverings $E(G/H)$. 
\end{proof}

The classification theorems for coverings of spaces and coverings of groupoids are
nicely related. In fact, the following diagram of functors commutes up to natural
isomorphism:
$$\diagram
 & \sO(\pi_1(B,b))\dlto_{E(-)} \drto^{\sE(-)} &  \\
\text{Cov}(B) \rrto_{\PI} & & \text{Cov}(\PI(B)). \\
\enddiagram$$

\begin{cor} $\PI: \text{\em Cov}(B) \rtarr \text{\em Cov}(\PI(B))$ is an equivalence of categories.
\end{cor} 

\vspace{.1in}

\begin{center}
PROBLEMS
\end{center}

In the following two problems, let $G$ be a connected and locally path connected topological 
group\index{topological group} with identity element $e$,  let $p: H\rtarr G$  be a covering, 
and fix $f\in H$ such 
that $p(f)=e$.  Prove the following. (Hint: Make repeated use of the fundamental theorem 
for covering spaces.)
\begin{enumerate}
\item 
\begin{enumerate}
\item[(a)] $H$ has a unique continuous product $H\times H\rtarr H$  with identity element $f$ 
such that $p$ is a homomorphism.
\item[(b)] $H$ is a topological group under this product, and $H$ is Abelian if $G$ is.
\end{enumerate}
\item
\begin{enumerate}
\item[(a)] The kernel $K$ of $p$ is a discrete normal subgroup of $H$.
\item[(b)] In general, any discrete normal subgroup $K$ of a connected topological group $H$  
is contained in the center of $H$.
\item[(c)] For $k\in K$,  define $t(k): H\rtarr H$  by $t(k)(h) = kh$.  Then $k\rtarr t(k)$
specifies an isomorphism between $K$ and the group Aut$(H)$.
\end{enumerate}
\end{enumerate}

Let $X$ and $Y$ be connected, locally path connected, and Hausdorff. A map $f: X\rtarr Y$ 
is said to be a local homeomorphism\index{local homeomorphism}
if every point of $X$ has an open neighborhood that maps 
homeomorphically onto an open set in $Y$. 

\begin{enumerate}
\item[3.] Give an example of a surjective local homeomorphism that is not a covering.
\item[4.]* Let $f: X\rtarr Y$ be a local homeomorphism, where $X$ is compact. Prove that $f$ 
is a (surjective!) covering with finite fibers.
\end{enumerate}

Let $X$ be a $G$-space, where $G$ is a (discrete) group.  For a subgroup $H$ of $G$, 
define 
$$X^H=\sset{x|hx = x\ \text{for all}\ \ h\in H}\subset X;$$
$X^H$ is the $H$-fixed point subspace\index{fixed point space} of $X$.
Topologize the set of functions $G/H\rtarr X$ as the product of copies
of $X$ indexed on the elements of $G/H$, and give the set of $G$-maps 
$G/H\rtarr X$ the subspace topology.

\begin{enumerate}
\item[5.]  Show that the space of $G$-maps $G/H\rtarr X$ 
is naturally homeomorphic to $X^H$. In particular, $\sO(G/H,G/K)\iso (G/K)^H$.
\item[6.] Let $X$ be a $G$-space. Show that passage to fixed point spaces,
$G/H \longmapsto X^H$, is the object function of a {\em contravariant}
functor $X^{(-)}: \sO(G)\rtarr \sU$.
\end{enumerate}

\chapter{Graphs}

We define graphs, describe their homotopy types, and use them to
show that a subgroup of a free group is free and that any group
is the fundamental group of some space.

\section{The definition of graphs}

We give the definition in a form that will later make it clear that a 
graph is exactly a one-dimensional CW complex. Note that the zero-sphere 
$S^0$ is a discrete space with two points. We think of $S^0$ as
the boundary of $I$ and so label the points $0$ and $1$.

\begin{defn} A graph\index{graph} $X$ is a space that is obtained from a (discrete) set $X^0$
of points, called vertices\index{vertex}, and a (discrete) set $J$ of functions 
$j: S^0\rtarr X^0$
as the quotient space of the disjoint union $X^0\amalg (J\times I)$ that is
obtained by identifying $(j,0)$ with $j(0)$ and $(j,1)$ with $j(1)$.
The images of the intervals $\{ j\}\times I$ are called edges\index{edge}. A graph is
finite if it has only finitely many vertices and edges or, equivalently, if
it is a compact space. A graph is locally finite if each vertex is a boundary
point of only finitely many edges or, equivalently, if it is a locally compact
space. A subgraph $A$ of $X$ is a graph $A\subset X$ with $A^0\subset X^0$. 
That is, $A$ is the union of some of the vertices and edges of $X$.
\end{defn}

Observe that a graph is a locally contractible\index{locally contractible space}
space: any neighborhood of any point contains a contractible neighborhood of 
that point. Therefore a connected graph has all possible covers.

\section{Edge paths and trees}

An oriented edge $k:I\rtarr X$ in a graph $X$ is the traversal of an edge in
either the forward or backward direction. An edge path\index{edge path} is a finite 
composite of 
oriented edges $k_n$ with $k_{n+1}(0)=k_n(1)$. Such a path is reduced\index{edge path!reduced} 
if it is never the case that $k_{n+1}$ is $k_n$ with the opposite orientation. An edge
path is closed\index{edge path!closed} if it starts and ends at the same vertex 
(and is thus a loop).

\begin{defn} A tree\index{tree} is a connected graph with no closed reduced edge paths.
\end{defn}

A subspace $A$ of a space $X$ is a deformation retract\index{deformation retract} if 
there is a homotopy 
$h:X\times I\rtarr X$ such that $h(x,0)=x$, $h(a,t)=a$, and $h(x,1)\in A$
for all $x\in X$, $a\in A$, and $t\in I$. Such a homotopy is called a 
deformation\index{deformation} of $X$ onto $A$.

\begin{lem}
Any vertex $v_0$ of a tree $T$ is a deformation retract of $T$.
\end{lem}
\begin{proof}
This is true by induction on the number of edges when $T$ is finite since
we can prune the last branch. For the general case, observe that each vertex
$v$ lies in some finite connected subtree $T(v)$ that also contains $v_0$.
Choose an edge path $a(v):I\rtarr T(v)$ connecting $v$ to $v_0$. For an edge 
$j$ from $v$ to $v'$, $T(v)\cup T(v')\cup j$ is a finite connected subtree of $T$. 
On the square $j\times I$, we define 
$$h: j\times I\rtarr T(v)\cup T(v')\cup j$$
by requiring $h=a(v)$ on $\sset{v}\times I$, $h=a(v')$ on 
$\sset{v'}\times I$, $h(x,0)=x$ and $h(x,1)=v_0$ for all 
$x\in j$, and extending over the interior of the square by use of the
simple connectivity of $T(v)\cup T(v')\cup j$. As $j$ runs over the edges, 
these homotopies glue together to specify a deformation $h$ of $T$ onto $v_0$.
\end{proof}

A subtree of a graph $X$ is maximal{\index{tree!maximal} if it is contained in no 
strictly larger tree.

\begin{lem} If a tree $T$ is a subgraph of a graph $X$, then $T$ is contained in
a maximal tree. If $X$ is connected, then a tree in $X$ is maximal if and only if 
it contains all vertices of $X$.
\end{lem}
\begin{proof}
Since the union of an increasing family of trees in $X$ is a tree, the first
statement holds by Zorn's lemma. If $X$ is connected, then a tree containing all 
vertices is maximal since addition of an edge would result in a subgraph that
contains a closed reduced edge path and, conversely, a tree $T$ that does not 
contain all vertices is not maximal since a vertex not in $T$ can be connected 
to a vertex in $T$ by a reduced edge path consisting of edges not in $T$.
\end{proof}

\section{The homotopy types of graphs}

Graph theory is a branch of combinatorics. The homotopy theory of graphs is
essentially trivial, by the following result.

\begin{thm}
Let $X$ be a connected graph with maximal tree $T$. Then the quotient space 
$X/T$ is the wedge of one circle for each edge of $X$ not in $T$, and the
quotient map $q: X\rtarr X/T$ is a homotopy equivalence.
\end{thm}
\begin{proof}
The first clause is evident. The second is a direct consequence of a later
result (that will be left as an exercise): for a suitably nice inclusion,
called a ``cofibration,'' of a contractible space $T$ in a space $X$, the
quotient map $X\rtarr X/T$ is a homotopy equivalence. A direct proof in the
present situation is longer and uglier. With the notation in our proof that
a vertex $v_0$ is a deformation retract of $T$ via a deformation $h$, 
define a loop $b_j=a(v')\cdot j\cdot a(v)^{-1}$ at $v_0$ for each edge $j: v\rtarr v'$ not 
in $T$. The $b_j$ together specify a map $b$ from $X/T\iso\bigvee_j S^1$ to $X$. 
The composite $q\com b: X/T\rtarr X/T$ is the wedge over $j$ of copies of the loop 
$c_{v_0}\cdot \id \cdot\, c_{v_0}^{-1}:S^1\rtarr S^1$ and is therefore homotopic to the identity. 
To prove that $b\com q$ is homotopic to the identity, observe that $h$
is a homotopy $\id\htp b\com q$ on $T$. This homotopy extends to a homotopy
$H:\id\htp b\com q$ on all of $X$. To see this, we need only construct $H$ on
$j\times I$ for an edge $j:v\rtarr v'$ not in $T$. The following schematic
description of the prescribed behavior on the boundary of the square makes it
clear that $H$ exists:
$$\diagram
\xline[0,3]^<(0.2){a(v)^{-1}} ^<(0.5){j} ^<(0.8){a(v')} \xline[3,0]_{a(v)} 
& \xline[3,-1]^{c_v} & \xline[3,1]_{c_{v'}} &  \xline[3,0]^{a(v')} \\
& & &  \\
& & &  \\
\xline[0,3]_{j} & & & \\
\enddiagram$$
\renewcommand{\qed}{}\end{proof}

\section{Covers of graphs and Euler characteristics}

Define the Euler characteristic\index{Euler characteristic! of a finite graph} $\ch (X)$ of a 
finite graph $X$ to be $V-E$, where
$V$ is the number of vertices of $X$ and $E$ is the number of edges. By induction
on the number of edges, $\ch (T)=1$ for any finite tree. The determination of the 
homotopy types of graphs has the following immediate implication. 

\begin{cor} If $X$ is a connected graph, then $\pi_1(X)$ is a free group with one
generator for each edge not in a given maximal tree. If $X$ is finite, then $\pi_1(X)$
is free on $1-\ch (X)$ generators; in particular, $\ch (X)\leq 1$, with equality if and
only if $X$ is a tree.
\end{cor}

\begin{thm}
If $B$ is a connected graph with vertex set $B^0$ and $p:E\rtarr B$ is a covering,
then $E$ is a connected graph with vertex set $E^0=p^{-1}(B^0)$ and with one edge for 
each edge $j$ of $B$ and point $e\in F_{j(0)}$. Therefore, if $B$ is finite 
and $p$ is a finite cover whose fibers have cardinality $n$, then $E$ is finite and 
$\ch (E)= n\ch (B)$.
\end{thm}
\begin{proof} Regard an edge $j$ of $B$ as a path $I\rtarr B$ and let $k(e):I\rtarr E$
be the unique path such that $p\com k = j$  and $k(e)(0)=e$, where $e\in F_{j(0)}$. We
claim that $E$ is a graph with $E^0$ as vertex set and the $k(e)$ as edges. An easy
path lifting argument shows that each point of $E-E^0$ is an interior point of exactly
one edge, hence we have a continuous bijection from the graph $E^0\amalg (K\times I)/(\sim )$
to $E$, where $K$ is the evident set of ``attaching maps'' $S^0\rtarr E^0$ for the 
specified edges. This map is a homeomorphism since it is a local homeomorphism over $B$.
\end{proof}

\section{Applications to groups}

The following purely algebraic result is most simply proved by topology.

\begin{thm}
A subgroup $H$ of a free group\index{free group} $G$ is free. If $G$ is free on 
$k$ generators and $H$ has
finite index $n$ in $G$, then $H$ is free on $1-n+nk$ generators.
\end{thm}
\begin{proof}
Realize $G$ as $\pi_1(B)$, where $B$ is the wedge of one circle for each generator of $G$
in a given free basis. Construct a covering $p: E\rtarr B$ such that $p_*(\pi_1(E))=H$.
Since $E$ is a graph, $H$ must be free. If $G$ has $k$ generators, then $\ch (B)=1-k$. If
$[G:H]=n$, then $F_b$ has cardinality $n$ and $\ch (E)= n\ch(B)$. Therefore 
$1-\ch(E)=1-n+nk$.
\end{proof}

We can extend the idea to realize any group as the fundamental group of some connected
space.\index{fundamental group}

\begin{thm}
For any group $G$, there is a connected space $X$ such that $\pi_1(X)$ is
isomorphic to $G$.
\end{thm}
\begin{proof}
We may write $G=F/N$ for some free group $F$ and normal subgroup $N$. As above, we
may realize the inclusion of $N$ in $F$ by passage to fundamental groups from a cover 
$p:E\rtarr B$. Define the (unreduced) cone on $E$ to be $CE = (E\times I)/(E\times\sset{1})$
and define
$$X=B\cup_pCE/(\sim),$$
where $(e,0)\sim p(e)$. Let $U$ and $V$ be the images in $X$ of $B\amalg (E\times [0,3/4))$
and $E\times (1/4,1]$, respectively, and choose a basepoint in $E\times \sset{1/2}$. Since
$U$ and $U\cap V$ are homotopy equivalent to $B$ and $E$ via evident deformations and $V$
is contractible, a consequence of the van Kampen theorem gives the conclusion.
\end{proof} 

The space $X$ constructed in the proof is called the ``homotopy cofiber''\index{homotopy cofiber} 
of the map $p$. It is an important general construction to which we shall return shortly.

\vspace{.1in}

\begin{center}
PROBLEMS
\end{center}
\begin{enumerate}
\item Let $F$ be a free group on two generators $a$ and $b$. How many subgroups of $F$ have
index $2$? Specify generators for each of these subgroups. 
\item Prove that a non-trivial normal subgroup $N$ with infinite index in a free group $F$ 
cannot be finitely generated.
\item* Essay: Describe a necessary and sufficient condition for a graph to be embeddable
in the plane.
\end{enumerate}

\chapter{Compactly generated spaces}

We briefly describe the category of spaces in which algebraic topologists
customarily work. The ordinary category of spaces allows pathology that
obstructs a clean development of the foundations. The homotopy and homology 
groups of spaces are supported on compact subspaces, and it turns out that if
one assumes a separation property that is a little weaker than the Hausdorff 
property, then one can refine the point-set topology of spaces to eliminate 
such pathology without changing these invariants. We shall leave the proofs 
to the reader, but the wise reader will simply take our word for it, at least 
on a first reading: we do not want to overemphasize this material, the 
importance of which can only become apparent in retrospect.

\section{The definition of compactly generated spaces}

We shall understand compact spaces\index{compact space} to be both compact and Hausdorff, following
Bourbaki. A space $X$ is said to be ``weak Hausdorff''\index{weak Hausdorff space} if $g(K)$ is closed 
in $X$ for every map $g: K\rtarr X$ from a compact space $K$ into $X$. When this holds,
the image $g(K)$ is Hausdorff and is therefore a compact subspace of $X$. This separation 
property lies between $T_1$ (points are closed) and Hausdorff, but it is not much weaker 
than the latter.

A subspace $A$ of $X$ is said to be ``compactly closed''\index{compactly closed subspace} if 
$g^{-1}(A)$ is closed in $K$
for any map $g:K\rtarr X$ from a compact space $K$ into $X$. When $X$ is weak Hausdorff, 
this holds if and only if the intersection of $A$ with each compact subset of $X$ is
closed. A space $X$ is a ``$k$-space''\index{kspace@$k$-space} if every compactly closed 
subspace is closed. 

A space $X$ is ``compactly generated''\index{compactly generated space} if it is a weak Hausdorff 
$k$-space. For example,
any locally compact space and any weak Hausdorff space that satisfies the first axiom of 
countability\index{first axiom of countability} 
(every point has a countable neighborhood basis) is compactly generated. We 
have expressed the definition in a form that should make the following statement clear.

\begin{lem} If $X$ is a compactly generated space and $Y$ is any space, then a function
$f:X\rtarr Y$ is continuous if and only if its restriction to each compact
subspace $K$ of $X$ is continuous.
\end{lem}

We can make a space $X$ into a $k$-space by giving it a new topology in which a
space is closed if and only if it is compactly closed in the original topology.
We call the resulting space $kX$. Clearly the identity function $kX\rtarr X$ is 
continuous. If $X$ is weak Hausdorff, then so is $kX$, hence $kX$ is compactly 
generated. Moreover, $X$ and $kX$ then have exactly the same compact subsets. 

Write $X\times_c Y$ for the product of $X$ and $Y$ with its usual topology and 
write $X\times Y=k(X\times_cY)$. If $X$ and $Y$ are weak Hausdorff, then 
$X\times Y=kX\times kY$. If $X$ is locally compact and $Y$ is compactly generated, 
then $X\times Y=X\times_cY$.

By definition, a space $X$ is Hausdorff if the diagonal subspace $\DE X=\sset{(x,x)}$ 
is closed in $X\times_c X$. The weak Hausdorff property admits a similar characterization.

\begin{lem} If $X$ is a $k$-space, then $X$ is weak Hausdorff if and only if
$\DE X$ is closed in $X\times X$.
\end{lem}


\section{The category of compactly generated spaces}

One major source of point-set level pathology can be passage to quotient spaces\index{quotient
space}. Use
of compactly generated topologies alleviates this. 

\begin{prop}
If $X$ is compactly generated and $\pi: X\rtarr Y$ is a quotient map, then $Y$ is 
compactly generated if and only if $(\pi\times\pi)^{-1}(\DE Y)$ is closed in $X\times X$.
\end{prop}

The interpretation is that a quotient space of a compactly generated space by a
``closed equivalence relation'' is compactly generated. We are particularly interested
in the following consequence.

\begin{prop}
If $X$ and $Y$ are compactly generated spaces, $A$ is a closed subspace of $X$, and
$f: A\rtarr Y$ is any continuous map, then the pushout $Y\cup_f X$ is compactly generated.
\end{prop}


Another source of pathology is passage to colimits over sequences of maps $X_i\rtarr X_{i+1}$.
When the given maps are inclusions, the colimit is the union of the sets $X_i$ with the 
``topology of the union;''\index{topology of the union} a set is closed if and 
only if its intersection with each $X_i$ is closed.

\begin{prop}
If $\sset{X_i}$ is a sequence of compactly generated spaces and inclusions 
$X_i\rtarr X_{i+1}$ with closed images, then\, {\em colim}$\,X_i$ is compactly 
generated.
\end{prop}

We now adopt a more categorical point of view. We redefine $\sU$\index{U@$\sU$} to be the category 
of compactly generated spaces and continuous maps, and we redefine $\sT$\index{T@$\sT$} to be its 
subcategory of based spaces and based maps. 

Let $w\sU$ be the category of weak Hausdorff spaces. We have the functor $k: w\sU\rtarr \sU$, 
and we have the forgetful functor $j:\sU\rtarr w\sU$, which embeds $\sU$ as a full subcategory 
of $w\sU$. Clearly
$$\sU(X,kY)\iso w\sU(jX,Y)$$
for $X\in\sU$ and $Y\in w\sU$ since the identity map $kY\rtarr Y$ is continuous and
continuity of maps defined on compactly generated spaces is compactly determined.
Thus $k$ is right adjoint to $j$.

We can construct colimits and limits of spaces by performing these constructions
on sets: they inherit topologies that give them the universal properties of 
colimits\index{colimit} and 
limits\index{limit} in the classical category of spaces. Limits of weak Hausdorff spaces are weak
Hausdorff, but limits of $k$-spaces need not be $k$-spaces. We construct limits of compactly 
generated spaces by applying the functor $k$ to their limits as spaces. It is 
a categorical fact that functors which are right adjoints preserve limits, so this does give 
categorical limits in $\sU$. This is how we defined $X\times Y$, for example. 

Point-set level colimits of weak Hausdorff spaces need not be weak Hausdorff.
However, if a point-set level colimit of compactly generated spaces is weak Hausdorff, 
then it is a $k$-space and therefore compactly generated. We shall only be interested in 
colimits in those cases where this holds. The propositions above give examples. 
In such cases, these constructions give categorical colimits in $\sU$.

From here on, we agree that all given spaces are to be compactly generated, and we
agree to redefine any construction on spaces by applying the functor $k$ to it. For
example, for spaces $X$ and $Y$ in $\sU$, we understand the function space\index{function space} 
$\text{Map}(X,Y)=Y^X$ to mean the set of continuous maps from $X$ to $Y$ with the 
$k$-ification of the standard compact-open topology;\index{compact-open topology} the latter 
topology has as basis the 
finite intersections of the subsets of the form $\{ f|f(K)\subset U\}$ for some compact subset 
$K$ of $X$ and open subset $U$ of $Y$. This leads to the following adjointness homeomorphism, 
which holds without restriction when we work in the category of compactly generated spaces.

\begin{prop}
For spaces $X$, $Y$, and $Z$ in $\sU$, the canonical bijection
$$Z^{(X\times Y)} \iso (Z^Y)^X$$
is a homeomorphism.
\end{prop}

Observe in particular that a homotopy $X\times I\rtarr Y$ can equally well be
viewed as a map $X\rtarr Y^I$. These adjoint, or ``dual,'' points of view will 
play an important role in the next two chapters.

\vspace{.1in}

\begin{center}
PROBLEMS
\end{center}
\begin{enumerate}
\item
\begin{enumerate}
\item[(a)] Any subspace of a weak Hausdorff space is weak Hausdorff.
\item[(b)] Any closed subspace of a $k$-space is a $k$-space.
\item[(c)] An open subset $U$ of a compactly generated space $X$ is compactly
generated if each point has an open neighborhood in $X$ with closure contained 
in $U$.
\end{enumerate}
\item* A Tychonoff (or completely regular) space $X$ is a $T_1$-space (points are
closed) such that for each point $x\in X$ and each closed subset $A$ such that 
$x\notin A$, there is a function $f: X\rtarr I$ such that $f(x)=0$ and $f(a)=1$ if 
$a\in A$. Prove the following (e.g., Kelley, {\em General Topology}).
\begin{enumerate}
\item[(a)] A space is Tychonoff\index{Tychonoff space} if and only if it can be 
embedded in a cube (a product of copies of $I$). 
\item[(b)] There are Tychonoff spaces that are not $k$-spaces, but every cube is
a compact Hausdorff space.
\end{enumerate}
\item Brief essay: In view of Problems 1 and 2, what should we mean by a ``subspace'' of a 
compactly generated space. (We do {\em not} want to restrict the allowable set of subsets.)
\end{enumerate}

\clearpage

\thispagestyle{empty}

\chapter{Cofibrations}

Exact sequences that feature in the study of homotopy, homology, and cohomology 
groups all can be derived homotopically from the theory of cofiber and fiber 
sequences that we present in this and the following two chapters. Abstractions of 
these ideas are at the heart of modern axiomatic treatments of homotopical algebra 
and of the foundations of algebraic $K$-theory. 

The theories of cofiber and fiber sequences illustrate an important, but 
informal, duality theory, known as Eckmann-Hilton duality.\index{Eckmann-Hilton duality} It 
is based on the 
adjunction between Cartesian products and function spaces. Our standing 
hypothesis that all spaces in sight are compactly generated allows the 
theory to be developed without further restrictions on the given spaces.
We discuss ``cofibrations'' here and the ``dual'' notion of ``fibrations'' 
in the next chapter.

\section{The definition of cofibrations}

\begin{defn}
A map $i: A\rtarr X$ is a cofibration\index{cofibration} if it satisfies the homotopy extension
property (HEP).\index{homotopy extension property}\index{HEP} This means that if $h\com i_0=f\com i$ in the 
diagram
$$\diagram
A \rrto^{i_0} \ddto_{i} & & A\times I \dlto_{h}  \ddto^{i\times\id} \\
& Y &  \\
X \rrto_{i_0}  \urto^{f} & & X\times I,  \uldashed_{\tilde{h}}|>\tip\\
\enddiagram$$
then there exists $\tilde{h}$ that makes the diagram commute.
\end{defn}

Here $i_0(x)=(x,0)$. We do not require $\tilde{h}$ to be unique, and it 
usually isn't. Using our alternative way of writing homotopies, we see that
the ``test diagram'' displayed in the definition can be rewritten in the
equivalent form
$$\diagram
A \dto_i \rto^{h} & Y^I \dto^{p_0} \\
X \urdashed^{\tilde{h}}|>\tip \rto_f & Y, \\
\enddiagram$$
where $p_0(\xi)=\xi(0)$.

Pushouts\index{pushout} of cofibrations are cofibrations, in the sense 
of the following result. We generally write $B\cup_g X$ for the pushout 
of a given cofibration $i:A\rtarr X$ and a map $g:A\rtarr B$.

\begin{lem}
If $i:A\rtarr X$ is a cofibration and $g :A\rtarr B$ is any map, then
the induced map $B\rtarr B\cup_g X$ is a cofibration.
\end{lem}
\begin{proof}
Notice that $(B\cup_g X)\times I\iso (B\times I)\cup_{g\times \id}(X\times I)$
and consider a typical test diagram for the HEP.  The proof is a formal 
chase of the following diagram:
$$\diagram
A \xto[0,4]^{i_0} \xto[4,0]^{\ \ \text{pushout}}_{i} \drto^g &  &  &  & 
A\times I \xto[4,0]^{i\times\id}_{\text{pushout}\ \ \ \, }  \dlto_{g\times \id}\\
& B \ddto \rrto & & B\times I \dlto_h \ddto & \\
& & Y & & \\
& B\cup_g X \urto^{f} \rrto 
& &(B\cup_g X)\times I \uldashed_{\tilde{h}}|>\tip & \\
X \urto \xto[0,4]_{i_0} &  &  &  & X\times I. \ulto 
\xdashed '[-1,-2]^{\bar{h}} '[-2,-2]|>\tip \\
\enddiagram$$
We first use that $A\rtarr X$ is a cofibration to obtain a homotopy 
$\bar{h}: X\times I\rtarr Y$ and then use the right-hand pushout to 
see that $\bar{h}$ and $h$ induce the required homotopy $\tilde{h}$.
\end{proof}

\section{Mapping cylinders and cofibrations}

Although the HEP is expressed in terms of general test diagrams, there is a 
certain universal test diagram. Namely, we can let $Y$ in our original test 
diagram be the ``mapping cylinder''\index{mapping cylinder}
$$Mi \equiv X\cup_i(A\times I),$$
which is the pushout of $i$ and $i_0$. Indeed, suppose that we can construct 
a map $r$ that makes the following diagram commute:
$$\diagram
A \rrto^{i_0} \ddto_{i} & & A\times I \dlto  \ddto^{i\times\id} \\
& Mi &  \\
X \rrto_{i_0}  \urto &  & X\times I.  \uldashed_{r}|>\tip\\
\enddiagram$$
By the universal property of pushouts, the given maps $f$ and $h$ in our original
test diagram induce a map $Mi\rtarr Y$, and its composite with $r$ gives a homotopy 
$\tilde{h}$ that makes the test diagram commute. 

A map $r$ that makes the previous diagram commute satisfies $r\com j=\id$, where 
$j: Mi\rtarr X\times I$ is the map that restricts to $i_0$ on $X$ and to $i\times \id$ 
on $A\times I$. As a matter of point-set topology, left as an exercise, it follows that
a cofibration is an inclusion with closed image. 

\section{Replacing maps by cofibrations}

We can use the mapping cylinder construction to decompose an arbitrary map
$f: X\rtarr Y$ as the composite of a cofibration and a homotopy equivalence.
That is, up to homotopy, any map can be replaced by a cofibration. To see this, 
recall that $Mf = Y\cup_f (X\times I)$ and observe that $f$ 
coincides with the composite
$$X \overto{j} Mf \overto{r} Y,$$
where $j(x)=(x,1)$ and where $r(y)=y$ on $Y$ and $r(x,s)=f(x)$ on $X\times I$.
If $i: Y\rtarr Mf$ is the inclusion, then $r\com i=\id$ and $\id\htp i\com r$.
In fact, we can define a deformation $h: Mf\times I\rtarr Mf$ of $Mf$ onto $i(Y)$
by setting
$$h(y,t) = y \ \ \tand \ \ h((x,s),t)=(x,(1-t)s).$$
It is not hard to check directly that $j: X\rtarr Mf$ satisfies the HEP, and this
will also follow from the general criterion for a map to be a cofibration to which
we turn next.

\section{A criterion for a map to be a cofibration}

We want a criterion that allows us to recognize cofibrations when we see 
them. We shall often consider pairs $(X,A)$ consisting of a space $X$ and
a subspace $A$. Co\-fibration pairs will be those pairs that ``behave
homologically'' just like the associated quotient spaces $X/A$. 

\begin{defn}
A pair $(X,A)$ is an NDR-pair\index{NDR-pair} (= neighborhood deformation retract pair) if
there is a map $u: X\rtarr I$ such that $u^{-1}(0)=A$ and a homotopy
$h:X\times I\rtarr X$ such that $h_0=\id$, $h(a,t)=a$ for $a\in A$ and 
$t\in I$, and $h(x,1)\in A$ if $u(x)<1$; $(X,A)$ is a DR-pair\index{DR-pair} if $u(x)<1$
for all $x\in X$, in which case $A$ is a deformation retract of $X$.
\end{defn}

\begin{lem}
If $(h,u)$ and $(j,v)$ represent $(X,A)$ and $(Y,B)$ as NDR-pairs, then
$(k,w)$ represents the ``product pair'' $(X\times Y,X\times B\cup A\times Y)$
as an NDR-pair, where $w(x,y)=\text{\em min}(u(x),v(y))$ and
$$k(x,y,t)=
\begin{cases}
(h(x,t),j(y,tu(x)/v(y))) \ \ \text{if} \ v(y)\geq u(x)\\
(h(x,tv(y)/u(x)),j(y,t)) \ \ \text{if} \ u(x)\geq v(y).
\end{cases}$$
If $(X,A)$ or $(Y,B)$ is a DR-pair, then so is $(X\times Y,X\times B\cup A\times Y)$.
\end{lem}
\begin{proof}
If $v(y)=0$ and $v(y)\geq u(x)$, then $u(x)=0$ and both $y\in B$ and $x\in A$; therefore
we can and must understand $k(x,y,t)$ to be $(x,y)$. It is easy to check from this and 
the symmetric observation that $k$ is a well defined continuous homotopy as desired.
\end{proof}

\begin{thm}
Let $A$ be a closed subspace of $X$. Then the following are equivalent:
\begin{enumerate}
\item[(i)] $(X,A)$ is an NDR-pair.
\item[(ii)] $(X\times I, X\times\sset{0}\cup A\times I)$ is a DR-pair.
\item[(iii)] $X\times\sset{0}\cup A\times I$ is a retract of $X\times I$.
\item[(iv)] The inclusion $i:A\rtarr X$ is a cofibration.
\end{enumerate}
\end{thm}
\begin{proof}
The lemma gives that (i) implies (ii), (ii) trivially implies (iii), and we
have already seen that (iii) and (iv) are equivalent. Assume given a retraction
$r: X\times I\rtarr X\times\sset{0}\cup A\times I$. Let 
$\pi_1: X\times I\rtarr X$ and $\pi_2: X\times I\rtarr I$ be the projections
and define $u: X\rtarr I$ by
$$u(x) = \text{sup}\{ t-\pi_2r(x,t)|t\in I\}$$
and $h: X\times I\rtarr X$ by
$$h(x,t)=\pi_1r(x,t).$$
Then $(h,u)$ represents $(X,A)$ as an NDR-pair. Here $u^{-1}(0)=A$ since
$u(x)=0$ implies that $r(x,t)\in A\times I$ for $t>0$ and thus also for
$t=0$ since $A\times I$ is closed in $X\times I$.
\end{proof}

\section{Cofiber homotopy equivalence}

It is often important to work in the category of spaces under a given space $A$, and we 
shall later need a basic result about homotopy equivalences in this category. We shall
also need a generalization concerning homotopy equivalences of pairs. The reader is warned 
that the results of this section, although easy enough to understand, have fairly lengthy 
and unilluminating proofs. 

A space 
under $A$ is a map $i: A\rtarr X$. A map of spaces under $A$ is a commutative diagram
$$\diagram
& A \dlto_i \drto^{j}\\
X\rrto_f & &  Y \\
\enddiagram$$
A homotopy between maps under $A$ is a homotopy that at each time $t$ is a map under $A$.
We then write $h: f\htp f'\ \text{rel}\ A$ and have $h(i(a),t)=j(a)$ for all $a\in A$ and 
$t\in I$. There results a notion of a homotopy equivalence under $A$. Such an equivalence
is called a ``cofiber homotopy equivalence.''\index{cofiber homotopy equivalence} The 
name is suggested by the following result,
whose proof illustrates a more substantial use of the HEP than we have seen before.

\begin{prop} Let $i: A\rtarr X$ and $j: A\rtarr Y$ be cofibrations and let $f: X\rtarr Y$
be a map such that $f\com i = j$. Suppose that $f$ is a homotopy equivalence. Then $f$ is a
cofiber homotopy equivalence.
\end{prop} 
\begin{proof} It suffices to find a map $g: Y\rtarr X$ under $A$ and a homotopy 
$g\com f\htp \id \ \text{rel}\ A$. Indeed, $g$ will then be a homotopy equivalence, and
we can repeat the argument to obtain $f': X\rtarr Y$ such that $f'\com g\htp \id \ \text{rel}\ A$; 
it will follow formally that $f'\htp f\ \text{rel}\ A$.
By hypothesis, there is a map $g'': Y\rtarr X$ that is a homotopy inverse to $f$. Since
$g''\com f\htp\id$, $g''\com j\htp i$. Since $j$ satisfies the HEP, it follows directly
that $g''$ is homotopic to a map $g'$ such that $g'\com j=i$. It suffices to prove that
$g'\com f: X\rtarr X$ has a left homotopy inverse $e: X\rtarr X$ under $A$, since
$g=e\com g'$ will then satisfy $g\com f\htp \id \ \text{rel}\ A$. Replacing our original
map $f$ with $g'\com f$, we see that it suffices to obtain a left homotopy inverse under $A$ 
to a map $f:X\rtarr X$ such that $f\com i=i$ and $f\htp\id$. Choose a homotopy $h:f\htp \id$.
Since $h_0\com i=f\com i=i$ and $h_1=\id$, we can apply the HEP to 
$h\com(i\times\id): A\times I\rtarr X$ and the identity map of $X$ to obtain a homotopy 
$k:\id \htp k_1\equiv e$ such that $k\com(i\times\id)=h\com(i\times\id)$. Certainly $e\com i=i$. 
Now apply the HEP to the following diagram:
$$\diagram
A\times I \rrto^{i_0} \ddto_{i\times\id} & 
& A\times I\times I \ddto^{i\times\id\times\id} \dlto_K\\
& X & \\
X\times I  \urto^J \rrto_{i_0} & & X\times I\times I. \uldashed_{L}|>\tip \\
\enddiagram$$
Here $J$ is the homotopy $e\com f\htp \id$ specified by
$$J(x,s)=\begin{cases}
k(f(x),1-2s)\ \ \text{if}\ \ s\leq 1/2 \\
h(x,2s-1)\ \ \ \ \ \  \text{if}\ \ 1/2\leq s.\\
\end{cases}$$
The homotopy between homotopies $K$ is specified by
$$K(a,s,t)=\begin{cases}
k(i(a),1-2s(1-t))\ \ \ \ \ \ \ \ \ \ \text{if}\ \ s\leq 1/2 \\
h(i(a),1-2(1-s)(1-t))\ \ \, \text{if}\ \ s\geq 1/2.\\
\end{cases}$$
Traversal of $L$ around the three faces of $I\times I$ other than that specified by $J$
gives a homotopy 
$$e\com f = J_0 = L_{0,0} \htp L_{0,1}\htp L_{1,1}\htp L_{1,0} = J_1=\id \ \text{rel}\ A. \qed$$
\renewcommand{\qed}{}\end{proof}

The proposition applies to the following previously encountered situation.

\begin{exmp}
Let $i: A\rtarr X$ be a cofibration. We then have the commutative diagram
$$\diagram
& A \dlto_j \drto^i \\
Mi \rrto_r & & X, \\
\enddiagram$$
where $j(a)=(a,1)$. The obvious homotopy inverse $\io: X\rtarr Mi$ has $\io(x)=(x,0)$ 
and is thus very far from being a map under $A$. The proposition ensures that $\io$ is
homotopic to a map under $A$ that is homotopy inverse to $r$ under $A$.
\end{exmp}

The following generalization asserts that, for inclusions that are cofibrations, 
a pair of homotopy equivalences is a homotopy equivalence of 
pairs.\index{homotopy equivalence!of pairs} It is often
used implicitly in setting up homology and cohomology theories on pairs of spaces.

\begin{prop} Assume given a commutative diagram
$$\diagram
A \rto^{d} \dto_{i} & B \dto^{j} \\
X \rto_{f} & Y\\
\enddiagram$$
in which $i$ and $j$ are cofibrations and $d$ and $f$ are homotopy equivalences.
Then $(f,d):(X,A)\rtarr (Y,B)$ is a homotopy equivalence of pairs.
\end{prop}
\begin{proof}
The statement means that there are homotopy inverses $e$ of $d$ and $g$
of $f$ such that $g\com j=i\com e$ together with homotopies $H: g\com f\htp\id$ and
$K: f\com g\htp \id$ that extend homotopies $h: e\com d\htp\id$ and $k: d\com e\htp\id$.
Choose any homotopy inverse $e$ to $d$, together with homotopies
$h: e\com d\htp \id$ and $\ell: d\com e\htp\id$. By HEP for $j$, there is a homotopy
inverse $g'$ for $f$ such that $g'\com j = i\com e$. Then, by HEP for $i$, there is 
a homotopy $m$ of $g'\com f$ such that $m\com (i\times \id)=i\com h$. Let $\ph = m_1$. Then
$\ph\com i = i$ and $\ph$ is a cofiber homotopy equivalence by the previous result. 
Let $\ps: X\rtarr X$ be a homotopy inverse under $i$ and let $n: \ps\com\ph\htp \id$ 
be a homotopy under $i$. Define $g= \ps\com g'$. Clearly $g\com j = i\com e$. Using that
the pairs $(I\times I, I\times \sset{0})$ and $(I\times I, I\times \sset{0}\cup \pa I\times I)$
are homeomorphic, we can construct a homotopy between homotopies $\LA$ by applying HEP to the
diagram
$$\diagram (A\times I \times {0})\cup (A\times \pa I\times I)  \ddto_{i\times \id}
\rrto^{\subset} & & A\times I\times I \ddto^{i\times \id} \dlto_{\GA}\\
& X & \\
(X\times I \times {0})\cup (X\times \pa I\times I)  \urto^{\ga}
\rrto_{\subset} & & X\times I\times I. \ulto_{\LA}  \\
\enddiagram$$
Here
$$\ga(x,s,0) = \left\{ \begin{array}{ll}
            \ps (m(x,2s)) & \mbox{if $s\leq 1/2$} \\
             n(x, 2s-1) & \mbox{if $s\geq 1/2$,}
            \end{array}
            \right. $$
$$\ga(x,0,t) = (g\com f)(x) = (\ps\com g'\com f)(x),$$
and
$$\ga(x,1,t) = x,$$
while 
$$\GA(a,s,t) = \left\{ \begin{array}{ll}
               i(h(a,2s/(1+t))) & \mbox{if $2s\leq 1+t$} \\
               i(a) & \mbox{if $2s\geq 1+t$}
            \end{array}
            \right. $$
Define $H(x,s)= \LA(x,s,1)$. Then $H: g\com f\htp\id$ and $H\com(i\times\id)=i\com h$.
Application of this argument with $d$ and $f$ replaced by $e$ and $g$ gives a left
homotopy inverse $f'$ to $g$ and a homotopy $L: f'\com g\htp \id$ such that 
$f'\com i = j\com d$ and $L\com(j\times \id) = j\com \ell$. Adding homotopies by
concentrating them on successive fractions  of the unit interval and letting
the negative of a homotopy be obtained by reversal of direction, define
$$k=(-\ell)(de\times \id) + dh(e\times \id) + \ell$$ 
and
$$K=(-L)(fg\times \id) + f'H(g\times \id)+ L.$$ 
Then $K: f\com g\htp\id$ and $K\com(j\times \id)= j\com k$.
\end{proof}

\vspace{.1in}

\begin{center}
PROBLEMS
\end{center}
\begin{enumerate}
\item Show that a cofibration $i: A\rtarr X$ is an inclusion with closed image.
\item Let $i: A\rtarr X$ be a cofibration, where $A$ is a contractible space.
Prove that the quotient map $X\rtarr X/A$ is a homotopy equivalence.
\end{enumerate}

\chapter{Fibrations}

We ``dualize'' the definitions and theory of the previous chapter to the study
of fibrations, which are ``up to homotopy'' generalizations of covering spaces.

\section{The definition of fibrations}

\begin{defn}
A surjective map $p: E\rtarr B$ is a fibration\index{fibration} if it satisfies the 
covering homotopy property (CHP).\index{CHP}\index{covering homotopy property} This 
means that if $h\com i_0=p\com f$ in the diagram
$$\diagram
Y \rto^f \dto_{i_0} & E \dto^p \\
Y\times I \rto_h \urdashed^{\tilde{h}}|>\tip & B,\\
\enddiagram$$
then there exists $\tilde{h}$ that makes the diagram commute.
\end{defn}

This notion of a fibration is due to Hurewicz. There is a more general notion of a 
Serre fibration\index{fibration!Serre}, in which the test spaces $Y$ are restricted 
to be cubes $I^n$. Serre fibrations are more appropriate for many purposes, but we 
shall make no use of them. The test diagram in the definition can be rewritten in the 
equivalent form
$$\diagram
E \ddto_{p} & & E^I \llto_{p_0}  \ddto^{p^I} \\
& Y \urdashed^{\tilde{h}}|>\tip \ulto^f \drto_h &  \\
B  & & B^I. \llto^{p_0}\\
\enddiagram$$
Here $p_0(\be)=\be(0)$ for $\be\in B^I$. With this formulation, we can 
``dualize'' the proof that pushouts of cofibrations are cofibrations to 
show that pullbacks of fibrations are fibrations. We often write 
$A\times_g E$ for the pullback\index{pullback} of a given fibration 
$p:E\rtarr B$ and a map $g:A\rtarr B$.

\begin{lem}
If $p:E\rtarr B$ is a fibration and $g :A\rtarr B$ is any map, then
the induced map $A\times_g E\rtarr A$ is a fibration.
\end{lem}

\section{Path lifting functions and fibrations}

Although the CHP is expressed in terms of general test diagrams, there is a 
certain universal test diagram. Namely, we can let $Y$ in our original test 
diagram be the ``mapping path space''\index{mapping path space}
$$Np \equiv E\times_p B^I =\{ (e,\be)| \be(0)=p(e)\} \subset E\times B^I.$$
That is, $Np$ is the pullback of $p$ and $p_0$ in the second form of the test 
diagram and, with $Y=Np$, $f$ and $h$ in that diagram are the evident projections. 
A map $s: Np\rtarr E^I$
such that $k\com s=\id$, where $k: E^I\rtarr Np$ has coordinates $p_0$ and
$p^I$, is called a path lifting function.\index{path lifting function} Thus
$$s(e,\be)(0)=e \ \ \tand \ \ p\com s(e,\be)=\be.$$
Given a general test diagram, there results a map $g: Y\rtarr Np$ determined
by $f$ and $h$, and we can take $\tilde{h}= s\com g$. 

In general, path lifting functions are not unique. In fact, we have already 
studied the special kinds of fibrations for which they are unique.

\begin{lem} If $p: E\rtarr B$ is a covering, then $p$ is a fibration with
a unique path lifting function $s$.
\end{lem}
\begin{proof}
The unique lifts of paths with a given initial point specify $s$.
\end{proof}

Fibrations and cofibrations are related by the following useful observation.

\begin{lem}
If $i: A\rtarr X$ is a cofibration and $B$ is a space, then the induced map
$$p= B^i: B^X \rtarr B^A$$
is a fibration.
\end{lem}
\begin{proof}
It is an easy matter to check that we have a homeomorphism
$$B^{Mi} = B^{X\times\sset{0}\cup A\times I} \iso B^X\times_p (B^A)^I = Np.$$
If $r: X\times I\rtarr Mi$ is a retraction, then
$$B^r: Np \iso B^{Mi}\rtarr B^{X\times I}\iso (B^X)^I$$
is a path lifting function.
\end{proof}

\section{Replacing maps by fibrations}

We can use the mapping path space construction to decompose an arbitrary map
$f: X\rtarr Y$ as the composite of a homotopy equivalence and a fibration.
That is, up to homotopy, any map can be replaced by a fibration. To see this, 
recall that $Nf = X\times_f Y^I$ and observe that $f$ 
coincides with the composite
$$X \overto{\nu} Nf \overto{\rh} Y,$$
where $\nu(x)=(x,c_{f(x)})$ and $\rh (x,\ch)=\ch(1)$. Let $\pi: Nf\rtarr X$
be the projection. Then $\pi\com\nu=\id$ and $\id\htp \nu\com\pi$ since 
we can define a deformation $h: Nf\times I\rtarr Nf$ of $Nf$ onto $\nu(X)$ by
setting
$$h(x,\ch)(t)=(x,\ch_t), \ \text{where}\ \ch_t(s)=\ch((1-t)s).$$
We check directly that $\rh: Nf\rtarr Y$ satisfies the CHP. Consider a test 
diagram
$$\diagram
A \rto^g \dto_{i_0} & Nf \dto^{\rh} \\
A\times I \rto_h \urdashed^{\tilde{h}}|>\tip & Y.\\
\enddiagram$$
We are given $g$ and $h$ such that $h\com i_0=\rh\com g$ and must construct 
$\tilde{h}$ that makes the diagram commute. We write $g(a)=(g_1(a),g_2(a))$
and set
$$\tilde{h}(a,t)=(g_1(a),j(a,t)),$$
where
$$j(a,t)(s)=
\begin{cases} 
g_2(a)(s+st) \ \ \ \ \ \text{if}\ \  0\leq s \leq 1/(1+t)\\
h(a,s+ts-1) \ \  \text{if} \ \  1/(1+t) \leq s\leq 1.
\end{cases}$$

\section{A criterion for a map to be a fibration}

Again, we want a criterion that allows us to recognize fibrations when we see
them. Here the idea of duality fails, and we instead think of fibrations as
generalizations of coverings. When restricted to the spaces $U$ in a well 
chosen open cover $\sO$ of the base space $B$, a covering is homeomorphic 
to the projection $U\times F\rtarr U$, where $F$ is a fixed discrete set. 

The obvious generalization of this is the notion of a bundle. A map $p: E\rtarr B$ 
is a bundle\index{bundle} if, when restricted to the spaces $U$ in a well chosen open 
cover $\sO$ of $B$, there are homeomorphisms $\ph: U\times F\rtarr p^{-1}(U)$ such 
that $p\com \ph= \pi_1$, where $F$ is a fixed topological space. We require of a ``well 
chosen'' open cover that it be numerable.\index{numerable open cover} 
This means that there are continuous maps $\la_U: B\rtarr I$ such that $\la_U^{-1}(0,1]=U$
and that the cover is locally finite, in the sense that each $b\in B$ has a neighborhood
that intersects only finitely many $U\in \sO$. Any open cover of a paracompact space has a 
numerable refinement. With this proviso on the open covers allowed in the 
definition of a bundle, the following result shows in particular that every
bundle is a fibration.

\begin{thm} Let $p: E\rtarr B$ be a map and let $\sO$ be a numerable open
cover of $B$. Then $p$ is a fibration if and only if $p: p^{-1}(U)\rtarr U$
is a fibration for every $U\in \sO$.
\end{thm}
\begin{proof}
Since pullbacks of fibrations are fibrations, necessity is obvious. Thus 
assume that $p|p^{-1}(U)$ is a fibration for each $U\in \sO$. We shall construct a
path lifting function for $B$ by patching together path lifting functions
for the $p|p^{-1}(U)$, but we first set up the scaffolding of the patching argument.
Choose maps $\la_U: B\rtarr I$ such that $\la_U^{-1}(0,1]=U$. For a finite ordered
subset $T=\sset{U_1,\ldots\!,U_n}$ of sets in $\sO$, 
define $c(T)=n$ and define $\la_T: B^I\rtarr I$ by
$$\la_T(\be)= 
\text{inf}\{ (\la_{U_i}\com\be)(t)|(i-1)/n\leq t\leq i/n,\ 1\leq i\leq n \}.$$
Let $W_T=\la_T^{-1}(0,1]$. Equivalently,
$$W_T=\{ \be| \be(t)\in U_i \ \text{if}\  t\in [(i-1)/n,i/n] \} \subset B^I.$$
The set $\sset{W_T}$ is an open cover of $B^I$, but it need not be locally
finite. However, $\{ W_T|c(T)<n\}$ is locally finite for each fixed $n$. If
$c(T)=n$, define $\ga_T: B^I\rtarr I$ by
$$\ga_T(\be)=\text{max}\{ 0,\la_T(\be)-n \textstyle{\sum}_{c(S)<n}\, \la_S(\be)\},$$
and define
$$V_T=\{ \be|\ga_T(\be)>0\}\subset W_T.$$
Then $\sset{V_T}$ is a locally finite open cover of $B^I$. We choose a total
ordering of the set of all finite ordered subsets $T$ of $\sO$.

With this scaffolding in place, choose path lifting functions
$$s_U:p^{-1}(U)\times_pU^I\rtarr p^{-1}(U)^I$$
for $U\in\sO$, so that $(p\com s_U)(e,\be)=\be$ and $s_U(e,\be)(0)=e$. For a given
$T=\sset{U_1,\ldots\!,U_n}$, consider paths $\be\in V_T$.
For $0\leq u <v\leq 1$, let $\be[u,v]$ be the restriction of $\be$ to $[u,v]$.
If $u\in [(i-1)/n,i/n]$ and  $v\in [(j-1)/n,j/n]$, where $0\leq i \leq j\leq n$,
and if $e\in p^{-1}(\be (u))$, define $s_T(e,\be[u,v]): [u,v]\rtarr E$ to be the
path that starts at $e$ and covers $\be[u,v]$ that is obtained by applying 
$s_{U_i}$ to lift over $[u,i/n]$ (or over $[u,v]$ if $i=j$), using $s_{U_{i+1}}$,
starting at the point where the first lifted path ends, to lift over $[i/n,(i+1)/n]$
and so on inductively, ending with use of $s_{U_{j}}$ to lift over $[(j-1)/n,v]$.
(Technically, since we are lifting over partial intervals and the $s_U$ lift
paths defined on $I$ to paths defined on $I$, this involves a rescaling: we must 
shrink $I$ linearly onto our subinterval, then apply the relevant part of $\be$, next 
lift the resulting path, and finally apply the result to the linear expansion of our 
subinterval onto $I$.) For a point $(e,\be)$ in $Np$, define $s(e,\be)$ to be the 
concatenation of the paths $s_{T_j}(e_{j-1},\be[u_{j-1},u_j])$, $1\leq j\leq q$, where
the $T_i$, in order, run through the set of all $T$ such that $\be\in V_T$, where
$u_0=0$ and $u_j=\sum_{i=1}^j \ga_{T_i}(\be)$ for $1\leq j\leq q$, and where $e_0=e$
and $e_j$ is the endpoint of $s_{T_j}(e_{j-1},\be[u_{j-1},u_j])$ for $1\leq j< q$.
Certainly $s(e,\be)=e$ and $(p\com s)(e,\be)=\be$. It is not hard to check that
$s$ is well defined and continuous, hence it is a path lifting function for $p$.
\end{proof}

\section{Fiber homotopy equivalence}

It is often important to study fibrations over a given base space $B$, working in the 
category of spaces over $B$. A space over $B$ is a map $p: E\rtarr B$. A map of spaces 
over $B$ is a commutative diagram
$$\diagram
D\rrto^f \drto_p & &  E \dlto^q \\
& B & \\
\enddiagram$$
A homotopy between maps over $B$ is a homotopy that at each time $t$ is a map over $B$.
There results a notion of a homotopy equivalence over $B$. Such an equivalence
is called a ``fiber homotopy equivalence.''\index{fiber homotopy equivalence} The name is 
suggested by the following result,
whose proof is precisely dual to the corresponding result for cofibrations and is left as
an exercise.

\begin{prop} Let $p: D\rtarr B$ and $q: E\rtarr B$ be fibrations and let $f: D\rtarr E$
be a map such that $q\com f = p$. Suppose that $f$ is a homotopy equivalence. Then $f$ is a
fiber homotopy equivalence.
\end{prop} 

\begin{exmp}
Let $p: E\rtarr B$ be a fibration. We then have the commutative diagram
$$\diagram
E \drto_{p} \drto_p \rrto^{\nu}& & Np \dlto^{\rh} \\
& B & \\
\enddiagram$$
where $\nu(e)=(e,c_{p(e)})$ and $\rh(e,\ch)=\ch(1)$. The obvious homotopy inverse 
$\pi: Np\rtarr E$ is not a map over $B$, but the proposition ensures that it is 
homotopic to a map over $B$ that is homotopy inverse to $\nu$ over $B$.
\end{exmp}

The result generalizes as follows, the proof again being dual to the proof of the
corresponding result for cofibrations.

\begin{prop} Assume given a commutative diagram
$$\diagram
D \rto^{f} \dto_{p} & E \dto^{q} \\
A \rto_{d} & B\\
\enddiagram$$
in which $p$ and $q$ are fibrations and $d$ and $f$ are homotopy equivalences.
Then $(f,d):p\rtarr q$ is a homotopy equivalence of fibrations.
\end{prop}

The statement means that there are homotopy inverses $e$ of $d$ and $g$
of $f$ such that $p\com g=e\com q$ together with homotopies $H: g\com f\htp\id$ and
$K: f\com g\htp \id$ that cover homotopies $h: e\com d\htp\id$ and $k: d\com e\htp\id$.

\section{Change of fiber}

Translation of fibers along paths in the base space played a fundamental
role in our study of covering spaces. Fibrations admit an up to homotopy
version of that theory that well illustrates the use of the CHP and will
be used later.

Let $p:E\rtarr B$ be a fibration with fiber $F_b$ over $b\in B$
and let $i_b: F_b\rtarr E$ be the inclusion.
For a path $\be: I\rtarr B$ from $b$ to $b'$, the CHP gives a lift $\tilde{\be}$
in the diagram
$$\diagram
F_b\times \sset{0} \dto \rrto^{i_b} & & E \dto^p\\
F_b\times I \urrdashed^{\tilde{\be}}|>\tip \rto_{\pi_2} & I \rto_{\be} & B.\\
\enddiagram$$
At time $t$, $\tilde{\be}$ maps $F_b$ to the fiber $F_{\be(t)}$. In particular, at $t=1$, 
this gives a map 
$$\ta[\be]\equiv [\tilde{\be_1}]: F_b\rtarr F_{b'},$$
which we call the translation of fibers along the path class $[\be]$. 

We claim that, as indicated by our choice of notation, the homotopy class of the 
map $\tilde{\be_1}$ is independent of the choice of $\be$ in its path class. Thus 
suppose that $\be$ and $\be'$ are equivalent paths from $b$ to $b'$, let
$h: I\times I\rtarr B$ be a homotopy $\be\htp\be'$ through paths from $b$ to 
$b'$, and let $\tilde{\be}': F_b\times I\rtarr E$ cover $\be'\pi_2$. Observe that if
$$J^2=I\times\pa I\cup \sset{0}\times I\subset I^2,$$
then the pairs $(I^2,J^2)$ and $(I\times I,I\times\sset{0})$ are homeomorphic. 
Define $f: F_b\times J^2 \rtarr E$ to be $\tilde{\be}$ on $F_b\times I\times\sset{0}$,
$\tilde{\be'}$ on $F_b\times I\times\sset{1}$, and $i_b\com \pi_1$ on 
$F_b \times\sset{0}\times I$. Then another application of the CHP gives a lift 
$\tilde{h}$ in the diagram
$$\diagram
F_b\times J^2 \dto \rrto^f & & E \dto^p \\
F_b\times I^2 \urrdashed^{\tilde{h}}|>\tip \rto_{\pi_2} & I^2 \rto_{h} & B.\\
\enddiagram$$
Thus $\tilde{h}:\tilde{\be}\htp\tilde{\be'}$ through maps $F_b\times I\rtarr E$,
each of which starts at the inclusion of $F_b$ in $E$. At time $t=1$, this gives
a homotopy $\tilde{\be_1}\htp\tilde{\be'_1}$. Thus $\ta[\be] = [\tilde{\be_1}]$
is a well defined {\em homotopy class} of maps $F_b\rtarr F_{b'}$.

We think of $\ta[\be]$ as a map in the homotopy category $h\sU$. It is clear 
that, in the homotopy category,
$$\ta[c_b]=[\id] \ \ \ \tand \ \ \ \ta[\ga\cdot\be]=\ta[\ga]\com\ta[\be]$$
if $\ga(0)=\be(1)$. It follows that $\ta[\be]$ is an isomorphism with inverse
$\ta[\be^{-1}]$. This can be stated formally as follows.

\begin{thm} Lifting of equivalence classes of paths in $B$ to homotopy classes
of maps of fibers specifies a functor $\la: \PI(B)\rtarr h\sU$. Therefore, if 
$B$ is path connected, then any two fibers of $B$ are homotopy equivalent.
\end{thm}

Just as the fundamental group $\pi_1(B,b)$ of the base space of a covering acts 
on the fiber $F_b$, so the fundamental group $\pi_1(B,b)$ of the base space of a 
fibration acts ``up to homotopy'' on the fiber, in a sense made precise by the
following corollary. For a space $X$, let $\pi_0(X)$ denote the set of path components of $X$. 
The set of homotopy equivalences of $X$ is denoted
$\text{Aut}(X)$ and is topologized as a subspace of the function space of maps $X\rtarr X$.
The composite of homotopy equivalences is a homotopy equivalence, and composition defines
a continuous product on $\text{Aut}(X)$. With this product, $\text{Aut}(X)$ is a 
``topological monoid,'' namely a space with a continuous and associative multiplication
with a two-sided identity element, but it is not a group. However, the path components
of $\text{Aut}(X)$ are the homotopy classes of homotopy equivalences of $X$, and these
do form a group under composition.

\begin{cor}
Lifting of equivalence classes of loops specifies a
homomorphism $\pi_1(B,b)\rtarr \pi_0(${\em Aut}$(F_b))$.
\end{cor}

We have the following naturality statement with respect to maps of fibrations.

\begin{thm}
Let $p$ and $q$ be fibrations in the commutative diagram
$$\diagram
D \rto^g \dto_q & E \dto^p\\
A \rto_f & B.\\
\enddiagram$$
For a path $\al: I\rtarr A$ from $a$ to $a'$, the following diagram commutes 
in $h\sU$:
$$\diagram
F_a \rto^g \dto_{\ta[\al]} & F_{f(a)} \dto^{\ta[f\com\al]} \\
F_{a'}\rto_g & F_{f(a')}. \\
\enddiagram$$
If, further, $h:f\htp f'$ and $H:g\htp g'$ in the commutative diagram
$$\diagram
D\times I \rto^H \dto_{q\times\id} & E \dto^p\\
A\times I \rto_h & B,\\
\enddiagram$$
then the following diagram in $h\sU$ also commutes, where $h(a)(t)=h(a,t)$:
$$\diagram
& F_a  \dlto_g \drto^{g'} &  \\
F_{f(a)} \rrto_{\ta[h(a)]} & & F_{f'(a)}. \\
\enddiagram$$
\end{thm}
\begin{proof}
Let $\tilde{\al}: F_a\times I\rtarr D$ lift $\al$ and 
$\tilde{\be}:F_{f(a)}\times I\rtarr E$
lift $f\com\al$. Define $j: F_a\times J^2 \rtarr E$ to be $g\com \tilde{\al}$ on 
$F_a \times I\times\sset{0}$, $\tilde{\be}\com (g\times\id)$ on 
$F_a \times I\times\sset{1}$, and $g\com \pi_1$ on $F_a \times \sset{0}\times I$. Define 
$k: I^2\rtarr B$ to be the constant homotopy which at each time $t$ is $f\com \al$. 
Another application of the CHP gives a lift $\tilde{k}$ in the diagram:
$$\diagram
F_a\times J^2 \dto \rrto^j & & E \dto^p\\
F_a\times I^2 \urrdashed^{\tilde{k}}|>\tip \rto_{\pi_2} & I^2 \rto_{k} & B.\\
\enddiagram$$
Here $\tilde{k}$ is a homotopy $g\com\tilde{\al}\htp\tilde{\be}\com(g\times\id)$ through
homotopies starting at $g\com\pi_1: F_a\times I\rtarr E$. This gives the diagram 
claimed in the first statement. For the second statement, define 
$\al: I\rtarr A\times I$ by $\al(t)=(a,t)$,
so that $h(a)=h\com\al$. Define $\tilde{\al}: F_a\rtarr F_a\times I$ by 
$\tilde{\al}(f)=(f,t)$. Then $\tilde{\al}$ lifts $\al$ and 
$$\ta[\al]=[\id]:F_a=F_a\times \sset{0}\rtarr F_a\times \sset{1}=F_a.$$
We conclude that the second statement is a special case of the first.
\end{proof}

\vspace{.1in}

\begin{center}
PROBLEM
\end{center}
\begin{enumerate}
\item Prove the proposition stated in \S5.
\end{enumerate}

\clearpage

\thispagestyle{empty}

\chapter{Based cofiber and fiber sequences}

We use cofibrations and fibrations in the category $\sT$ of based spaces
to generate two ``exact sequences of spaces'' from a given map of based
spaces. We shall write $*$ generically for the basepoints of based spaces.
Much that we do for cofibrations can be done equally well in the unbased
context of the previous chapter. However, the dual theory of fibration
sequences only makes sense in the based context.

\section{Based homotopy classes of maps}

For based spaces $X$ and $Y$, we let $[X,Y]$ denote the set of based
homotopy classes of based maps $X\rtarr Y$. This set has a natural
basepoint, namely the homotopy class of the constant map from $X$ to
the basepoint of $Y$. 

The appropriate analogue of the Cartesian product in the category of based
spaces is the ``smash product''\index{smash product} $X\sma Y$ defined by
$$ X\sma Y = X\times Y/X\wed Y.$$
Here $X\wed Y$ is viewed as the subspace of $X\times Y$ consisting of
those pairs $(x,y)$ such that either $x$ is the basepoint of $X$ or 
$y$ is the basepoint of $Y$. 

The appropriate based analogue of the function space is the subspace $F(X,Y)$ 
of $Y^X$ consisting of the based maps, with the constant based map as 
basepoint.\index{function space!based} With these definitions, we have a natural
homeomorphism of based spaces
$$F(X\sma Y,Z)\iso F(X,F(Y,Z))$$
for based spaces $X$ and $Y$.

Recall that $\pi_0(X)$ denotes the set of path components of $X$.
When $X$ is based, so is this set, and we sometimes denote it by $\pi_0(X,*)$.
Observe that $[X,Y]$ may be identified with $\pi_0(F(X,Y))$.

\section{Cones, suspensions, paths, loops}

Let $X$ be a based space. We define the cone\index{cone} on $X$ to be $CX=X\sma I$,
where $I$ is given the basepoint $1$. That is, 
$$CX = X\times I/(\sset{*}\times I\cup X\times\sset{1}).$$
We view $S^1$ as $I/\pa I$, denote its basepoint by $1$, and define the 
suspension\index{suspension} of $X$ to be $\SI X= X\sma S^1$. That is,
$$\SI X= X\times S^1/(\sset{*}\times S^1\cup X\times\sset{1}).$$
These are sometimes called the reduced cone and suspension,\index{cone!reduced}
\index{suspension!reduced} to distinguish them 
from the unreduced constructions, in which the line $\sset{*}\times I$ 
through the basepoint of $X$ is not identified to a point. We shall make use of
both constructions in our work, but we shall not distinguish them notationally.

Dually, we define the path space\index{path space} of $X$ to be $PX=F(I,X)$, where $I$ is
given the basepoint $0$. Thus the points of $PX$ are the paths in $X$ that 
start at the basepoint. We define the loop space\index{loop space} of $X$ to be $\OM X=F(S^1,X)$.
Its points are the loops at the basepoint. 

We have the adjunction
$$F(\SI X,Y) \iso F(X,\OM Y).$$
Passing to $\pi_0$, this gives that
$$[\SI X,Y]\iso [X,\OM Y].$$

Composition of loops defines a multiplication on this set. Explicitly, for
$f,g:\SI X\rtarr Y$, we write
$$(g+f)(x\sma t) = (g(x)\cdot f(x))(t)=
\begin{cases}
f(x\sma 2t) \ \ \ \ \ \ \ \ \ \, \text{if} \ 0\leq t\leq 1/2 \\
g(x\sma(2t-1)) \ \  \text{if} \ 1/2\leq t\leq 1.
\end{cases}$$

\begin{lem} $[\SI X,Y]$ is a group and $[\SI^2 X,Y]$ is an Abelian group.
\end{lem}
\begin{proof}
The first statement is proved just as for the fundamental group. For the
second, think of maps $f,g: \SI^2 X\rtarr Y$ as maps $S^2\rtarr F(X,Y)$
and think of $S^2$ as the quotient $I^2/\pa I^2$. Then a homotopy between
$g+f$ and $f+g$ can be pictured schematically as follows:
$$\diagram
\rrline \ddline & & \ddline     & 
\rrline \ddline & \ddline  & \ddline     & 
\rrline \ddline & \ddline  & \ddline     & 
\rrline \ddline & & \ddline  \\
\rrline^{^f} & & \rto & \rrline^<(0.25){^*} ^<(0.75){^f} & & \rto 
& \rrline^<(0.25){^g} ^<(0.75){^*} & & \rto & \rrline^{^g} & & \\
\rrline^{^g} & &      & \rrline^<(0.25){^g} ^<(0.75){^*} & &      
& \rrline^<(0.25){^*} ^<(0.75){^f} & &      & \rrline^{^f} & & \\
\enddiagram$$
\end{proof}

\section{Based cofibrations}

The definition of a cofibration\index{cofibration!based} has an evident based variant, in which
all given and constructed maps in our test diagrams are required to be
based. A based map $i: A\rtarr X$ that is a cofibration in the unbased 
sense is necessarily a cofibration in the based sense since the basepoint
of $X$ must lie in $A$.

We say that $X$ is ``nondegenerately based,'' or ``well pointed,''\index{nondegenerately
based space}\index{well pointed space} if the inclusion of its basepoint is a cofibration in the unbased 
sense. If $A$ and $X$ are nondegenerately based and $i: A\rtarr X$ is a based cofibration, 
then $i$ is necessarily an unbased cofibration.

We refer to based cofibrations simply as cofibrations in the rest of this
chapter.

Write $Y_+$ for the union of a space $Y$ and a disjoint basepoint and observe
that we can identify $X\sma Y_+$ with $X\times Y/\sset{*}\times Y$. 

The space $X\sma I_+$ is called the reduced cylinder\index{cylinder!reduced} on $X$, and a based 
homotopy\index{homotopy!based}  $X\times I\rtarr Y$ is the same thing as a based map 
$X\sma I_+\rtarr Y$. We change notations and write $Mf$ for the based
mapping cylinder $Y\cup_f (X\sma I_+)$ of a based map $f$. 

As in the unbased case, we conclude that a based map $i:A\rtarr X$ is a 
cofibration if and only if $Mi$ is a retract of $X\sma I_+$.  

\section{Cofiber sequences}

For a based map $f:X\rtarr Y$, define the ``homotopy cofiber''\index{homotopy cofiber} $Cf$ to be
$$Cf=Y\cup_f CX= Mf/j(X),$$
where $j: X\rtarr Mf$ sends $x$ to $(x,1)$. As in the unbased case, our
original map $f$ is the composite of the cofibration $j$ and the evident
retraction $r: Mf\rtarr Y$. Thus $Cf$ is constructed by first replacing
$f$ by the cofibration $j$ and then taking the associated quotient space.

Let $i: Y\rtarr Cf$ be the inclusion. It is a cofibration since it is the 
pushout of $f$ and the cofibration $X\rtarr CX$ that sends $x$ to $(x,0)$.
Let
$$\pi:Cf\rtarr Cf/Y\iso \SI X$$
be the quotient map. The sequence
$$ X\overto{f} Y\overto{i} Cf\overto{\pi} \SI X 
\overto{-\SI f} \SI Y \overto{-\SI i} \SI Cf \overto{-\SI \pi} \SI^2 X
\overto{\SI^2f} \SI^2 Y\rtarr\cdots$$
is called the cofiber sequence\index{cofiber sequence} generated by the map $f$; here 
$$(-\SI f)(x\sma t)=f(x)\sma(1-t).$$

These ``long exact sequences of based spaces''\index{long exact sequence!of based spaces}
give rise to long exact
sequences of pointed sets,\index{long exact sequence!of pointed sets} where a sequence 
$$S'\overto{f} S\overto{g} S''$$
of pointed sets is said to be exact if $g(s)=*$ if and only if $s=f(s')$ 
for some $s'$.

\begin{thm}
For any based space $Z$, the induced sequence
$$\cdots \rtarr [\SI Cf,Z] \rtarr [\SI Y,Z]\rtarr [\SI X,Z]
\rtarr [Cf,Z]\rtarr [Y,Z]\rtarr [X,Z]$$
is an exact sequence of pointed sets, or of groups to the left of $[\SI X,Z]$,
or of Abelian groups to the left of $[\SI^2 X,Z]$.
\end{thm}

Exactness is clear at the first stage, where we are considering the composite of
$f: X\rtarr Y$ and the inclusion $i$ of $Y$ in the cofiber $Cf$. To see this, consider 
the diagram
$$\diagram
X\rto^f & Y \dto_g \rto^(0.2)i & Cf= Y\cup_f CX \dldashed^{\tilde{g}=g\cup h}|>\tip \\
& Z. &\\
\enddiagram$$
Here $h:g\com f\htp c_*$, and we view $h$ as a map $CX\rtarr Z$. Thus we 
check exactness by using any given homotopy to extend $g$ over the cofiber. We emphasize
that this applies to any composite pair of maps of the form $(f,i)$, where $i$ is the
inclusion of the target of $f$ in the cofiber of $f$.

We claim that, up to homotopy equivalence, each consecutive pair of maps 
in our cofiber sequence is the composite of a map and the inclusion of its 
target in its cofiber. This will imply the theorem. We observe that, for any
map $f$, interchange of the cone and suspension coordinate gives a homeomorphism
$$\SI Cf\iso C(\SI f)$$
such that the following diagram commutes:
$$\diagram
\SI X\rto^{\SI f} \ddouble & \SI Y \ddouble \rto^{\SI i(f)} & \SI Cf \rto^{\SI \pi(f)} \dto^{\iso}
& \SI^2X\dto^{\tau} \\
\SI X\rto_{\SI f} & \SI Y    \rto_(0.4){ i(\SI f)} & C (\SI f) \rto_{\pi(\SI f)} & \SI^2X.\\
\enddiagram$$
Here $\ta: \SI^2X\rtarr \SI^2X$ is the homeomorphism obtained by interchanging the two 
suspension coordinates; we shall see later, and leave as an exercise here, that $\ta$ is
homotopic to $-\id$. We have written $i(f)$, $\pi(f)$, etc., to indicate the maps to which the 
generic constructions $i$ and $\pi$ are applied. Using this inductively, we see that we need only
verify our claim for the two pairs of maps $(i(f),\pi(f))$ and $(\pi(f),-\SI f)$. The 
following two lemmas will imply the claim in these two cases. More precisely, they will
imply the claim directly for the first pair and will imply that the second pair is equivalent 
to a pair of the same form as the first pair.

\begin{lem}
If $i: A\rtarr X$ is a cofibration, then the quotient map
$$\ps: Ci\rtarr Ci/CA\iso X/A$$
is a based homotopy equivalence.
\end{lem}
\begin{proof}
Since $i$ is a cofibration, there is a retraction
$$r: X\sma I_+\rtarr Mi = X\cup_i (A\sma I_+).$$
We embed $X$ as $X\times\sset{1}$ in the source and collapse out 
$A\times\sset{1}$ from the target. The resulting composite $X\rtarr Ci$
maps $A$ to $\sset{*}$ and so induces a map $\ph: X/A\rtarr Ci$. The
map $r$ restricts to the identity on $A\sma I_+$, and if we collapse out
$A\sma I_+$ from its source and target, then $r$ becomes a homotopy 
$\id\htp \ps\com\ph$. The map $r$ on $X\sma I_+$ glues together with the 
map $h: CA\sma I_+\rtarr CA$ specified by
$$h(a,s,t)=(a,\text{max}(s,t))$$
to give a homotopy $Ci\sma I_+\rtarr Ci$ from the identity to $\phi\com\ps$.
\end{proof}

\begin{lem}
The left triangle commutes and the right triangle commutes up to homotopy
in the diagram
$$\diagram
 X\rto^{f} & Y\rto^{i(f)} & Cf\rto^{\pi(f)} \drto_{i(i(f))} & \SI X 
\rto^{-\SI f} & \SI Y \rto & \cdots \\
 & & & Ci(f) \uto_{\ps} \urto_{\pi(i(f))}& & \\
\enddiagram$$
\end{lem}
\begin{proof}
Observe that $Ci(f)$ is
obtained by gluing the cones $CX$ and $CY$ along their bases via the map
$f: X\rtarr Y$. The left triangle commutes since collapsing out $CY$ from 
$Ci(f)$ is the same as collapsing out $Y$ from $Cf$. A homotopy
$h: Ci(f)\sma I_+\rtarr \SI Y$ from $\pi$ to $(-\SI f)\com\ps$ is given by
$$h(x,s,t)=(f(x),t-st) \ \ \text{on} \ \  CX$$
and
$$h(y,s,t)=(y,s+t-st) \ \ \ \text{on} \ \  CY. \qed$$
\renewcommand{\qed}{}\end{proof}

\section{Based fibrations}

Similarly, the definition of a fibration\index{fibration!based} has an evident based variant, 
in which all given and constructed maps in our test diagrams are required 
to be based. A based fibration $p: E\rtarr B$ is necessarily a fibration 
in the unbased sense, as we see by restricting to spaces of the form $Y_+$ 
in test diagrams and noting that $Y_+\sma I_+\iso (Y\times I)_+$. Less 
obviously, if $p$ is a based map that is an unbased fibration, then it
satisfies the based CHP for test diagrams in which  $Y$ is nondegenerately 
based.

We refer to based fibrations simply as fibrations in the rest of this
chapter.

Observe that a based homotopy\index{homotopy!based}  $X\sma I_+\rtarr Y$ is the same thing as a 
based map $X\rtarr F(I_+,Y)$. Here $F(I_+,Y)$ is the same space as $Y^I$,
but given a basepoint determined by the basepoint of $Y$. Therefore the based version 
of the mapping path space $Nf$\index{mapping path space} of a based map $f:X\rtarr Y$ 
is the same space 
as the unbased version, but given a basepoint determined by the given basepoints
of $X$ and $Y$. However, because path spaces are always defined with $I$ having 
basepoint $0$ rather than $1$, we find it convenient to redefine 
$Nf$ correspondingly, setting
$$Nf=\{ (x,\ch)|\ch (1)=f(x)\} \subset X\times Y^I.$$

As in the unbased case, we easily check that a based map $p:E\rtarr B$ is a 
fibration if and only if there is a based path lifting function\index{path lifting function!based} 
$$s: Np\rtarr F(I_+,E).$$

\section{Fiber sequences}

For a based map $f:X\rtarr Y$, define the ``homotopy fiber''\index{homotopy fiber} $Ff$ to be
$$Ff=X\times_f PY= \{ (x,\ch)| f(x)=\ch(1)\}\subset X\times PY.$$
Equivalently, $Ff$ is the pullback displayed in the diagram
$$\diagram
Ff\rto \dto_{\pi} & PY \dto^{p_1}\\
X\rto_f & Y, \\
\enddiagram$$
where $\pi(x,\ch)=x$. As a pullback of a fibration, $\pi$ is a fibration.

If $\rh: Nf\rtarr Y$ is defined by $\rh(x,\ch)=\ch(0)$, then $f=\rh\com\nu$, 
where $\nu(x)=(x,c_{f(x)})$, and $Ff$ is the fiber $\rh^{-1}(*)$. Thus the 
homotopy fiber $Ff$ is constructed by first replacing $f$ by the fibration 
$\rh$ and then taking the actual fiber.

Let $\io: \OM Y\rtarr Ff$ be the inclusion specified by $\io(\ch)=(*,\ch)$.
The sequence

$$ \cdots \rtarr \OM^2X \overto{\OM^2f} \OM^2Y \overto{-\OM\io} \OM Ff
\overto{-\OM \pi} \OM X 
\overto{-\OM f} \OM Y \overto{\io} Ff \overto{\pi} X
\overto{f} Y$$
is called the fiber sequence generated by the map $f$; here 
$$(-\OM f)(\ze)(t)=(f\com \ze)(1-t) \ \ \text{for}\ \ze\in \OM X.$$

These ``long exact sequences of based spaces''\index{long exact sequence!of based spaces}
also give rise to long exact
sequences of pointed sets,\index{long exact sequence!of pointed sets} this time covariantly.

\begin{thm}
For any based space $Z$, the induced sequence
$$\cdots \rtarr [Z,\OM Ff]\rtarr [Z,\OM X] \rtarr [Z,\OM Y]\rtarr [Z,Ff]
\rtarr [Z,X]\rtarr [Z,Y]$$
is an exact sequence of pointed sets, or of groups to the left of $[Z,\OM Y]$,
or of Abelian groups to the left of $[Z,\OM^2 Y]$.
\end{thm}

Exactness is clear at the first stage. To see this, consider the diagram
$$\diagram
& Z \dldashed_{\tilde{g}=g\times h}|>\tip \dto^g &\\
Ff=X\times_f PY\rto_(0.7){\pi} & X \rto_f & Y\\
\enddiagram$$
Here $h:c_*\htp f\com g$, and we view $h$ as a map $Z\rtarr PY$. Thus we 
check exactness by using any given homotopy to lift $g$ to the fiber. 

We claim that, up to homotopy equivalence, each consecutive pair of maps 
in our fiber sequence is the composite of a map and the projection from its
fiber onto its source. This will imply the theorem. We observe that, for any
map $f$, interchange of coordinates gives a homeomorphism
$$\OM Ff\iso F(\OM f)$$
such that the following diagram commutes:
$$\diagram
\OM^2 Y\rto^{\OM \io(f)} \dto_{\ta} & \OM Ff \dto^{\iso}\rto^(0.55){\OM \pi(f)} & \OM X \rto^{\OM f} 
\ddouble 
& \OM Y\ddouble \\
\OM^2 Y\rto_(0.45){\io(\OM f)} & F(\OM f)  \rto_(0.55){ \pi(\OM f)} & \OM X \rto_{\OM f} & \OM Y.\\
\enddiagram$$
Here $\ta$ is obtained by interchanging the loop coordinates and is homotopic to $-\id$. 
We have written $\io(f)$, $\pi(f)$, etc., to indicate the maps to which the generic 
constructions $\io$ and $\pi$ are applied. Using this inductively, we see that we need only
verify our claim for the two pairs of maps $(\io(f),\pi(f))$ and $(-\OM f,\io(f))$. The 
following two lemmas will imply the claim in these two cases. More precisely, they will
imply the claim directly for the first pair and will imply that the second pair is equivalent 
to a pair of the same form as the first pair. The proofs of the lemmas are left as exercises.

\begin{lem}
If $p: E\rtarr B$ is a fibration, then the inclusion
$$\ph: p^{-1}(*)\rtarr Fp$$
specified by $\ph (e)=(e,c_*)$ is a based homotopy equivalence.
\end{lem}

\begin{lem}
The right triangle commutes and the left triangle commutes up to homotopy
in the diagram
$$\diagram
\cdots \rto & \OM X \rto^{-\OM f} \drto_{\io(\pi (f))} & \OM Y 
\dto^{\ph}  \rto^{\io (f)} & Ff \rto^{\pi (f)} & X \rto^f & Y.\\
 & & F\pi (f) \urto_{\pi (\pi (f))} & & & \\
\enddiagram$$
\end{lem}

\section{Connections between cofiber and fiber sequences}

It is often useful to know that cofiber sequences and fiber sequences can be 
connected to one another. The adjunction between loops and suspension has 
``unit'' and ``counit'' maps
$$\et: X\rtarr \OM\SI X\ \  \tand \ \ \epz: \SI\OM X\rtarr X.$$
Explicitly, $\et(x)(t)=x\sma t$ and $\epz(\ch\sma t)= \ch(t)$ for $x\in X$, $\ch\in\OM X$, 
and $t\in S^1$. For a map $f: X\rtarr Y$, we define 
$$\et: Ff\rtarr \OM Cf \ \ \tand \ \ \epz: \SI Ff\rtarr Cf$$
by 
$$\et(x,\ga)(t)=\epz(x,\ga,t)=\begin{cases}
\ga(2t) \ \ \ \ \ \ \ \ \ \ \text{if}\ \ t\leq 1/2 \\
(x,2t-1) \ \ \ \, \text{if}\ \ t\geq 1/2\\
\end{cases}$$
for $x\in X$ and $\ga\in PY$ such that $\ga(1)=f(x)$. Thus $\epz$ is just the
adjoint of $\et$. 

\begin{lem} Let $f: X\rtarr Y$ be a map of based spaces. Then the following
diagram, in which the top row is the suspension of part of the fiber sequence 
of $f$ and the bottom row is the loops on part of the cofiber sequence of $f$,
is homotopy commutative:
$$\diagram
 & \SI\OM Ff \dto_{\epz} \rto^{\SI\OM p} & \SI\OM X \dto_{\epz} \rto^{\SI\OM f} 
& \SI \OM Y \rto^{\SI\io} \dto^{\epz} 
& \SI Ff \dto^{\epz} \rto^{\SI p} & \SI X \ddouble \\
\OM Y\rto^{\io} \ddouble & Ff \rto^{p} \dto_{\eta} & X \rto^{f} \dto_{\et} & 
Y \rto^{i} \dto^{\et} & Cf \rto^{\pi} \dto^{\et} & \SI X \\
\OM Y\rto_{\OM i} & \OM Cf \rto_{\OM\pi} & 
\OM\SI X \rto_{\OM\SI f} & \OM\SI Y \rto_{\OM\SI i} & \OM\SI Cf.\\
\enddiagram$$
\end{lem}
\begin{proof}
Four of the squares commute by naturality and the remaining four squares 
consist of two pairs that are adjoint to each other. To see that the
two bottom left squares commute up to homotopy one need only write down
the relevant maps explicitly.
\end{proof}

Another easily verified result along the same lines relates the quotient
map $(Mf,X)\rtarr (Cf,*)$ to $\et: Ff\rtarr \OM Cf$. Here in the based 
context we let $Mf$ be the reduced mapping cylinder, in which the line 
through the basepoint of $X$ is collapsed to a point.

\begin{lem}
Let $f: X\rtarr Y$ be a map of based spaces. Then the following diagram is 
homotopy commutative, where $j: X\rtarr Mf$ is the inclusion, $r:Mf\rtarr Y$
is the retraction, and $\pi$ is induced by the quotient map $Mf\rtarr Cf$:
$$\diagram
Fj=X\times_jPMf\rrto^{Fr=\id\times Pr} \drto_{\pi} & &X\times_f PY=Ff \dlto^{\et}\\
& \OM Cf. & \\
\enddiagram$$
\end{lem}

\vspace{.1in}

\begin{center}
PROBLEM
\end{center}
\begin{enumerate}
\item Prove the two lemmas stated at the end of \S6.
\end{enumerate}

\clearpage

\thispagestyle{empty}

\chapter{Higher homotopy groups}

The most basic invariants in algebraic topology are the homotopy groups.
They are very easy to define, but very hard to compute. We give the
basic properties of these groups here.

\section{The definition of homotopy groups}

For $n\geq 0$ and a based space $X$, define\index{homotopy groups}
$$\pi_n(X)=\pi_n(X,*)=[S^n,X],$$
the set of homotopy classes of based maps $S^n\rtarr X$. This is a 
group if $n\geq 1$ and an Abelian group if $n\geq 2$. When $n=0$ and
$n=1$, this agrees with our previous definitions. Observe that
$$\pi_n(X)=\pi_{n-1}(\OM X)=\cdots=\pi_0(\OM^nX).$$

For $*\in A\subset X$, the (homotopy) fiber of the inclusion $A\rtarr X$
may be identified with the space $P(X;*,A)$ of paths in $X$ that begin at
the basepoint and end in $A$. For $n\geq 1$, define 
$$\pi_n(X,A)=\pi_n(X,A,*)=\pi_{n-1}P(X;*,A).$$
This is a group if $n\geq 2$ and an Abelian group if $n\geq 3$. Again,
$$\pi_n(X,A)=\pi_0(\OM^{n-1}P(X;*,A)).$$
These are called relative homotopy groups.\index{homotopy groups!relative}

\section{Long exact sequences associated to pairs}

With $Fi=P(X;*,A)$, we have the fiber sequence
$$\cdots\rtarr \OM^2A\rtarr \OM^2X \rtarr \OM Fi\rtarr \OM A
\rtarr \OM X \overto{\io} Fi\overto{p_1} A \overto{i}X$$
associated to the inclusion $i: A\rtarr X$, where $p_1$ is the endpoint
projection and $\io$ is the inclusion. Applying the functor 
$\pi_0(-)=[S^0,-]$ to this sequence, we obtain the long exact sequence
$$\cdots\rtarr \pi_n(A) \rtarr \pi_n(X) \rtarr \pi_n(X,A)
\overto{\pa} \pi_{n-1}(A)
\rtarr \cdots \rtarr \pi_0(A)\rtarr \pi_0(X).$$

Define 
$$J^n=\pa I^{n-1}\times I\cup I^{n-1}\times\sset{0}\subset I^n,$$
with $J^1=\sset{0}\subset I$. We can write 
$$\pi_n(X,A,*)=[(I^n,\pa I^n,J^n),(X,A,*)],$$
where the notation indicates the homotopy classes of maps of triples:\index{triple}
maps and homotopies carry $\pa I^n$ into A and $J^n$ to the basepoint. Then
$$\pa: \pi_n(X,A)\rtarr \pi_{n-1}(A)$$
is obtained by restricting maps 
$$(I^n,\pa I^n,J^n)\rtarr (X,A,*)$$ 
to maps
$$(I^{n-1}\times\sset{1},\pa I^{n-1}\times\sset{1})\rtarr (A,*),$$
while $\pi_n(A)\rtarr \pi_n(X)$ and $\pi_n(X)\rtarr \pi_n(X,A)$ are
induced by the inclusions
$$(A,*)\subset (X,*) \ \ \ \tand \ \ \ (X,*,*)\subset (X,A,*).$$

\section{Long exact sequences associated to fibrations}

Let $p: E\rtarr B$ be a fibration, where $B$ is path connected. Fix a 
basepoint $*\in B$, let $F=p^{-1}(*)$, and fix a basepoint $*\in F\subset E$.
The inclusion $\ph: F\rtarr Fp$ is a homotopy equivalence, and, being
pedantically careful to choose signs appropriately, we obtain the following
diagram, in which two out of each three consecutive squares commute and the
third commutes up to homotopy:
$$\diagram
\cdots\rto & \OM^2E \rto^{-\OM\io} \dto^{\id} 
& \OM Fi \rto^{-\OM p_1} \dto^{-\OM p} & \OM F \dto^{\OM\ph}
\rto^{-\OM i} & \OM E \dto^{\id} \rto^{\io} & Fi \rto^{p_1} \dto^{-p} 
& F \rto^{i} \dto^{\ph} & E \dto^{\id}  \\
\cdots\rto & \OM^2E \rto_{\OM^2p} & \OM^2B \rto_{-\OM \io} & \OM Fp
\rto_{-\OM \pi} & \OM E \rto_{-\OM p} & \OM B\rto_{\io} & Fp \rto_{\pi}
 & E.\\
\enddiagram$$
Here $Fi=P(E;*,F)$, $p(\xi)=p\com \xi\in\OM B$ for $\xi\in Fi$, and the next to 
last square commutes up to the homotopy $h: \io\com(-p)\htp\ph\com p_1$ 
specified by
$$h(\xi,t)=(\xi(t),p(\xi[1,t])),$$
where $\xi[1,t](s)=\xi(1-s+st)$. 

Passing to long exact sequences of homotopy groups and using the five lemma, 
together with a little extra argument in the case $n=1$, we conclude that
$$p_*: \pi_n(E,F)\rtarr \pi_n(B)$$
is an isomorphism for $n\geq 1$. This can also be derived directly from the
covering homotopy property.

Using $\ph_*$ to identify $\pi_*F$ with $\pi_*(Fp)$, we may rewrite the long
exact sequence of the bottom row of the diagram as
$$\cdots\rtarr \pi_n(F) \rtarr \pi_n(E) \rtarr \pi_n(B)
\overto{\pa} \pi_{n-1}(F)
\rtarr \cdots \rtarr \pi_0(E)\rtarr \sset{*}.$$
(At the end, a little path lifting argument shows that $\pi_0(F)\rtarr \pi_0(E)$ is
a surjection.) This is one of the main tools for the computation of homotopy groups.

\section{A few calculations}

We observe some easily derived calculational facts about homotopy groups.

\begin{lem}  If $X$ is contractible, then $\pi_n(X)=0$ for all $n\geq 0$.
\end{lem}

\begin{lem}  If $X$ is discrete, then $\pi_n(X)=0$ for all $n > 0$.
\end{lem}

\begin{lem}  If $p: E\rtarr B$ is a covering, then 
$p_*: \pi_n(E)\rtarr \pi_n(B)$ is an isomorphism for all $n\geq 2.$
\end{lem}

\begin{lem} $\pi_1(S^1)=\bZ$ and $\pi_n(S^1)=0$ if $n\neq 1$.
\end{lem}

\begin{lem} If $i\geq 2$, then $\pi_1(\bR P^i)=\bZ_2$ and 
$\pi_n(\bR P^i)\iso \pi_n(S^i)$ for $n\neq 1$.
\end{lem}

\begin{lem} For all spaces $X$ and $Y$ and all $n$, 
$$\pi_n(X\times Y)\iso \pi_n(X)\times \pi_n(Y).$$
\end{lem}

\begin{lem} If $i<n$, then $\pi_i(S^n)=0$.
\end{lem}
\begin{proof}
For any based map $f: S^i\rtarr S^n$, we can apply smooth (or simplicial)
approximation to obtain a based homotopy from $f$ to a map that misses a 
point $p$, and we can then deform $f$ to the trivial map by contracting
$S^n-\sset{p}$ to the basepoint.
\end{proof}

There are three standard bundles, called the Hopf bundles,\index{Hopf bundles} that 
can be used to obtain a bit more information about the homotopy groups of
spheres. Recall that $\bC P^1$ is the space of complex lines in $\bC^2$.
That is, $\bC P^1=(\bC\times\bC -\sset{0})/(\sim)$, where 
$(z_1,z_2)\sim (\la z_1,\la z_2)$ for complex numbers $\la$, $z_1$, and
$z_2$. Write $[z_1,z_2]$ for the equivalence class of $(z_1,z_2)$. We
obtain a homeomorphism $\bC P^1\rtarr S^2$ by identifying $S^2$ with the
one-point compactification of $\bC$ and mapping $[z_1,z_2]$ to $z_2/z_1$
if $z_1\neq 0$ and to the point at $\infty$ if $z_1=0$. The Hopf map
$\et: S^3\rtarr S^2$ is specified by $\et(z_1,z_2) = [z_1,z_2]$, where
$S^3$ is identified with the unit sphere in the complex plane $\bC^2$. It is
a worthwhile exercise to check that $\et$ is a bundle with fiber $S^1$.
By use of the quaternions and Cayley numbers, we obtain analogous Hopf
maps $\nu: S^7\rtarr S^4$ and $\si: S^{15}\rtarr S^8$. Then $\nu$ is
a bundle with fiber $S^3$ and $\si$ is a bundle with fiber $S^7$. Since
we have complete information on the homotopy groups of $S^1$, the long
exact sequence of homotopy groups associated to $\et$ has the following
direct consequence.

\begin{lem}
$\pi_2(S^2)\iso \bZ$ and $\pi_n(S^3)\iso \pi_n(S^2)$ for $n\geq 3$. 
\end{lem}

We shall later prove the following more substantial result.

\begin{thm}
For all $n\geq 1$, $\pi_n(S^n)\iso \bZ$.
\end{thm}

It is left as an exercise to show that the long exact sequence associated
to $\nu$ implies that $\pi_7(S^4)$ contains an element of infinite order,
and $\si$ can be used similarly to show the same for $\pi_{15}(S^8)$.

In fact, the homotopy groups $\pi_q(S^n)$ for $q>n>1$ are all finite
except for $\pi_{4n-1}(S^{2n})$, which is the direct sum of $\bZ$ and
a finite group.

The difficulty of computing homotopy groups is well illustrated by the fact 
that there is no non-contractible simply connected compact manifold (or
finite CW complex) all of whose homotopy groups are known. We shall find
many non-compact spaces whose homotopy groups we can determine completely.
Such computations will rely on the following observation.

\begin{lem}
If $X$ is the colimit of a sequence of inclusions $X_i\rtarr X_{i+1}$
of based spaces, then the natural map 
$$ \colim_i\pi_n(X_i)\rtarr \pi_n(X)$$
is an isomorphism for each $n$.
\end{lem}
\begin{proof}
This follows directly from the point-set topological fact that if $K$ is a 
compact space, then a map $K\rtarr X$ has image in one of the $X_i$.
\end{proof}

\section{Change of basepoint}

We shall use our results on change of fibers to generalize our results on 
change of basepoint from the fundamental group to the higher absolute and
relative homotopy groups. In the absolute case, we have the identification
$$\pi_n(X,x)=[(S^n,*),(X,x)],$$
where we assume that $n\geq 1$. Since the inclusion of the basepoint in 
$S^n$ is a cofibration, evaluation
at the basepoint gives a fibration $p: X^{S^n}\rtarr X$. We may identify
$\pi_n(X,x)$ with $\pi_0(F_x)$ since a path in $F_x$ is just a based homotopy
$h: S^n\times I\rtarr X$ with respect to the basepoint $x$. Another way to see
this is to observe that $F_x$ is the $n$th loop space $\OM^nX$, specified with 
respect to the basepoint $x$. A path class
$[\xi]: I\rtarr X$ from $x$ to $x'$ induces a homotopy equivalence 
$\ta[\xi]: F_x\rtarr F_{x'}$, and we continue to write $\ta[\xi]$ for
the induced bijection
$$\ta[\xi]: \pi_n(X,x)\rtarr \pi_n(X,x').$$

This bijection is an isomorphism of groups. One conceptual way to see this
is to observe that addition is induced from the ``pinch map'' $S^n\rtarr S^n\wed S^n$
that is obtained by collapsing an equator to the basepoint. That is, the sum of maps
$f,g: S^n\rtarr X$ is the composite
$$S^n \rtarr S^n\wed S^n \overto{f\wed g} X\wed X\overto{\triangledown} X,$$
where $\triangledown$ is the folding map, which restricts to the identity map
$X\rtarr X$ on each wedge summand. Evaluation at the basepoint of $S^n\wed S^n$ gives
a fibration $X^{S^n\wed S^n}\rtarr X$, and the pinch map induces a map of fibrations
$$\diagram
X^{S^n\wed S^n} \dto \rto & X^{S^n} \dto\\
X \rdouble & X.\\
\enddiagram$$
The fiber over $x$ in the left-hand fibration is the product $F_x\times F_x$,
where $F_x$ is the fiber over $x$ in the right-hand fibration. In fact, the induced map of
fibers can be identified as the map $\OM^nX\times \OM^nX\rtarr \OM^nX$ given by composition
of loops (using the first loop coordinate say). By the naturality of
translations of fibers with respect to maps of fibrations, we have a homotopy commutative 
diagram
$$\diagram
F_x\times F_x \rto \dto_{\ta[\xi]\times \ta[\xi]} & F_x \dto^{\ta[\xi]}\\
F_{x'}\times F_{x'} \rto & F_{x'}
\enddiagram$$
in which the horizontal arrows induce addition on passage to $\pi_0$.

We can argue similarly in the relative case. The triple $(I^n,\pa I^n,J^n)$ 
is homotopy equivalent to the triple $(CS^{n-1},S^{n-1},*)$, as we see by 
quotienting out $J^n$. Therefore, for $a\in A$, we have the identification
$$\pi_n(X,A,a)\iso [(CS^{n-1},S^{n-1},*),(X,A,a)].$$
Using that the inclusions $\sset{*}\rtarr S^{n-1}$ and $S^{n-1}\rtarr CS^{n-1}$
are both cofibrations, we can check that evaluation at $*$ specifies a fibration
$$p: (X,A)^{(CS^{n-1},S^{n-1})}\rtarr A,$$
where the domain is the subspace of $X^{CS^{n-1}}$ consisting of the
indicated maps of pairs. We may identify $\pi_n(X,A,a)$ with $\pi_0(F_a)$. A path 
class $[\al]: I\rtarr A$ from $a$ to $a'$ induces a homotopy equivalence 
$\ta[\al]: F_a\rtarr F_{a'}$, and we continue to write $\ta[\al]$ for
the induced isomorphism
$$\ta[\al]: \pi_n(X,A,a)\rtarr \pi_n(X,A,a').$$
Our naturality results on change of fibers now directly imply the desired results
on change of basepoint.

\begin{thm}
If $f: (X,A)\rtarr (Y,B)$ is a map of pairs and $\al: I\rtarr A$ is a path
from $a$ to $a'$, then the following diagram commutes:
$$\diagram
\pi_n(X,A,a) \dto_{\ta[\al]} \rto^{f_*} & \pi_n(Y,B,f(a)) \dto^{\ta[f\com\al]}\\
\pi_n(X,A,a')  \rto^{f_*} & \pi_n(Y,B,f(a'))\\
\enddiagram$$
If $h:f\htp f'$ is a homotopy of maps of pairs and $h(a)(t)=h(a,t)$, then the 
following diagram commutes:
$$\diagram
& \pi_n(X,A,a) \dlto_{f_*} \drto^{f'_*} &\\
\pi_n(Y,B,f(a)) \rrto_{\ta[h(a)]} & & \pi_n(Y,B,f'(a)).\\
\enddiagram$$
The analogous conclusions hold for the absolute homotopy groups.
\end{thm}

Therefore, up to non-canonical isomorphism, the homotopy groups of $(X,A)$ are
independent of the choice of basepoint in a given path component of $A$. 

\begin{cor}
A homotopy equivalence of spaces or of pairs of spaces induces an isomorphism
on all homotopy groups.
\end{cor}

We shall soon show that the converse holds for a quite general class of spaces, 
namely the class of CW complexes, but we first need a few preliminaries. 

\section{$n$-Equivalences, weak equivalences, and a technical lemma}

\begin{defn}
A map $e: Y\rtarr Z$ is an $n$-equivalence\index{nequivalence@$n$-equivalence} if, for all $y\in Y$, the map
$$e_*:\pi_q(Y,y)\rtarr \pi_q(Z,e(y))$$
is an injection for $q<n$ and a surjection for $q\leq n$; $e$ is said to be
a weak equivalence\index{weak equivalence} if it is an $n$-equivalence for all $n$. 
\end{defn}

Thus any homotopy equivalence is a weak equivalence. The following technical lemma 
will be at the heart of our study of CW complexes, but it will take some getting 
used to. It gives a useful criterion for determining when a given map is an 
$n$-equivalence. 

It is convenient to take $CX$ to be the unreduced\index{cone!unreduced} cone 
$X\times I/X\times \sset{1}$ 
here. If $f,f': (X,A)\rtarr (Y,B)$ are maps of pairs such that $f=f'$ on $A$, then we 
say that $f$ and $f'$ are homotopic\index{homotopy!relative to a subspace} relative to $A$ 
if there is a homotopy $h: f\htp f'$ 
such that $h$ is constant on $A$, in the sense that $h(a,t)=f(a)$ for all $a\in A$ and 
$t\in I$; we write $h:f\htp f'\ \text{rel}\ A$. Observe that $\pi_{n+1}(X,x)$ can be
viewed as the set of relative homotopy classes of maps $(CS^n,S^n)\rtarr (X,x)$.

\begin{lem}
The following conditions on a map $e: Y\rtarr Z$ are equivalent.
\begin{enumerate}
\item[(i)] For any $y\in Y$, $e_*: \pi_q(Y,y)\rtarr \pi_q(Z,e(y))$ is an injection for 
$q=n$ and a surjection for $q=n+1$.
\item[(ii)] Given maps $f:CS^n \rtarr Z$, $g:S^n \rtarr Y$, 
and $h:S^n\times I \rtarr Z$ such that $f|S^n=h\com i_0$ and $e\com g=h\com i_1$ 
in the following diagram, there are maps $\tilde g$ and $\tilde h$ that make the 
entire diagram commute.
$$
\diagram
S^n \ddto \rrto^{i_0} & & S^n \times I \dlto_h \ddto & & S^n \llto_{i_1} \dlto_{g}
\ddto \\
& Z & & Y \llto_<(0.4){e} & \\
CS^n \rrto_{i_0} \urto^{f} && CS^n \times I \uldashed^{\tilde{h}}|>\tip & & 
CS^n \uldashed^{\tilde{g}}|>\tip \llto^{i_1} \\
\enddiagram $$
\item[(iii)] The conclusion of (ii) holds when $f|S^n=e\com g$ and $h$ is the 
constant homotopy at this map.
\end{enumerate}
\end{lem}
\begin{proof}
Trivially (ii) implies (iii). We first show that (iii) implies (i). If $n=0$, (iii)
says (in part) that if $e(y)$ and $e(y')$ can be connected by a path in $Z$, then 
$y$ and $y'$ can be connected by a path in $Y$. If $n>0$, then (iii) says (in part) 
that if $e\com g$ is null homotopic, then $g$ is null homotopic. Therefore 
$\pi_n(e)$ is injective. If we specialize (iii) by letting $g$ be the constant map
at a point $y\in Y$, then $f$ is a map $(CS^n,S^n)\rtarr (Z,e(y))$, $\tilde{g}$ is a map
$(CS^n,S^n)\rtarr (Y,y)$, and $\tilde{h}: f\htp e\com \tilde{g}\ \text{rel}\ S^n$. 
Therefore $\pi_{n+1}(e)$ is surjective.

Thus assume (i). We must prove (ii), and we assume given $f$, $g$, and $h$ making
the solid arrow part of the diagram commute. The idea is to use (i) to show that
the $n$th homotopy group of the fiber $F(e)$ is zero, to use the given part of the
diagram to construct a map $S^n\rtarr F(e)$, and to use a null homotopy of that 
map to construct $\tilde{g}$ and $\tilde{h}$. However, since homotopy groups involve
choices of basepoints and the diagram makes no reference to basepoints, the details
require careful tracking of basepoints. Thus fix a basepoint $*\in S^n$, let $\bullet$ be 
the cone point of $CS^n$, and define
$$y_1=g(*),\ \ z_1=e(y_1),\ \ z_0=f(*,0),\ \ \tand\ \ z_{-1}=f(\bullet).$$
For $x\in S^n$, let $f_x: I\rtarr Z$ and $h_x: I\rtarr Z$ be the paths 
$f_x(s)=f(x,s)$ from $f(x,0)=h(x,0)$ to $z_{-1}$ and $h_x(t)=h(x,t)$ from $h(x,0)$ to 
$h(x,1)=(e\com g)(x)$. Consider the homotopy fiber
$$F(e;y_1)=\sset{(y,\ze)|\ze(0)=z_1 \, \tand \, e(y)=\ze(1)}\subset Y\times Z^I.$$
This has basepoint $w_1=(y_1,c_{z_1})$. By (i) and the exact sequence
$$\pi_{n+1}(Y,y_1)\overto{e_*}\pi_{n+1}(Z,z_1)\rtarr \pi_n(F(e;y_1),w_1)\rtarr
\pi_n(Y,y_1)\overto{e_*}\pi_n(Z,z_1),$$ 
we see that $\pi_n(F(e;y_1),w_1)=0$. Define $k_0: S^n\rtarr F(e;y_1)$ by
$$k_0(x)=(g(x),h_x\cdot f_x^{-1}\cdot f_*\cdot h_*^{-1}).$$
While $k_0$ is not a based map, $k_0(*)$ is connected to the basepoint since
$h_*\cdot f_*^{-1}\cdot f_*\cdot h_*^{-1}$ is equivalent to $c_{z_1}$. By HEP for
the cofibration $\sset{*}\rtarr S^n$, $k_0$ is homotopic to a based map. This based map is
null homotopic in the based sense, hence $k_0$ is null homotopic in the unbased
sense. Let $k: S^n\times I\rtarr F(e;y_1)$ be a homotopy from $k_0$ to the
trivial map at $w_1$. Write 
$$k(x,t)=(\tilde{g}(x,t),\ze(x,t)).$$
Then $\tilde{g}(x,1)=y_1$ for all $x\in S^n$, so that $\tilde{g}$ factors through
a map $CS^n\rtarr Y$, and $\tilde{g}=g$ on $S^n$. We have a map 
$j: S^n\times I\times I$ given by $j(x,s,t)=\ze(x,t)(s)$ that behaves 
as follows on the boundary of the square for each fixed $x\in S^n$, where 
$\tilde{g}_x(t)=\tilde{g}(x,t)$:
$$\diagram
\rrline^{c_{z_1}}|\tip & & \\
& & \\
\uuline^{c_{z_1}}|\tip \rrline_{h_x\cdot f_x^{-1}\cdot f_*\cdot h_*^{-1}}|\tip
& & \uuline_{e\com \tilde{g}_x}|\tip \\
\enddiagram$$
The desired homotopy $\tilde{h}$, written $\tilde{h}(x,s,t)$ where $s$ is the cone
coordinate and $t$ is the interval coordinate, should behave as follows on the boundary
of the square:
$$\diagram
\rrline^{e\com \tilde{g}_x}|\tip & & \\
& & \\
\uuline^{h_x}|\tip \rrline_{f_x}|\tip
& & \uuline_{h_*\cdot f_*^{-1}}|\tip \\
\enddiagram$$
Thus we can obtain $\tilde{h}$ by composing $j$ with a suitable
reparametrization $I^2\rtarr I^2$ of the square.
\end{proof}

\vspace{.1in}

\begin{center}
PROBLEMS
\end{center}
\begin{enumerate}
\item Show that, if $n\geq 2$, then $\pi_n(X\wed Y)$ is isomorphic to
$$\pi_n(X)\oplus\pi_n(Y)\oplus \pi_{n+1}(X\times Y, X\wed Y).$$
\item Compute $\pi_n(\bR P^n,\bR P^{n-1})$ for $n\geq 2$. Deduce that the quotient map 
$$(\bR P^n,\bR P^{n-1})\to (\bR P^n/\bR P^{n-1},*)$$ 
does not induce an isomorphism of homotopy groups.
\item Compute the homotopy groups of complex projective space $\bC P^n$ in terms 
of the homotopy groups of spheres.
\item Verify that the ``Hopf bundles'' are in fact bundles.
\item Show that $\pi_7(S^4)$ contains an element of infinite order. 
\item Compute all of the homotopy groups of $\bR P^{\infty}$ and $\bC P^{\infty}$.
\end{enumerate}

\clearpage

\thispagestyle{empty}

\chapter{CW complexes}

We introduce a large class of spaces, called CW complexes, between which a
weak equivalence is necessarily a homotopy equivalence. Thus,
for such spaces, the homotopy groups are, in a sense, a complete set of 
invariants. Moreover, we shall see that every space is weakly 
equivalent to a CW complex.

\section{The definition and some examples of CW complexes}

Let $D^{n+1}$ be the unit disk\index{disk} $\{ x\, |\, |x|\leq 1\} \subset \bR^{n+1}$ 
with boundary $S^n$. 

\begin{defn} (i) A CW complex\index{CW complex} $X$ is a space $X$ which is the union of an
expanding sequence of subspaces $X^n$ such that, inductively, $X^0$ is
a discrete set of points (called vertices)\index{vertex} and $X^{n+1}$ is the pushout
obtained from $X^n$ by attaching disks $D^{n+1}$ along ``attaching maps''\index{attaching map}
$j: S^n\rtarr X^n$. Thus $X^{n+1}$ is the quotient space obtained from 
$X^n\cup (J_{n+1}\times D^{n+1})$ by identifying $(j,x)$ with $j(x)$ for 
$x\in S^n$, where $J_{n+1}$ is the discrete set of such attaching maps $j$.
Each resulting map
$D^{n+1}\rtarr X$ is called a ``cell.''\index{cell} The subspace $X^n$ is called the
$n$-skeleton of $X$.\index{skeleton!of a CW complex} 

(ii) More generally, given any space $A$, we 
define a relative 
CW complex\index{CW complex!relative} $(X,A)$ in the same fashion, but with 
$X^0$ replaced by the union of 
$A$ and a (possibly empty) discrete set of points; we write $(X,A)^n$, or $X^n$ 
when $A$ is clear from the context, for the relative $n$-skeleton, and we say
that $(X,A)$ has dimension $\leq n$ if $X=X^n$. 

(iii) A subcomplex\index{subcomplex!of a CW complex} $A$ of a CW
complex $X$ is a subspace and a CW complex such that the composite of each
cell $D^n\rtarr A$ of $A$ and the inclusion of $A$ in $X$ is a cell of $X$.
That is, $A$ is the union of some of the cells of $X$. The pair $(X,A)$ can
then be viewed as a relative CW complex. 

(iv) A map of pairs $f:(X,A)\rtarr (Y,B)$ 
between relative CW complexes is said to be ``cellular'' if $f(X^n)\subset Y^n$
for all $n$.\index{cellular map}
\end{defn}

Of course, pushouts and unions are understood in the topological sense, with the
compactly generated topologies. A subspace of $X$ is closed if and only if its 
intersection with each $X^n$ is closed.

\begin{exmps} (i) A graph is a one-dimensional CW complex.

(ii) Via a homeomorphism $I\times I\iso D^2$, the standard presentations of the 
torus\index{torus}
$T=S^1\times S^1$, the projective plane\index{projective plane} $\bR P^2$, and the 
Klein bottle\index{Klein bottle} $K$ as quotients 
of a square display these spaces as CW complexes with
one or two vertices, two edges, and one $2$-cell:

\begin{small}
$$\diagram
& T & \\
v \rrline^{e_1}|\tip \ddline_{e_2}|{\rotate\tip} & & v \ddline^{e_2}|{\rotate\tip} \\
& & \\
v \rrline_{e_1}|\tip      & & v         \\
\enddiagram
\hspace{.2in}
\diagram
& \bR P^2 & \\
v_1 \rrline^{e_1}|\tip \ddline_{e_2}|{\rotate\tip} & & v_2 \ddline^{e_2}|\tip \\
& & \\
v_2 \rrline_{e_1}|{\rotate\tip}         & & v_1         \\
\enddiagram
\hspace{.2in}
\diagram
& K & \\
v \rrline^{e_1}|\tip \ddline_{e_2}|{\rotate\tip} & & v \ddline^{e_2}|\tip \\
& & \\
v \rrline_{e_1}|\tip         & & v         \\
\enddiagram$$
\end{small}

(iii) For $n\geq 1$, $S^n$ is a CW complex with one vertex $\sset{*}$ and
one $n$-cell, the attaching map $S^{n-1}\rtarr \sset{*}$ being the only possible
map. Note that this entails a choice of homeomorphism $D^n/S^{n-1}\iso S^n$. If
$m<n$, then the only cellular map $S^m\rtarr S^n$ is the trivial map. If
$m\geq n$, then every based map $S^m\rtarr S^n$ is cellular.

(iv) $\bR P^n$\index{projective space!real} is a CW complex with $m$-skeleton $\bR P^m$ and with one 
$m$-cell for each $m\leq n$. The attaching map $j: S^{n-1}\rtarr \bR P^{n-1}$ is
the standard double cover. That is, $\bR P^n$ is homeomorphic to 
$\bR P^{n-1}\cup_j D^n$. Explicitly, write $\bar{x}=[x_1,\ldots\!,x_{n+1}]$, 
$\sum x_i^2=1$, for a typical point of $\bR P^n$. Then $\bar{x}$ is in $\bR P^{n-1}$
if and only if $x_{n+1}=0$. The required homeomorphism is obtained by identifying
$D^n$ and its boundary sphere with the upper hemisphere 
$$E^n_+=\{ (x_1,\ldots\!,x_{n+1})\,|\, \textstyle{\sum} {x_i^2} =1 \ \tand\ x_{n+1}\geq 0\}$$
and its boundary sphere.

(v) $\bC P^n$\index{projective space!complex} is a CW complex whose $2m$-skeleton 
and $(2m+1)$-skeleton
are both $\bC P^m$ and which has one $2m$-cell for each $m\leq n$. The attaching map 
$S^{2n-1}\rtarr \bC P^{n-1}$ is the standard bundle with fiber $S^1$, where
$S^{2n-1}$ is identified with the unit sphere in $\bC ^n$. We leave the specification
of the required homeomorphism as an exercise.
\end{exmps}

\section{Some constructions on CW complexes}

We need to know that various constructions on spaces preserve CW complexes. We
leave most of the proofs as exercises in the meaning of the definitions.

\begin{lem} If $(X,A)$ is a relative CW complex, then the quotient space $X/A$
is a CW complex with a vertex corresponding to $A$ and one $n$-cell for each
relative $n$-cell of $(X,A)$.
\end{lem}

\begin{lem} For CW complexes $X_i$ with basepoints that are vertices, the
wedge of the $X_i$ is a CW complex which contains each $X_i$ as a subcomplex.
\end{lem}

\begin{lem}
If $A$ is a subcomplex of a CW complex $X$, $Y$ is a CW complex, and $f: A\rtarr Y$
is a cellular map, then the pushout $Y\cup_f X$  is a CW complex that contains $Y$ as
a subcomplex and has one cell for each cell of $X$ that is not in $A$. The quotient
complex $(Y\cup_f X)/Y$ is isomorphic to $X/A$.
\end{lem}

\begin{lem}
The colimit of a sequence of inclusions of subcomplexes $X_n\rtarr X_{n+1}$ in CW
complexes is a CW complex that contains each of the $X_i$ as a subcomplex. 
\end{lem}

\begin{lem}
The product $X\times Y$ of CW complexes $X$ and $Y$ is a CW complex with an $n$-cell
for each pair consisting of a $p$-cell of $X$ and $q$-cell of $Y$, where $p+q=n$.
\end{lem}
\begin{proof}
For $p+q=n$, there are canonical homeomorphisms
$$(D^n,S^{n-1})\iso (D^p\times D^q,D^p\times S^{q-1}\cup S^{p-1}\times D^q).$$
This allows us to define product cells.
\end{proof}

We shall look at the general case more closely later, but we point out one important 
special case for immediate use. Of course, the unit interval is a graph with two vertices
and one edge.

\begin{lem}
For a CW complex $X$, $X\times I$ is a CW complex that contains $X\times \pa I$ as a
subcomplex and, in addition, has one $(n+1)$-cell for each $n$-cell of $X$. 
\end{lem}

A ``cellular homotopy''\index{cellular homotopy} $h:f\htp f'$ between 
cellular maps $X\rtarr Y$ of CW complexes 
is a homotopy that is itself a cellular map $X\times I\rtarr Y$.

\section{HELP and the Whitehead theorem}

The following ``homotopy extension and lifting property'' is a powerful organizational 
principle for proofs of results about CW complexes. In the case 
$$(X,A)=(D^n,S^{n-1})\iso (CS^{n-1},S^{n-1}),$$ 
it is the main point of the technical lemma proved at the end of the last chapter.

\begin{thm}[HELP]\index{HELP}\index{homotopy extension and lifting property} Let 
$(X,A)$ be a relative CW complex of dimension $\leq n$ and let 
$e:Y \rtarr Z$ be an $n$-equivalence. Then, given maps $f:X \rtarr Z$, $g:A \rtarr Y$, 
and $h:A\times I \rtarr Z$ such that $f|A=h\com i_0$ and $e\com g=h\com i_1$ 
in the following diagram, there are maps $\tilde g$ and $\tilde h$ that make the 
entire diagram commute:
$$
\diagram
A \ddto \rrto^{i_0} & & A \times I \dlto_h \ddto & & A \llto_{i_1} \dlto_{g}
\ddto \\
& Z & & Y \llto_<(0.4){e} & \\
X \rrto_{i_0} \urto^{f} && X \times I \uldashed^{\tilde{h}}|>\tip & & 
X \uldashed^{\tilde{g}}|>\tip \llto^{i_1} \\
\enddiagram $$
\end{thm}
\begin{proof}
Proceed by induction over skeleta, applying the case $(D^n,S^{n-1})$ one
cell at a time to the $n$-cells of $X$ not in $A$. 
\end{proof}

In particular, if we take $e$ to be the identity map of $Y$, we see that the
inclusion $A\rtarr X$ is a cofibration. Observe that, by passage to colimits,
we are free to take $n=\infty$ in the theorem.

We write $[X,Y]$ for homotopy classes of unbased maps in this chapter, and we
have the following direct and important application of HELP.

\begin{thm}[Whitehead]\index{Whitehead theorem} If $X$ is a CW complex and $e:Y\rtarr Z$ is an 
$n$-equivalence, then $e_*:[X,Y]\rtarr [X,Z]$ is
a bijection if \text{\em dim}\,$X <n$ and a surjection if \text{\em dim}\,$X=n$.
\end{thm}
\begin{proof}
Apply HELP to the pair $(X,\emptyset)$ to see the surjectivity. Apply HELP to the pair
$(X\times I, X\times\pa I)$, taking $h$ to be a constant homotopy, to see the
injectivity.
\end{proof}

\begin{thm}[Whitehead]\index{Whitehead theorem}  An $n$-equivalence between CW complexes of 
dimension less than $n$ is a homotopy equivalence. A weak equivalence between CW complexes 
is a homotopy equivalence.
\end{thm}
\begin{proof}
Let $e: Y\rtarr Z$ satisfy either hypothesis. Since $e_*:[Z,Y]\rtarr [Z,Z]$ is a bijection, 
there is a map $f: Z\rtarr Y$ such that $e\com f\htp\id$. Then $e\com f\com e\htp e$, and, 
since $e_*:[Y,Y]\rtarr [Y,Z]$ is also a bijection, this implies that $f\com e\htp \id$.
\end{proof}

If $X$ is a finite CW complex, in the sense that it has finitely many cells, and if 
dim$\,X>1$ and $X$ is not contractible,  then it is known that $X$ has infinitely
many non-zero homotopy groups. The Whitehead theorem is thus surprisingly strong: in its
first statement, if low dimensional homotopy groups are mapped isomorphically, then so are 
all higher homotopy groups.

\section{The cellular approximation theorem}

Cellular maps are under much better algebraic control than general maps, as will become
both clear and important later. Fortunately, any map between CW complexes is homotopic to
a cellular map. We need a lemma.

\begin{defn}
A space $X$ is said to be $n$-connected\index{nconnected@$n$-connected space} if $\pi_q(X,x)=0$ 
for $0\leq q\leq n$ and all $x$.
A pair $(X,A)$ is said to be $n$-connected if $\pi_0(A)\rtarr \pi_0(X)$ is surjective and
$\pi_q(X,A,a)=0$ for $1\leq q\leq n$ and all $a$. It is equivalent that the inclusion 
$A\rtarr X$ be an $n$-equivalence.
\end{defn}

\begin{lem}
A relative CW complex $(X,A)$ with no $m$-cells for $m\leq n$ is $n$-connected. In particular,
$(X,X^n)$ is $n$-connected for any CW complex $X$.
\end{lem}
\begin{proof}
Consider $f:(I^q,\pa I^q,J^q)\rtarr (X,A,a)$, where $q\leq n$. Since the image of $f$
is compact, we may assume that $(X,A)$ has finitely many cells. By induction on the number
of cells, we may assume that $X=A\cup_jD^r$, where $r>n$. By smooth (or simplicial) 
approximation, there is a map $f': I^q\rtarr X$ such that $f'=f$ on $\pa I^q$, 
$f'\htp f \ \text{rel}\ \pa I^q$  and $f'$ misses a point $p$ in the interior of $D^r$. Clearly
we can deform $X-\sset{p}$ onto $A$ and so deform $f'$ to a map into $A$. 
\end{proof}

\begin{thm}[Cellular approximation]\index{cellular approximation theorem}
Any map $f:(X,A)\rtarr (Y,B)$ between relative CW complexes is homotopic relative to $A$ to 
a cellular map.  
\end{thm}
\begin{proof}
We proceed by induction over skeleta. To start the induction, note that any point of $Y$ 
is connected by a path to a point in $Y^0$ and apply this to the images of points of 
$X^0-A$ to obtain a homotopy of $f|X^0$ to a map into $Y^0$. Assume given 
$g_n: X^n\rtarr Y^n$ and $h_n: X^n\times I\rtarr Y$ such 
that $h_n: f|X^n\htp \io_n\com g_n$, where $\io_n: Y^n\rtarr Y$
is the inclusion. For an attaching map $j: S^n\rtarr X^n$ of a cell 
$\tilde{j}: D^{n+1}\rtarr X$, we apply HELP to the following diagram:
$$\diagram
S^n \ddto \rrto^{i_0} & & S^n\times I \dlto_{h_n\com (j\times\id)} \ddto & & S^n \llto_{i_1}  
\ddto \dlto_{g_n\com j}\\
& Y & & Y^{n+1} \llto_<(0.25){\io_{n+1}}\\
D^{n+1} \urto^{f\com\tilde{j}} \rrto_{i_0} & & D^{n+1}\times I \uldashed_{h_{n+1}}|>\tip 
& &  D^{n+1} \llto^{i_1} \uldashed_{g_{n+1}}|>\tip \\
\enddiagram$$
where $g_n\com j: S^n\rtarr Y^n$ is composed with the inclusion $Y^n\rtarr Y^{n+1}$;
HELP applies since $\io_{n+1}$ is an $(n+1)$-equivalence.
\end{proof}

\begin{cor}
For CW complexes $X$ and $Y$, any map $X \rtarr Y$ is homotopic to a 
cellular map, and any two homotopic cellular maps are cellularly homotopic.
\end{cor}

\section{Approximation of spaces by CW complexes}

The following result says that there is a functor $\GA: h\sU \rtarr h\sU$ and
a natural transformation $\ga: \GA\rtarr \Id$ that assign a CW complex $\GA X$ 
and a weak equivalence $\ga: \GA X\rtarr X$ to a space $X$.

\begin{thm}[Approximation by CW complexes] For any space $X$, there
is a CW complex\, $\GA X$ and a weak equivalence $\ga:\GA X \rtarr X$. 
For a map $f:X\rtarr Y$ and another such CW approximation $\ga: \GA Y\rtarr Y$,
there is a map $\GA f: \GA X\rtarr \GA Y$, unique up to homotopy, such that
the following diagram is homotopy commutative:
$$\diagram
\GA X \rto^{\GA f} \dto_{\ga} & \GA Y \dto^{\ga} \\
X \rto_{f} & Y.\\
\enddiagram$$
If $X$ is $n$-connected, $n\geq 1$, then $\GA X$ can be chosen to have a unique vertex and no
$q$-cells for $1\leq q\leq n$.
\end{thm}
\begin{proof} The existence and uniqueness up to homotopy of $\GA f$ will be immediate
since the Whitehead theorem will give a bijection
$$\ga_*: [\GA X,\GA Y]\rtarr [\GA X,Y].$$
Proceeding one path component at a time, we may as well assume that $X$ is path
connected, and we may then work with based spaces and based maps. We construct $\GA X$ as the 
colimit of a sequence of cellular inclusions
$$\diagram
X_1\rto^{i_1} \ddrrto_{\ga_1} & X_2\rto^{i_2} \ddrto^{\ga_2} 
& \cdots \rto & X_n\rto^{i_n} \ddlto_{\ga_n} \rto & X_{n+1} \ddllto^{\ga_{n+1}} \rto & \cdots \\
& & & & &  \\
& & X. & & & \\
\enddiagram$$
Let $X_1$ be a wedge of spheres $S^q$, $q\geq 1$, one for each pair $(q,j)$, where 
$j: S^q\rtarr X$ represents a generator of the group $\pi_q(X)$. On the $(q,j)$th
wedge summand, the map $\ga_1$ is the given map $j$. Clearly $\ga_1: X_1\rtarr X$ 
induces an epimorphism on all homotopy groups. We give $X_1$ the CW structure induced 
by the standard CW structures on the spheres $S^q$. Inductively, suppose that we have 
constructed CW complexes $X_m$, cellular inclusions $i_{m-1}$, and maps $\ga_m$ for 
$m\leq n$ such that $\ga_m\com i_{m-1}=\ga_{m-1}$ and $(\ga_m)_*:\pi_q(X_m)\rtarr \pi_q(X)$ 
is a surjection for all $q$ and a bijection for $q<m$. We construct 
$$X_{n+1}=X_n\cup(\bigvee_{(f,g)}(S^n\sma I_+)),$$
where the wedge is taken over cellular representatives $f,g: S^n\rtarr X_n$ in
each pair of homotopy classes $[f],[g]\in \pi_n(X_n)$ such that $[f]\neq [g]$ but
$[\ga_n\com f]=[\ga_n\com g]$. We attach the $(f,g)$th reduced cylinder $S^n\sma I_+$
to $X_n$ by identifying $(s,0)$ with $f(s)$ and $(s,1)$ with $g(s)$ for $s\in S^n$. 
Let $i_n: X_n\rtarr X_{n+1}$ be the inclusion and observe that $(i_n)_*[f]=(i_n)_*[g]$. Define 
$\ga_{n+1}: X_{n+1}\rtarr X$ by means of $\ga_n$ on $X_n$ and a chosen homotopy 
$h:S^n\sma I_+\rtarr X$ from $\ga_n\com f$ to $\ga_n\com g$ on the $(f,g)$th cylinder. Then 
$(\ga_{n+1})_*:\pi_q(X_{n+1})\rtarr \pi_q(X)$ is a surjection for all $q$, because 
$(\ga_n)_*$ is so, and a bijection for $q\leq n$ by construction. We have not changed the
homotopy groups in dimensions less than $n$ since we have not changed the $n$-skeleton. Since
$f$ and $g$ are cellular and since, as is easily verified, $S^n\sma I_+$ admits a CW structure 
with $S^n\sma (\pa I)_+$ as a subcomplex, we conclude from the pushout property of CW
complexes that $X_{n+1}$ is a CW complex that contains $X_n$ as a subcomplex. Then the
colimit $\GA X$ of the $X_n$ is a CW complex that contains all of the $X_i$ as subcomplexes,
and the induced map  $\ga: \GA X\rtarr X$ induces an isomorphism on all homotopy groups
since the homotopy groups of $\GA X$ are the colimits of the homotopy groups of the $X_n$.
If $X$ is $n$-connected, then we have used no $q$-cells for $q\leq n$ in the construction.
\end{proof}

\section{Approximation of pairs by CW pairs}

We will need a relative generalization of the previous result, but the reader should
not dwell on the details: there are no new ideas.

\begin{thm}
For any pair of spaces $(X,A)$ and any CW approximation $\ga: \GA A\rtarr A$, there is 
a CW approximation $\ga: \GA X\rtarr X$ such that\, $\GA A$ is a subcomplex of $\GA X$ and $\ga$
restricts to the given $\ga$ on $\GA A$. If $f: (X,A)\rtarr (Y,B)$ is a map of pairs and
$\ga:(\GA Y,\GA B)\rtarr (Y,B)$ is another such CW approximation of pairs, there is a map
$\GA f: (\GA X,\GA A)\rtarr (\GA Y,\GA B)$, unique up to homotopy, such that the 
following diagram of pairs is homotopy commutative:
$$\diagram
(\GA X,\GA A) \rto^{\GA f} \dto_{\ga} & (\GA Y,\GA B) \dto^{\ga} \\
(X,A) \rto_{f} & (Y,B).\\
\enddiagram$$
If $(X,A)$ is $n$-connected, then $(\GA X,\GA A)$ can be chosen to have no relative
$q$-cells for $q\leq n$.
\end{thm}
\begin{proof}
We proceed as above. We may assume that $X$ has a basepoint in $A$ and that $X$, 
but not necessarily $A$, is path connected. We start with
$$ X_0 = \GA A \vee (\bigvee_{(q,j)}S^q),$$
where $\sset{(q,j)}$ runs over $q\geq 1$ and based maps $j: S^q\rtarr X$
that represent generators of $\pi_q(X)$. Here the chosen basepoint is in
$\Gamma A$. Construct $\ga_0: X_0\rtarr X$ using the maps $j$ and the given 
map $\ga: \Gamma A\rtarr A$. Construct $X_1$ from $X_0$ by attaching $1$-cells
connecting the vertices in the non-basepoint components of $\Gamma A$ to the
base vertex. Paths in $X$ that connect the images under $\ga$ of the non-basepoint 
vertices to the basepoint of $X$ give $\ga_1: X_1\rtarr X$ extending
$\ga_0$. From here, the construction continues as in \S5. 
If $(X,A)$ is $n$-connected, then $\pi_q(A)\rtarr \pi_q(X)$ is bijective for $q<n$ and 
surjective for $q=n$, hence we need only use spheres $S^q$ with $q>n$ to arrange the
surjectivity of $\pi_*(X_0)\rtarr \pi_*(X)$. To construct $\GA f$, we first construct
it on $\GA A$ and then use HELP to extend to $\GA X$:
$$\diagram
\GA A \xto[0,3] \xto[3,0] \drto^{\ga} & & & \GA A\times I 
\xto[3,0] \dlto_h & & & \GA A \xto[0,-3] \xto[3,0] \dlto_{\GA f}\\
& A\rto^f \dto & B \dto & & & \GA B \dto \xto[0,-3]_{\ga} \dto & \\
& X\rto_f      & Y      & & & \GA Y      \xto[0,-3]^{\ga}      & \\
\GA X \urto^{\ga} \xto[0,3]  & & & \GA X\times I 
\uldashed_{\tilde{h}}|>\tip & & & \GA X \xto[0,-3] \uldashed_{\GA f}|>\tip \\
\enddiagram$$
The uniqueness up to homotopy of $\GA f$ is proved similarly.
\end{proof}

\section{Approximation of excisive triads by CW triads}

We will need another, and considerably more subtle, relative approximation theorem.
A triad $(X;A,B)$\index{triad} is a space $X$ together with subspaces $A$ and $B$. This must
not be confused with a triple $(X,A,B)$,\index{triple} which would require $B\subset A\subset X$.
A triad $(X;A,B)$ is said to be excisive\index{triad!excisive} if $X$ is the union of the 
interiors of $A$ and $B$. Such triads play a fundamental role in homology and cohomology theory,
and some version of the arguments to follow must play a role in any treatment.
We prefer to use these arguments to prove a strong homotopical result, rather than
its pale homological reflection that is seen in standard treatments of the subject.

A CW triad $(X;A,B)$\index{triad!CW} is a CW complex $X$ with subcomplexes $A$ and $B$ such that
$X=A\cup B$.

\begin{thm}
Let $(X;A,B)$ be an excisive triad and let $C=A\cap B$. Then there is a CW triad 
$(\GA X;\GA A,\GA B)$ and a map of triads 
$$\ga: (\GA X;\GA A,\GA B)\rtarr (X;A,B)$$
such that, with $\GA C = \GA A\cap \GA B$, the maps 
$$ \ga: \GA C\rtarr C,\ \ \ga: \GA A\rtarr A,\ \  \ga: \GA B\rtarr B,\ \ \tand \ \ 
\ga: \GA X\rtarr X$$
are all weak equivalences. If $(A,C)$ is $n$-connected, then $(\GA A,\GA C)$ can be
chosen to have no $q$-cells for $q\leq n$, and similarly for $(B,C)$. Up to homotopy, 
CW approximation of excisive triads is functorial in such a way that $\ga$ is natural.
\end{thm}
\begin{proof}
Choose a CW approximation $\ga: \GA C\rtarr C$ and use the previous 
result to extend it to CW approximations
$$\ga: (\GA A,\GA C)\rtarr (A,C) \ \ \ \tand \ \ \ \ga: (\GA B,\GA C)\rtarr (B,C).$$
We then define $\GA X$ to be the pushout $\GA A\cup_{\GA C}\GA B$ and let $\ga: \GA X\rtarr X$
be given by the universal property of pushouts. Certainly $\GA C=\GA A\cap \GA B$. All of the
conclusions except for the assertion that $\ga: \GA X\rtarr X$ is a weak equivalence follow
immediately from the result for pairs, and the lemma and theorem below will complete the proof.
\end{proof}

A CW triad $(X;A,B)$ is not excisive, since $A$ and $B$ are closed in $X$, but it is equivalent 
to an excisive triad. To see this,
we describe a simple but important general construction. Suppose that maps $i: C\rtarr A$ 
and $j: C\rtarr B$ are given. Define the double mapping cylinder 
$$M(i,j)= A\cup (C\times I)\cup B$$
to be the space obtained from $C\times I$ by gluing $A$ to $C\times\sset{0}$ along $i$ and 
gluing $B$ to $C\times\sset{1}$ along $j$. Let $A\cup_C B$ denote the pushout of $i$ and $j$
and observe that we obtain a natural quotient map $q: M(i,j)\rtarr A\cup_C B$ by collapsing
the cylinder, sending $(c,t)$ to the image of $c$ in the pushout.

\begin{lem}
For a cofibration $i: C\rtarr A$ and any map $j: C\rtarr B$, the quotient map
$q: M(i,j)\rtarr A\cup_C B$ is a homotopy equivalence.
\end{lem}
\begin{proof}
Because $i$ is a cofibration, the retraction $r: Mi\rtarr A$ is a cofiber homotopy 
equivalence. That is, there is a homotopy inverse map and a pair of homotopies under $C$.
These maps and homotopies induce maps of the pushouts that are obtained by gluing $B$ to 
$Mi$ and to $C$, and $q$ is induced by $r$.
\end{proof}

When $i$ is a cofibration and $j$ is an inclusion, with $X=A\cup B$ and $C=A\cap B$, 
we can think of $q$ as giving a map of triads
$$ q: (M(i,j); A\cup(C\times[0,2/3)), (C\times (1/3,1])\cup B)\rtarr (A\cup_C B;A,B).$$
The domain triad is excisive, and $q$ restricts to homotopy equivalences from the domain 
subspaces and their intersection to the target subspaces $A$, $B$, and $C$. This applies
when $(X;A,B)$ is a CW triad with $C=A\cap B$. Now our theorem on the approximation of 
excisive triads is a consequence of the following result.

\begin{thm}
If $e: (X;A,B)\rtarr (X';A',B')$ is a map of excisive triads such that the maps
$$e: C\rtarr C',\ \ \ e: A\rtarr A',\ \ \ \tand \ \ \  e: B\rtarr B'$$
are weak equivalences, where $C=A\cap B$ and $C'=A'\cap B'$, then $e: X\rtarr X'$
is a weak equivalence.
\end{thm}
\begin{proof}
By our technical lemma giving equivalent conditions for a map $e$ to be a weak 
equivalence, it suffices to show that if $f|S^n=e\com g$ in the following diagram,
then there exists a map $\tilde{g}$ such that $\tilde{g}|S^n = g$ and 
$f\htp e\com \tilde{g}\ \text{rel} \ S^n$:
$$\diagram
X\rto^e & X' \\
S^n \uto^g \rto & D^{n+1} \uldashed_{\tilde{g}}|>\tip \uto_f. \\
\enddiagram$$
We may assume without loss of generality that $S^n\subset U\subset D^{n+1}$, where $U$ is
open in $D^{n+1}$ and $g$ is the restriction of a map $\hat{g}: U\rtarr X$ such that 
$f|U=e\com \hat{g}$. To see this, define a deformation $d: D^{n+1}\times I\rtarr D^{n+1}$ by
$$d(x,t)=\begin{cases}
2x/(2-t) \ \ \ \text{if}\ \  |x|\leq (2-t)/2\\
x/|x| \ \ \ \ \ \ \ \ \ \ \text{if}\ \ |x|\geq (2-t)/2.
\end{cases}$$
Then $d(x,0)=x$, $d(x,t)=x$ if $x\in S^n$, and $d_1$ maps the boundary collar
$\sset{x\ |\ |x|\geq 1/2}$ onto $S^n$. Let $U$ be the open boundary collar 
$\sset{x\ |\ |x|> 1/2}$. Define $\hat{g}=g\com d_1: U\rtarr X$ and define
$f'=f\com d_1: D^{n+1}\rtarr X'$. Then $\hat{g}|S^n = g$, $e\com \hat{g}=f'|U$,
and $f'\htp f \ \text{rel}\ S^n$. Thus the conclusion will hold for $f$ if it holds
with $f$ replaced by $f'$. 

With this assumption on $g$ and $f$, we claim first that the closed sets
$$C_A=g^{-1}(X-\text{int}\,A)\cup \overline{f^{-1}(X'-A')}$$
and 
$$C_B=g^{-1}(X-\text{int}\,B)\cup \overline{f^{-1}(X'-B')},$$
have empty intersection. Indeed, these sets are contained in the sets
$\hat{C}_A$ and $\hat{C}_B$ that are obtained by replacing $g$ by $\hat{g}$ in the 
definitions of $C_A$ and $C_B$, and we claim that $\hat{C}_A\cap\hat{C}_B=\emptyset$. Certainly
$$\hat{g}^{-1}(X-\text{int}\,A)\cap \hat{g}^{-1}(X-\text{int}\,B)=\emptyset$$
since $(X-\text{int}\,A)\cap (X-\text{int}\,B)=\emptyset$. Similarly, 
$$f^{-1}(X'-\text{int}\,A')\cap f^{-1}(X'-\text{int}\,B')=\emptyset.$$
Since $\overline{f^{-1}(X'-A')}\subset f^{-1}(X'-\text{int}\,A')$ and similarly
for $B$,  this implies that 
$$\overline{f^{-1}(X'-A')}\cap \overline{f^{-1}(X'-B')}=\emptyset.$$
Now suppose that $v\in \hat{C}_A\cap\hat{C}_B$. In view of the possibilities that we have ruled 
out, we may assume that  
$$v\in \hat{g}^{-1}(X-\text{int}\,A)\cap \overline{f^{-1}(X'-B')}\subset
\hat{g}^{-1}(\text{int}\,B)\cap \overline{f^{-1}(X'-B')}.$$
Since $\hat{g}^{-1}(\text{int}\,B)$ is an open subset of $D^n$, there must be a point 
$$u\in \hat{g}^{-1}(\text{int}\,B) \cap f^{-1}(X'-B').$$ 
Then $\hat{g}(u)\in \text{int}\,B\subset B$ but $f(u)\not\in B'$. This contradicts 
$f|U=e\com\hat{g}$.

We can subdivide $D^{n+1}$ sufficiently finely (as a simplicial or CW complex) that no cell
intersects both $C_A$ and $C_B$. Let $K_A$ be the union of those cells $\si$ such that
$$g(\si\cap S^n)\subset\ \text{int}\,A\ \  \tand \ \ f(\si)\subset\ \text{int}\,A'$$
and define $K_B$ similarly. If $\si$ does not intersect $C_A$, then $\si\subset K_A$, and if
$\si$ does not intersect $C_B$, then $\si\subset K_B$. Therefore $D^{n+1}=K_A\cup K_B$. By
HELP, we can obtain a map $\bar{g}$ such that the lower triangle in the diagram 
$$\diagram
A\cap B \rto^e & A'\cap B' \\
S^n\cap (K_A\cap K_B) \uto^g \rto & K_A\cap K_B \uto_f \uldashed_{\bar{g}}|>\tip \\
\enddiagram$$
commutes, together with a homotopy $\bar{h}: (K_A\cap K_B)\times I\rtarr A'\cap B'$
such that 
$$\bar{h}: f\htp e\com\bar{g}\ \text{rel}\,S^n\cap (K_A \cap K_B).$$
Define $\bar{g}_A: K_A\cap(S^n\cup K_B)\rtarr A$ to be $g$ on $K_A\cap S^n$ and $\bar{g}$ on
$K_A\cap K_B$. Since $f=e\com g$ on $K_A\cap S^n$ and $\bar{h}: f\htp e\com\bar{g}$ on
$K_A\cap K_B$, $\bar{h}$ induces a homotopy
$$\bar{h}_A: f|K_A\cap(S^n\cup B)\htp e\com g_A\ \text{rel}\,S^n\cap K_A .$$
Applying HELP again, we can obtain maps $\tilde{g}_A$ and $\tilde{h}_A$ such that the 
following diagram commutes:
\begin{small}
$$
\diagram
K_A\cap(S^n\cup K_B) \ddto \rrto^{i_0} & & K_A\cap(S^n\cup K_B) \times I \dlto_{\bar{h}_A} \ddto 
& & K_A\cap(S^n\cup K_B) \llto_{i_1} \dlto_{\bar{g}_A} \ddto \\
& A' & & A \llto_<(0.4){e} & \\
K_A \rrto_{i_0} \urto^{f} && K_A \times I \uldashed^{\tilde{h}_A}|>\tip & & 
K_A \uldashed^{\tilde{g}_A}|>\tip \llto^{i_1} \\
\enddiagram $$
\end{small}
We have a symmetric diagram with the roles of $K_A$ and $K_B$ reversed. The maps $\tilde{g}_A$
and $\tilde{g}_B$ agree on $K_A\cap K_B$ and together define the desired map 
$\tilde{g}: D^{n+1}\rtarr X$. The homotopies $\tilde{h}_A$ and $\tilde{h}_B$ agree on
$(K_A\cap K_B)\times I$ and together define the desired homotopy 
$\tilde{h}_A: f\htp e\com \tilde{g}\ \text{rel}\,S^n$.
\end{proof}

\vspace{.1in}

\begin{center}
PROBLEMS
\end{center}
\begin{enumerate}
\item  Show that complex projective space $\bC P^n$  is a CW complex with one $2q$-cell for each $q$,
$0\leq q\leq n$.
\item  Let $X = \sset{x| x = 0 \ \text{or}\ x = 1/n \ \text{for a positive integer $n$}}\subset \bR$.  
Show that $X$ does not have the homotopy type of a CW 	complex.
\item  Assume given maps $f: X\rtarr Y$  and  $g: Y\rtarr X$  such that $g\com f$ is homotopic to 
the identity.  (We say that $Y$ ``dominates'' $X$.)  Suppose that $Y$ is a CW complex.  Prove 
that $X$ has the homotopy type of a CW complex.
\end{enumerate}

Define the Euler characteristic\index{Euler characteristic!of a CW complex} $\ch (X)$ of a 
finite CW complex $X$ to be the alternating
sum $\sum(-1)^n\ga_n(X)$, 
where $\ga_n(X)$ is the number of $n$-cells of $X$. Let $A$ be a subcomplex of a CW complex $X$, 
let $Y$ be a CW complex, let $f: A\rtarr Y$ be a cellular map, and let $Y\cup_f X$ be the
pushout of $f$ and the inclusion $A\rtarr X$. 

\begin{enumerate}
\item[4.] Show that $Y\cup_f X$ is a CW complex with $Y$ as a subcomplex and 
$X/A$ as a quotient complex. Formulate and prove a formula relating the Euler characteristics 
$\ch(A)$, $\ch(X)$, $\ch(Y)$, and $\ch(Y\cup_fX)$ when $X$ and $Y$ are finite. 
\item[5.]* Think about proving from what we have done so far that $\ch(X)$ depends only on the 
homotopy type of $X$, not on its decomposition as a finite CW complex.
\end{enumerate}

\chapter{The homotopy excision and suspension theorems}

The fundamental obstruction to the calculation of homotopy groups is
the failure of excision: for an excisive triad $(X;A,B)$, the inclusion
$(A,A\cap B)\rtarr (X,B)$ fails to induce an isomorphism of homotopy
groups in general. It is this that distinguishes homotopy groups from
the far more computable homology groups. However, we do have such an
isomorphism in a range of dimensions. This implies the Freudenthal
suspension theorem, which gives that $\pi_{n+q}(\SI^n X)$ is independent
of $n$ if $q$ is small relative to $n$. We shall rely on the consequence 
$\pi_n(S^n)\iso\bZ$ in our construction of homology groups.

\section{Statement of the homotopy excision theorem}

We shall prove the following theorem later in this chapter, but we first
explain its consequences. 

\begin{defn}
A map $f:(A,C)\rtarr (X,B)$ 
of pairs is an $n$-equivalence,\index{nequivalence@$n$-equivalence} $n\geq 1$, if 
$$(f_*)^{-1}(\im(\pi_0(B)\rtarr \pi_0(X)))=\im(\pi_0(C)\rtarr \pi_0(A))$$
(which holds automatically when $A$ and $X$ are path connected) and, for all 
choices of basepoint in $C$,
$$f_*:\pi_q(A,C )\rtarr \pi_q(X,B)$$ 
is a bijection for $q<n$ and a surjection for $q=n$.
\end{defn}

Recall that a pair $(A,C)$ is $n$-connected, $n\geq 0$, if $\pi_0(C)\rtarr\pi_0(A)$ 
is surjective and $\pi_q(A,C)=0$ for $q\leq n$.

\begin{thm}[Homotopy excision]\index{homotopy excision theorem}
Let $(X;A,B)$ be an excisive triad such that $C=A\cap B$ is non-empty.
Assume that $(A,C)$ is $(m-1)$-connected and $(B,C)$ is $(n-1)$-connected,
where $m\geq 2$ and $n\geq 1$. Then the inclusion $(A,C)\rtarr (X,B)$ is
an $(m+n-2)$-equivalence.
\end{thm}

This specializes to give a relationship between the homotopy groups of pairs 
$(X,A)$ and of quotients $X/A$ and to prove the Freudenthal suspension theorem. 

\begin{thm}  Let $f: X\rtarr Y$ be an $(n-1)$-equivalence between $(n-2)$-connected 
spaces, where $n\geq 2$; thus $\pi_{n-1}(f)$ is an epimorphism. Then the quotient map 
$\pi: (Mf,X)\rtarr (Cf,*)$ is a $(2n-2)$-equivalence. In particular, $Cf$ is 
$(n-1)$-connected. If $X$ and $Y$ are $(n-1)$-connected, then 
$\pi: (Mf,X)\rtarr (Cf,*)$ is a $(2n-1)$-equivalence.
\end{thm}
\begin{proof}
We are writing $Cf$ for the unreduced cofiber $Mf/X$. We have the excisive triad 
$(Cf;A,B)$, where
$$A=Y\cup(X\times[0,2/3])\ \ \tand\ \ B=(X\times[1/3,1])/(X\times\sset{1}).$$
Thus $C\equiv A\cap B= X\times [1/3,2/3]$. It is easy to check that 
$\pi$ is homotopic to a composite
$$(Mf,X) \overto{\htp} (A,C)\rtarr (Cf,B) \overto{\htp} (Cf,*),$$
the first and last arrows of which are homotopy equivalences of pairs. 
The hypothesis on $f$ and the long exact sequence of the pair $(Mf,X)$ imply that
$(Mf,X)$ and therefore also $(A,C)$ are $(n-1)$-connected. In view of the connecting
isomorphism $\pa:\pi_{q+1}(CX,X)\rtarr \pi_q(X)$ and the evident homotopy equivalence of 
pairs $(B,C)\htp (CX,X)$, $(B,C)$ is also $(n-1)$-connected, and it is $n$-connected
if $X$ is $(n-1)$-connected. The homotopy excision theorem gives the conclusions.
\end{proof}

We shall later use the following bit of the result to prove the Hurewicz theorem relating 
homotopy groups to homology groups. 

\begin{cor} Let $f: X\rtarr Y$ be a based map between $(n-1)$-connected nondegenerately
based spaces, where $n\geq 2$. Then $Cf$ is $(n-1)$-connected and
$$\pi_n(Mf,X)\rtarr \pi_n(Cf,*)$$
is an isomorphism. Moreover, the canonical map $\et: Ff\rtarr \OM Cf$ induces
an isomorphism
$$\pi_{n-1}(Ff)\rtarr \pi_n(Cf).$$
\end{cor}
\begin{proof} Here in the based context, we are thinking of the reduced mapping cylinder and
cofiber, but the maps to them from the unreduced constructions are homotopy equivalences
since our basepoints are nondegenerate. Thus the first statement is immediate from the theorem. 
For the second, if $j: X\rtarr Mf$ is the inclusion, then we have a map 
$$Fr: Fj=P(Mf;*,X)\rtarr Ff$$ induced by the retraction $r: Mf\rtarr Y$. By a comparison
of long exact sequences, $(Fr)_*: \pi_{q}(Mf,X)\rtarr \pi_{q-1}(Ff)$ is an isomorphism
for all $q$. Moreover, $\et$ factors through a map $Fj\rtarr \OM Cf$, as we noted at 
the end of Chapter 8 \S7. Thus the second statement follows from the first.
\end{proof}

Specializing $f$ to be a cofibration and changing notation, we obtain the
following version of the previous theorem. 

\begin{thm} Let  $i: A\rtarr X$ be a cofibration and an $(n-1)$-equivalence
between $(n-2)$-connected spaces, where $n\geq 2$. Then the quotient map  
$(X,A)\rtarr (X/A,*)$ is a $(2n-2)$-equivalence, and it is a $(2n-1)$-equivalence
if $A$ and $X$ are $(n-1)$-connected.
\end{thm}
\begin{proof}
The vertical arrows are homotopy equivalences of pairs in the 
commutative diagram
$$\diagram
(Mi,A) \dto_r \rto^{\pi} & (Ci,*) \dto^{\ps}\\
(X,A) \rto & (X/A,*). \\
\enddiagram$$
\end{proof}

\section{The Freudenthal suspension theorem}

A specialization of the last result gives the Freudenthal suspension theorem. For a 
based space $X$, define the suspension homomorphism\index{suspension homomorphism}
$$\SI: \pi_q(X)\rtarr \pi_{q+1}(\SI X)$$
by letting 
$$\SI f=f\sma\id: S^{q+1}\iso S^q\sma S^1\rtarr X\sma S^1=\SI X.$$

\begin{thm}[Freudenthal suspension]\index{Freudenthal suspension theorem}
Assume that $X$ is nondegenerately based and $(n-1)$-connected, where $n\geq 1$. Then 
$\SI$ is a bijection if $q<2n-1$ and a surjection if $q=2n-1$.
\end{thm}
\begin{proof}
We give a different description of $\SI$. Consider the ``reversed'' cone
$C'X=X\sma I$, where $I$ is given the basepoint $0$ rather than $1$. Thus
$$C'X = X\times I/X\times\sset{0}\cup \sset{*}\times I.$$
For a map $f: (I^q,\pa I^q)\rtarr (X,*)$, the product 
$f\times \id: I^{q+1}\rtarr X\times I$ passes to quotients to give a map 
of triples
$$(I^{q+1},\pa I^{q+1}, J^q)\rtarr (C'X,X,*)$$
whose restriction to $I^q\times\sset{1}$ is $f$ and which induces $\SI f$
when we quotient out $X\times\sset{1}$. That is, the following diagram 
commutes, where $\rh: C'X\rtarr \SI X$ is the quotient map:
$$\diagram
& \pi_{q+1}(C'X,X,*) \dlto_{\pa} \drto^{\rh_*} & \\
\pi_q(X) \rrto_{\SI} & & \pi_{q+1}(\SI X).\\
\enddiagram$$
Since $C'X$ is contractible, $\pa$ is an isomorphism. Since the inclusion $X\rtarr C'X$ is
a cofibration and an $n$-equivalence between $(n-1)$-connected spaces, $\rh$ is 
a $2n$-equivalence by the last theorem of the previous section. The conclusion follows.
\end{proof}

This implies the promised calculation of $\pi_n(S^n)$.

\begin{thm} For all $n\geq 1$, $\pi_n(S^n)=\bZ$ and 
$\SI: \pi_n(S^n)\rtarr \pi_{n+1}(S^{n+1})$ is an
isomorphism.
\end{thm}
\begin{proof}
We saw by use of the Hopf bundle $S^3\rtarr S^2$ that $\pi_2(S^2)=\bZ$, and
the suspension theorem applies to give the conclusion for $n\geq 2$. A little extra
argument is needed to check that $\SI$ is an isomorphism for $n=1$; one
can inspect the connecting homomorphism of the Hopf bundle or refer ahead to the
observation that the Hurewicz homomorphism commutes with the corresponding
suspension isomorphism in homology.
\end{proof}

The dimensional range of the suspension theorem is sharp. We saw before that 
$\pi_3(S^2)=\pi_3(S^3)$, which is $\bZ$. The suspension theorem applies to show 
that 
$$\SI: \pi_3(S^2)\rtarr \pi_4(S^3)$$
is an epimorphism, and it is known that $\pi_4(S^3)=\bZ_2$. 

Applying suspension repeatedly, we can form a colimit
$$\pi_q^s(X)=\colim\,\pi_{q+n}(\SI^n X).$$
This group is called the $q$th stable homotopy 
group\index{stable homotopy groups}\index{homotopy groups!stable} of $X$. For $q<n-1$,
the maps of the colimit system are isomorphisms and therefore
$$ \pi_q^s(X)=\pi_{q+n}(\SI^n X) \ \ \text{if}\ \ q<n-1.$$
The calculation of the stable homotopy groups of spheres, $\pi_q^s(S^0)$,
is one of the deepest and most studied problems in algebraic topology.
Important problems of geometric topology, such as the enumeration of the
distinct differential structures on $S^q$ for $q\geq 5$, have been reduced 
to the determination of these groups.

\section{Proof of the homotopy excision theorem}

This is a deep result, and it is remarkable that a direct homotopical proof,
in principle an elementary one, is possible. Most standard texts, if they
treat this topic at all, give a far more sophisticated proof of a significantly
weaker result. However, the reader may prefer to skip this argument on a first 
reading. The idea is clear enough. We are trying to show that a certain map of
pairs induces an isomorphism in a range of dimensions. We capture the relevant
map as part of a long exact sequence, and we prove that the third term in the
long exact sequence vanishes in the required range.

However, we start with an auxiliary long exact sequence that we shall also need.
Recall that a triple $(X,A,B)$ consists of spaces $B\subset A\subset X$ and must not be
confused with a triad. 

\begin{prop} For a triple\index{triple!exact sequence of} $(X,A,B)$ and any basepoint 
in $B$, the following sequence is exact:
$$\cdots \rtarr \pi_q(A,B)\overto{i_*} \pi_q(X,B)\overto{j_*} \pi_q(X,A) 
\overto{k_*\com\pa} 
\pi_{q-1}(A,B)\rtarr \cdots.$$
Here $i:(A,B)\rtarr (X,B)$, $j:(X,B)\rtarr (X,A)$, and $k:(A,*)\rtarr (A,B)$ 
are the inclusions.
\end{prop}
\begin{proof}
The proof is a purely algebraic deduction from the long exact sequences of the
various pairs in sight and is left as an exercise for the reader.
\end{proof}

We now define the ``triad homotopy groups''\index{triad homotopy groups} that are 
needed to implement the idea of the proof sketched above.

\begin{defn}
For a triad $(X;A,B)$ with basepoint $*\in C = A\cap B$, define
$$\pi_q(X;A,B)=\pi_{q-1}(P(X;*,B),P(A;*,C)),$$
where $q\geq 2$. More explicitly, $\pi_q(X;A,B)$ is the set of homotopy classes of
maps  of tetrads
$$\diagram
(I^q;I^{q-2}\times\sset{1}\times I, I^{q-1}\times\sset{1},
J^{q-2}\times I\cup I^{q-1}\times\sset{0}) \dto \\
(X;A,B,*),\\
\enddiagram$$
where $J^{q-2}=\pa I^{q-2}\times I\cup I^{q-2}\times\sset{0}\subset I^{q-1}$.
The long exact sequence of the pair in the first form of the definition is
$$\cdots \rtarr \pi_{q+1}(X;A,B)\rtarr \pi_q(A,C)\rtarr \pi_q(X,B)\rtarr \pi_q(X;A,B)
\rtarr \cdots.$$
\end{defn}

Now we return to the homotopy excision theorem. Its conditions $m\geq 1$ and $n\geq 1$
merely give that $\pi_0(C)\rtarr \pi_0(A)$ and $\pi_0(C)\rtarr \pi_0(B)$ are
surjective, and any extraneous components of $A$ or $B$ would not affect the relevant
homotopy groups. The condition $m\geq 2$ implies that $(X,B)$ is $1$-connected. By the
long exact sequence just given, the theorem is equivalent to the following one.

\begin{thm} Under the hypotheses of the homotopy excision theorem,
$$\pi_q(X;A,B)=0\ \ \text{for}\ \  2\leq q\leq m+n-2$$
and all choices of basepoint $*\in C$.
\end{thm}

In this form, the conclusion is symmetric in $A$ and $B$ and vacuous if 
$m+n\leq 3$. Thus our hypotheses $m\geq 2$ and $n\geq 1$ are the minimal
ones under which our strategy can apply.

In order to have some hope of tackling the problem in direct terms, we first reduce
it to the case when $A$ and $B$ are each obtained from $C$ by attaching a single
cell. We may approximate our given excisive triad by a weakly equivalent CW triad.
This does not change the triad homotopy groups. More precisely, by our connectivity
hypotheses, we may assume that $X$ is a CW complex that is the union of subcomplexes
$A$ and $B$ with intersection $C$, where $(A,C)$ has no relative $q$-cells for 
$q < m$ and $(B,C)$ has no relative $q$-cells for $q < n$. Since any map $I^q\rtarr X$
has image contained in a finite subcomplex, we may assume that $X$ has finitely many
cells. We may also assume that $(A,C)$ and $(B,C)$ each have at least one cell since
otherwise the result holds trivially.

We claim first that, inductively, it suffices to prove the result when $(A,C)$ has
exactly one cell. Indeed, suppose that  $C\subset A'\subset A$, where $A$ is obtained 
from $A'$ by attaching a single cell and $(A',C)$ has one less cell than $(A,C)$.
Let $X'=A'\cup_C B$. If the result holds for the triads $(X';A',B)$ and $(X;A,X')$,
then the result holds for the triad $(X;A,B)$ by application of the five lemma to
the following diagram:
$$\diagram
\pi_{q+1}(A,A') \rto \dto & \pi_q(A',C) \rto \dto 
& \pi_q(A,C) \rto \dto & \pi_q(A,A') \rto \dto & \pi_{q-1}(A',C)\dto \\
\pi_{q+1}(X,X') \rto & \pi_q(X',B) \rto 
& \pi_q(X,B) \rto  & \pi_q(X,X') \rto  & \pi_{q-1}(X',B). \\
\enddiagram$$
The rows are the exact sequences of the triples $(A,A',C)$ and $(X,X',B)$. Note for 
the case $q=1$ that all pairs in the diagram are $1$-connected.

We claim next that, inductively, it suffices to prove the result when $(B,C)$ also
has exactly one cell. Indeed, suppose that  $C\subset B'\subset B$, where $B$ is 
obtained  from $B'$ by attaching a single cell and $(B',C)$ has one less cell than 
$(B,C)$ and let $X'=A\cup_C B'$. If the result holds for the triads $(X';A,B')$ and 
$(X;X',B)$, then the result holds for the triad $(X;A,B)$ since the inclusion
$(A,C)\rtarr (X,B)$ factors as the composite
$$(A,C)\rtarr (X',B')\rtarr (X,B).$$

Thus we may assume that $A=C\cup D^m$ and $B=C\cup D^n$, where $m\geq 2$ and $n\geq 1$,
and we fix a basepoint $*\in C$. Assume given a map of tetrads
$$\diagram
(I^q;I^{q-2}\times\sset{1}\times I, I^{q-1}\times\sset{1},
J^{q-2}\times I\cup I^{q-1}\times\sset{0}) \dto^f \\
(X;A,B,*),\\
\enddiagram$$
where $2\leq q\leq m+n-2$. We must prove that $f$ is null homotopic as a map of tetrads.
For interior points $x\in D^m$ and $y\in D^n$, we have inclusions of based triads
$$(A;A,A-{x})\subset(X-\sset{y};A,X-\sset{x,y})\subset(X;A,X-\sset{x})\supset(X;A,B).$$
The first and third of these induce isomorphisms on triad homotopy groups in view of the
radial deformation away from $y$ of $X-\sset{y}$ onto $A$ and the radial deformation away 
from $x$ of $X-\sset{x}$ onto $B$. It is trivial to check that $\pi_*(A;A,A')=0$ for any
$A'\subset A$. We shall show that, for well chosen points $x$ and $y$, $f$ regarded
as a map of based triads into $(X;A,X-\sset{x})$ is homotopic to a map $f'$ that has 
image in $(X-\sset{y};A,X-\sset{x,y})$. This will imply that $f$ is null homotopic.

Let $D^m_{1/2}\subset D^m$ and $D^n_{1/2}\subset D^n$ be the subdisks of radius $1/2$. We can 
cubically subdivide $I^q$ into subcubes $I^q_{\al}$ such that $f(I^q_{\al})$ is contained in 
the interior of $D^m$ if it intersects $D^m_{1/2}$ and $f(I^q_{\al})$ is contained in the 
interior of $D^n$ if it intersects $D^n_{1/2}$. By simplicial approximation, $f$ is homotopic 
as a map of tetrads to a map $g$ whose restriction to the $(n-1)$-skeleton of $I^q$ with its 
subdivided cell structure does not cover $D^n_{1/2}$ and whose restriction to the $(m-1)$-skeleton
of $I^q$ does not cover $D^m_{1/2}$. Moreover, we can arrange that the dimension of $g^{-1}(y)$ is 
at most $q-n$ for a point $y\in D^n_{1/2}$ that is not in the image under $g$ of the 
$(n-1)$-skeleton 
of $I^q$. This is the main point of the proof, and to be completely rigorous about it we would 
have to digress to introduce a bit of dimension theory. Alternatively, we could use smooth
approximation to arrive at $g$ and $y$ with appropriate properties. Since the intuition should 
be clear, we shall content ourselves with showing how the conclusion of the theorem follows. 

Let $\pi: I^q\rtarr I^{q-1}$ be the projection on the first $q-1$ coordinates and let $K$ 
be the prism $\pi^{-1}(\pi(g^{-1}(y)))$. Then $K$ can have dimension at most one more than 
the dimension of $g^{-1}(y)$, so that 
$$\text{dim}\,K \leq q-n+1\leq m-1.$$
Therefore $g(K)$ cannot cover $D^m_{1/2}$. Choose a point $x\in D^m_{1/2}$ such that 
$x\not\in g(K)$. 
Since $g(\pa I^{q-1}\times I)\subset A$, we see that $\pi(g^{-1}(x))\cup \pa I^{q-1}$ and
$\pi(g^{-1}(y))$ are disjoint closed subsets of $I^{q-1}$. By Uryssohn's lemma, we may
choose a map $v: I^{q-1}\rtarr I$ such that 
$$v(\pi(g^{-1}(x))\cup \pa I^{q-1})=0 \ \ \ \tand \ \ \ v(\pi(g^{-1}(y)))=1.$$
Define $h: I^{q+1}\rtarr I^q$ by
$$h(r,s,t)=(r,s-stv(r))\ \ \text{for}\ \ r\in I^{q-1} \  \tand \   s,t\in I.$$
Then let $f'=g\com h_1$, where $h_1(r,s)=h(r,s,1)$. We claim that $f'$ is as desired. 
Observe that
$$h(r,s,0)=(r,s), \ \ h(r,0,t)=(r,0), \ \ \tand\ \ h(r,s,t)=(r,s)\ \text{if}\ r\in\pa I^{q-1}.$$
Moreover,
$$ h(r,s,t)=(r,s)\ \  \text{if}\ \ h(r,s,t)\in g^{-1}(x)$$
since $r\in \pi(g^{-1}(x))$ implies $v(r)=0$ and
$$ h(r,s,t)=(r,s-st) \ \ \text{if}\ \ h(r,s,t)\in g^{-1}(y)$$
since $r\in \pi(g^{-1}(y))$ implies $v(r)=1$. Then $g\com h$ is a homotopy of maps of tetrads
$$\diagram
(I^q;I^{q-2}\times\sset{1}\times I, I^{q-1}\times\sset{1},
J^{q-2}\times I\cup I^{q-1}\times\sset{0}) \dto \\
(X;A,X-\sset{x},*)\\
\enddiagram$$
from $g$ to $f'$, and $f'$ has image in $(X-\sset{y};A,X-\sset{x,y})$, as required.

\clearpage

\thispagestyle{empty}

\chapter{A little homological algebra}

Let $R$ be a commutative ring. The main example will be $R=\bZ$. We develop some
rudimentary homological algebra in the category of $R$-modules. We shall say more
later. For now, we give the minimum that will be needed to develop cellular and
singular homology theory.

\section{Chain complexes}

A chain complex\index{chain complex} over $R$ is a sequence of maps of $R$-modules
$$\cdots \rtarr X_{i+1} \overto{d_{i+1}} X_i\overto{d_i} X_{i-1}\rtarr \cdots$$
such that $d_i\com d_{i+1} = 0$ for all $i$. We generally abbreviate $d=d_i$. A 
cochain complex\index{cochain complex} over $R$ is an analogous sequence 
$$\cdots \rtarr Y^{i-1} \overto{d^{i-1}} Y^i\overto{d^i} Y^{i+1}\rtarr \cdots$$
with $d^i\com d^{i-1}=0$.  In practice, we usually require chain complexes to satisfy
$X_i=0$ for $i<0$ and cochain complexes to satisfy $Y^i=0$ for $i<0$. Without these
restrictions, the notions are equivalent since a chain complex $\sset{X_i,d_i}$ can
be rewritten as a cochain complex $\sset{X^{-i},d^{-i}}$, and vice versa.

An element of the kernel of $d_i$ is called a cycle\index{cycle} and an element of the image of 
$d_{i+1}$ is called a boundary.\index{boundary} We say that two cycles 
are ``homologous''\index{homologous cycles} if their 
difference is a boundary. We write $B_i(X)\subset Z_i(X)\subset X_i$ for the 
submodules of boundaries and cycles, respectively, and we define the $i$th homology
group\index{homology group!of a chain complex} $H_i(X)$ to be the quotient module 
$Z_i(X)/B_i(X)$. We write $H_*(X)$ for the
sequence of $R$-modules $H_i(X)$. We understand 
``graded $R$-modules''\index{graded $R$-module} to be sequences of $R$-modules 
such as this (and we never take the sum of elements in different 
gradings).

\section{Maps and homotopies of maps of chain complexes}

A map $f: X\rtarr X'$ of chain complexes\index{chain map} is a sequence of maps of $R$-modules
$f_i: X_i\rtarr X'_i$ such that $d'_i\com f_i=f_{i-1}\com d_i$ for all $i$. That is,
the following diagram commutes for each $i$:
$$\diagram
X_i\rto^{f_i} \dto_{d_i} & X'_i \dto^{d'_i}\\
X_{i-1} \rto_{f_{i-1}} & X'_{i-1}.\\
\enddiagram$$
It follows that $f_i(B_i(X))\subset B_i(X')$ and $f_i(Z_i(X))\subset Z_i(X')$. Therefore
$f$ induces a map of $R$-modules $f_*=H_i(f): H_i(X)\rtarr H_i(X')$.

A chain homotopy\index{chain homotopy} $s: f\htp g$ between chain maps $f,g: X\rtarr X'$ is a 
sequence of homomorphisms $s_i: X_i\rtarr X'_{i+1}$ such that
$$d'_{i+1}\com s_i+s_{i-1}\com d_i = f_i -g_i$$
for all $i$. Chain homotopy is an equivalence relation since if $t: g\htp h$, then
$s+t=\sset{s_i+t_i}$ is a chain homotopy $f\htp h$. 

\begin{lem}
Chain homotopic maps induce the same homomorphism of homology groups.
\end{lem}
\begin{proof}
Let $s:f\htp g$, $f,g: X\rtarr X'$. If $x\in Z_i(X)$, then
$$f_i(x)-g_i(x)=d'_{i+1}s_i(x),$$
so that $f_i(x)$ and $g_i(x)$ are homologous.
\end{proof}

\section{Tensor products of chain complexes}

The tensor product\index{tensor product!of chain complexes} (over $R$) of chain 
complexes $X$ and $Y$ is specified by letting
$$(X\otimes Y)_n = \sum_{i+j=n} X_i\otimes Y_j.$$
When $X_i$ and $Y_i$ are zero for $i<0$, the sum is finite, but we don't need
to assume this. The differential is specified by
$$d(x\otimes y) = d(x)\otimes y +(-1)^i x\otimes d(y)$$
for $x\in X_i$ and $y\in Y_j$. The sign ensures that $d\com d=0$. We may write
this as
$$d=d\otimes \id+\id\otimes\, d.$$
The sign is dictated by the general rule that whenever two entities to which degrees 
$m$ and $n$ can be assigned are permuted, the sign $(-1)^{mn}$ should be inserted. In the 
present instance, when calculating $(\id\otimes\, d)(x\otimes y)$, we must permute the map
$d$ of degree $-1$ with the element $x$ of degree $i$. 

We regard $R$-modules $M$ as chain complexes concentrated in degree zero, and thus with
zero differential. For a chain complex $X$, there results a chain complex $X\otimes M$;
$H_*(X\otimes M)$ is called the homology of $X$ with coefficients in $M$. 

Define a chain complex $\sI$ by letting $\sI_0$ be the free Abelian group with two 
generators $[0]$ and $[1]$, letting $\sI_1$ be the free Abelian group with one generator
$[I]$ such that $d([I])=[0]-[1]$, and letting $\sI_i=0$ for all other $i$. 

\begin{lem}
A chain homotopy\index{chain homotopy} $s:f\htp g$ between chain maps $f,g: X\rtarr X'$ 
determines and is
determined by a chain map $h: X\ten \sI\rtarr X'$ such that $h(x,[0])=f(x)$ and
$h(x,[1])=g(x)$.
\end{lem}
\begin{proof}
Let $s$ correspond to $h$ via $(-1)^i s(x)=h(x\otimes [I])$ for $x\in X_i$. The relation
$$d'_{i+1}(s_i(x)) = f_i(x) -g_i(x)-s_{i-1}(d_i(x))$$
corresponds to the relation $d'h=hd$ by the definition of our differential on $\sI$. The
sign in the correspondence would disappear if we replaced by $X\otimes \sI$ by $\sI\otimes X$.
\end{proof}

\section{Short and long exact sequences}

A sequence $M'\overto{f} M\overto{g} M''$ of modules is exact\index{exact sequence} if 
$\im f=\ker g$. If $M'=0$, this means that $g$ is a monomorphism; if $M''=0$, it means that 
$f$ is an epimorphism. A 
longer sequence is exact if it is exact at each position. A short exact sequence\index{short
exact sequence!of chain complexes} of chain
complexes is a sequence
$$0\rtarr X'\overto{f} X \overto{g} X'' \rtarr 0$$
that is exact in each degree. Here $0$ denotes the chain complex that is the zero module
in each degree.

\begin{prop}
A short exact sequence of chain complexes naturally gives rise to a long exact sequence of
$R$-modules
$$\cdots \rtarr H_q(X') \overto{f_*} H_q(X) \overto{g_*} 
H_q(X'') \overto{\pa} H_{q-1}(X') \rtarr\cdots.$$
\end{prop}
\begin{proof}
Write $[x]$ for the homology class of a cycle $x$. We define the 
``connecting homomorphism''\index{connecting homomorphism}
$\pa: H_q(X'')\rtarr H_{q-1}(X')$ by $\pa[x'']=[x']$, where $f(x')=d(x)$ for some $x$ such that 
$g(x)=x''$. There is such an $x$ since $g$ is an epimorphism, and there is such an $x'$ since
$gd(x)=dg(x)=0$. It is a standard exercise in ``diagram chasing'' to verify that $\pa$ is 
well defined and the sequence is exact. Naturality means that a commutative diagram of short
exact sequences of chain complexes gives rise to a commutative diagram of long exact sequences
of $R$-modules. The essential point is the naturality of the connecting homomorphism, which 
is easily checked.
\end{proof}

\vspace{.1in}

\begin{center}
PROBLEMS
\end{center}

For a graded vector space $V=\sset{V_n}$ with $V_n = 0$ for all but finitely many $n$ and 
with all $V_n$ finite dimensional, define the 
Euler characteristic\index{Euler characteristic!of a graded vector space} $\ch(V)$ to be 
$\sum(-1)^n \text{dim}\,V_n$. 
\begin{enumerate}
\item Let $V'$, $V$, and $V''$ be such graded vector spaces and suppose there is a long 
exact sequence 
$$\cdots \rtarr V'_n\rtarr V_n\rtarr V''_n\rtarr V'_{n-1}\rtarr \cdots .$$
Prove that $\ch(V) = \ch(V') + \ch (V'')$. 
\item If $\sset{V_n, d_n}$  is a chain complex, show that $\ch(V) = \ch(H_*(V))$. 
\item Let $0\rtarr \pi\overto{f}\rh\overto{g}\si\rtarr 0$ be an exact sequence of
Abelian groups and let $C$ be a chain complex of flat (= torsion free) Abelian groups. Write
$H_*(C;\pi)=H_*(C\ten \pi)$. Construct a natural long exact sequence
$$ \cdots \rtarr H_q(C;\pi)\overto{f_*} H_q(C;\rh)\overto{g_*} H_q(C;\si)
\overto{\be} H_{q-1}(C;\pi)\rtarr \cdots.$$
The connecting homomorphism $\be$ is called a Bockstein operation.\index{Bockstein operation}
\end{enumerate}

\clearpage

\thispagestyle{empty}

\chapter{Axiomatic and cellular homology theory}

Homology groups are the basic computable invariants of spaces. Unlike homotopy
groups, these are stable invariants,\index{stable}\index{stable invariants} the 
same for a space and its suspension, 
and it is this that makes them computable. In this and the following two chapters,
we first give both an axiomatic and a cellular description of homology, next revert 
to an axiomatic development of the properties of homology, and then prove the
Hurewicz theorem and use it to prove the uniqueness of homology.

\section{Axioms for homology}

Fix an Abelian group $\pi$ and consider pairs of spaces $(X,A)$. We
shall see that $\pi$ determines a ``homology theory\index{homology theory} on pairs $(X,A)$.''
We say that a map $(X,A)\rtarr (Y,B)$ of pairs is a weak equivalence\index{weak equivalence}
if its maps $A\rtarr B$ and $X\rtarr Y$ are weak equivalences.

\begin{thm} For integers $q$, there exist functors $H_q(X,A;\pi)$ from the 
homotopy category of pairs of spaces to the category of Abelian groups together 
with natural transformations $\pa: H_q(X,A;\pi)\rtarr H_{q-1}(A;\pi)$, where
$H_q(X;\pi)$ is defined to be $H_q(X,\emptyset;\pi)$. These functors and natural
transformations satisfy and are characterized by the following axioms.
\begin{itemize}
\item DIMENSION\index{dimension axiom}\ \  If $X$ is a point, then $H_0(X;\pi) = \pi$ and $H_q(X;\pi)=0$
for all other integers.
\item EXACTNESS\index{exactness axiom}\ \ The following sequence is exact, where the unlabeled arrows
are induced by the inclusions $A\rtarr X$ and $(X,\emptyset)\rtarr (X,A)$:
$$\cdots\rtarr H_q(A;\pi)\rtarr H_q(X;\pi)\rtarr 
H_q(X,A;\pi)\overto{\pa}  H_{q-1}(A;\pi)\rtarr \cdots .$$
\item EXCISION\index{excision axiom}\ \ 
If $(X;A,B)$ is an excisive triad, so that $X$ is the union of the interiors
of $A$ and $B$, then the inclusion $(A,A\cap B)\rtarr (X,B)$ induces an
isomorphism
$$H_*(A,A\cap B;\pi)\rtarr H_*(X,B;\pi).$$
\item ADDITIVITY\index{additivity axiom}\ \ 
If $(X,A)$ is the disjoint union of a set of pairs $(X_i,A_i)$, then
the inclusions $(X_i,A_i)\rtarr (X,A)$ induce an isomorphism
$$\textstyle{\sum}_i H_*(X_i,A_i;\pi)\rtarr H_*(X,A;\pi).$$
\item WEAK EQUIVALENCE\index{weak equivalence axiom}\ \  If $f:(X,A)\rtarr (Y,B)$ is a 
weak equivalence, then
$$f_*: H_*(X,A;\pi)\rtarr H_*(Y,B;\pi)$$
is an isomorphism.
\end{itemize}
\end{thm}

Here, by a standard abuse, we write $f_*$ instead of $H_*(f)$ or $H_q(f)$. 
Our approximation theorems for spaces, pairs, maps, homotopies, and excisive 
triads imply that such a theory determines and is determined by an appropriate 
theory defined on CW pairs, as spelled out in the following CW version of the 
theorem.\index{homology theory}

\begin{thm} For integers $q$, there exist functors $H_q(X,A;\pi)$ from the 
homotopy category of pairs of CW complexes to the category of 
Abelian groups together with natural transformations $\pa: H_q(X,A)\rtarr H_{q-1}(A;\pi)$, 
where $H_q(X;\pi)$ is defined to be $H_q(X,\emptyset;\pi)$. These functors and natural
transformations satisfy and are characterized by the following axioms.
\begin{itemize}
\item DIMENSION\index{dimension axiom}\ \  If $X$ is a point, then $H_0(X;\pi) = \pi$ and $H_q(X;\pi)=0$
for all other integers.
\item EXACTNESS\index{exactness axiom}\ \  The following sequence is exact, where the unlabeled arrows
are induced by the inclusions $A\rtarr X$ and $(X,\emptyset)\rtarr (X,A)$:
$$\cdots\rtarr H_q(A;\pi)\rtarr H_q(X;\pi)\rtarr 
H_q(X,A;\pi)\overto{\pa}  H_{q-1}(A;\pi)\rtarr \cdots .$$
\item EXCISION\index{excision axiom}\ \ 
If $X$ is the union of subcomplexes $A$ and $B$, then the inclusion 
$(A,A\cap B)\rtarr (X,B)$ induces an isomorphism
$$H_*(A,A\cap B;\pi)\rtarr H_*(X,B;\pi).$$
\item ADDITIVITY\index{additivity axiom}\ \ 
If $(X,A)$ is the disjoint union of a set of pairs $(X_i,A_i)$, then
the inclusions $(X_i,A_i)\rtarr (X,A)$ induce an isomorphism
$$\textstyle{\sum}_i H_*(X_i,A_i;\pi)\rtarr H_*(X,A;\pi).$$
\end{itemize}
Such a theory determines and is determined by a theory as in the previous 
theorem.
\end{thm}
\begin{proof}
We prove the last statement and return to the rest later. Since a CW triad 
(which, we recall, was required to be the union of its given subcomplexes)
is homotopy equivalent to an excisive triad, it is immediate that the 
restriction to CW pairs of a theory on pairs of spaces gives a theory on pairs
of CW complexes. Conversely, given a theory on CW pairs, we may define a theory
on pairs of spaces by turning the weak equivalence axiom into a definition.
That is, we fix a CW approximation functor $\GA$ from the homotopy category 
of pairs of spaces to the homotopy category of CW pairs and we define
$$H_*(X,A;\pi) = H_*(\GA X,\GA A;\pi).$$
Similarly, we define $\pa$ for $(X,A)$ to be $\pa$ for $(\GA X,\GA A)$.
For a map $f:(X,A)\rtarr (Y,B)$ of pairs, we define $f_*=(\GA f)_*$. It is
clear from our earlier results that this does give a well defined homology
theory on pairs of spaces.
\end{proof}

Clearly, up to canonical isomorphism, this construction of a homology
theory on pairs of spaces is independent of the choice of our CW approximation 
functor $\GA$. The reader may have seen singular homology before. As we shall 
explain later, the classical construction of singular homology amounts to a 
choice of a particularly nice CW approximation functor, one that is actually
functorial on the point-set level, before passage to homotopy categories.

\section{Cellular homology}

We must still construct $H_*(X,A;\pi)$ on CW pairs. We shall give a seemingly
ad hoc construction, but we shall later see that precisely this construction
is in fact forced upon us by the axioms. We concentrate on the case $\pi=\bZ$,
and we abbreviate notation by setting $H_*(X,A)=H_*(X,A;\bZ)$. 

Let $X$ be a CW complex. We shall define the cellular chain complex\index{cellular
chain complex|(} $C_*(X)$.
We let $C_n(X)$ be the free Abelian group with one generator $[j]$ for each
$n$-cell $j$. We must define a differential $d_n:C_n(X)\rtarr C_{n-1}(X)$. We shall 
first give a direct definition in terms of
the cell structure and then give a more conceptual description in terms of cofiber
sequences. It will be convenient to work with unreduced cones, cofibers, and suspensions 
in this section; that is, we do not choose basepoints and so we do not collapse out 
lines through basepoints. (We shall discuss this difference more formally in the 
next chapter.)  We still have the basic result that if $i: A\rtarr X$ is a cofibration, 
then collapsing the cone on $A$ to a point gives a homotopy equivalence $\ps: Ci\rtarr X/A$. 
We shall use the notation $\ps^{-1}$ for any chosen homotopy inverse to such a homotopy 
equivalence. We again obtain $\pi: Ci\rtarr \SI A$ by collapsing the base $X$ of the 
cofiber to a point. 

Our first definition of $d_n$ involves the calculation of the degrees of maps between 
spheres. A map $f: S^n\rtarr S^n$ induces a homomorphism $f_*: \pi_n(S^n)\rtarr \pi_n(S^n)$,
which is given by multiplication by an integer called the degree of $f$. As in our
discussion earlier for $\pi_1$, $f_*$ is defined using a change of basepoint
isomorphism, but deg\,$(f)$ is independent of the choice of the path connecting
$*$ to $f(*)$. Of course, this only makes sense for $n\geq 1$. 

To define and calculate degrees, the domain and target of $f$ must both be $S^n$.
However, there are three models of $S^n$ that are needed in our discussion: the standard
sphere $S^n\subset D^{n+1}$, the quotient $D^n/S^{n-1}$, and the (unreduced) suspension $\SI S^{n-1}$. 
We must fix suitably compatible homeomorphisms relating these ``$n$-spheres.'' We define a 
homeomorphism
$$\nu_n: D^n/S^{n-1}\rtarr S^n$$
by
$$\nu_n(tx_1,\ldots\!,tx_n)=(ux_1,\ldots\!,ux_n,2t-1)$$
for $0\leq t\leq 1$ and $(x_1,\ldots\!,x_n)\in S^{n-1}$, where $u=(1-(2t-1)^2)^{1/2}$.
Thus $\nu_n$ sends the ray from $0$ to $(x_1,\ldots\!,x_n)$ to the longitude that
runs from the south pole $(0,\ldots\!,0,-1)$ through the equatorial point
$(x_1,\ldots\!,x_n,0)$ to the north pole $(0,\ldots\!,0,1)$. We define a homeomorphism
$$\io_n: S^n\rtarr \SI S^{n-1}$$ 
by
$$\io_n(x_1,\ldots\!,x_{n+1})=(vx_1,\ldots\!,vx_n)\sma (x_{n+1}+1)/2,$$
where $v=1/(\sum_{i=1}^{n}x_i^2)^{1/2}$. This makes sense since if $x_i=0$ for 
$1\leq i\leq n$,
then $x_{n+1}=\pm 1$, so that $(x_{n+1}+1)/2=0\ \text{or}\ 1$ and 
$\io_n(x_1,\ldots\!,x_{n+1})$ is a cone point. In effect, $\io_n$ makes the last coordinate
the suspension coordinate. We define a homeomorphism of pairs
$$\xi_n: (D^n, S^{n-1}) \rtarr (CS^{n-1},S^{n-1})$$
by 
$$\xi_n(tx_1,\ldots\!,tx_n) = (x_1,\ldots\!,x_n) \sma (1-t),$$
and we continue to write $\xi_n$ for the induced homeomorphism 
$$D^n/S^{n-1} \iso CS^{n-1}/S^{n-1} = \SI S^{n-1}.$$ 
Observe that 
$$\io_n\com\nu_n = - \xi_n: D^n/S^{n-1}\rtarr \SI S^{n-1},$$
where the minus is interpreted as the sign map $y\sma t\rtarr y\sma (1-t)$ on $\SI S^{n-1}$.
We saw in our treatment of cofiber sequences that, up to homotopy, the maps 
$$CS^{n-1}\cup_{S^{n-1}} CS^{n-1} \rtarr \SI S^{n-1}$$
obtained by collapsing out the first and second cone also differ by this sign map. 
By an easy diagram chase, these observations imply the following compatibility 
statement. It will be used to show that the two definitions of $d_n$ that we shall 
give are in fact the same.

\begin{lem}
The following diagram is homotopy commutative:
$$\diagram
D^n\cup_{S^{n-1}}CS^{n-1}  \rto^(0.65){\pi} \dto_{\ps} & \SI S^{n-1} \\
D^n/S^{n-1} \rto_(0.55){\nu_n} & S^n \uto_{\io_n}. \\
\enddiagram$$
\end{lem}

Returning to our CW complex $X$, we think of an $n$-cell $j$ as a map of pairs
$$j:(D^n,S^{n-1})\rtarr (X^n,X^{n-1}).$$
There results a homeomorphism
$$\al: \textstyle{\bigvee}_j\, D^n/S^{n-1} \rtarr X^n/X^{n-1}$$
whose restriction to the $j$th wedge summand is induced by $j$. 
Define 
$$\pi_j:X^n/X^{n-1}\rtarr S^n$$ 
to be the composite of $\al^{-1}$ with the map given by 
$\nu_n$ on the $j$th wedge summand and the constant map at the basepoint on all other 
wedge summands. If $n=0$, we interpret $D^0$ to be a point and interpret $S^{-1}$ and 
$X^{-1}$ to be empty. With our convention that $X/\emptyset=X_+$, we see that $X^0/X^{-1}$ 
can be identified with the wedge of one copy of $S^0$ for each vertex $j$, and $\pi_j$ 
is still defined. Here we take $S^0=\sset{\pm 1}$, with basepoint $1$.

For an $n$-cell $j$ and an $(n-1)$-cell $i$, where $n\geq 1$, we have a composite
$$S^{n-1}\overto{j} X^{n-1} \overto{\rh} X^{n-1}/X^{n-2} \overto{\pi_i} S^{n-1}.$$
When $n=1$, we interpret $\rh$ to be the inclusion $X^0\rtarr X^0_+$.
When $n\geq 2$, let $a_{i,j}$ be the degree of this composite and define
$$d_n[j]=\textstyle{\sum}_{i}a_{i,j}[i].$$
When $n=1$, specify coefficents $a_{i,j}$ implicitly by defining
$$d_1[j]=[j(1)]-[j(-1)].$$
We claim that $d_{n-1}\com d_n=0$, and we define
$$H_*(X)=H_*(C_*(X)).$$

To see that $d_{n-1}\com d_n=0$, we use the theory of cofiber sequences to obtain a more 
conceptual description of $d_n$. We define the 
``topological boundary map''\index{topological boundary map}
$$\pa_n: X^n/X^{n-1}\rtarr \SI(X^{n-1}/X^{n-2})$$
to be the composite 
$$X^n/X^{n-1}\overto{\ps^{-1}} Ci \overto{\pi} \SI X^{n-1}
\overto{\SI\rh}\SI(X^{n-1}/X^{n-2}),$$
where $i: X^{n-1}\rtarr X^n$ is the inclusion. 

We claim that $\pa_n$ induces $d_n$ upon application of a suitable functor, and we
need some preliminaries to show this. For certain based spaces $X$, we adopt 
the following provisional definition of the ``reduced $n$th homology group'' 
of $X$.\index{reduced homology!provisional definition}

\begin{defn} Let $X$ be a based $(n-1)$-connected space. Define
$\tilde{H}'_n(X)$ as follows.
\begin{enumerate}
\item[$n=0$:]  The free Abelian group generated by the set 
$\pi_0(X)-\sset{*}$ of non-basepoint components of $X$.
\item[$n=1$:]  The Abelianization $\pi_1(X)/[\pi_1(X),\pi_1(X)]$ of the fundamental
group of $X$.
\item[$n\geq 2$:]  The $n$th homotopy group of $X$.
\end{enumerate}
\end{defn}

Up to canonical isomorphism, $\tilde{H}'_n(X)$ is independent of the choice of basepoint 
in its path component. In fact, we can define $\tilde{H}'_n(X)$ in terms of unbased 
homotopy classes of maps $S^n\rtarr X$. We also need a suspension homomorphism, and
we adopt another provisional definition.

\begin{defn} Let $X$ be a based $(n-1)$-connected space. Define 
$$\SI: \tilde{H}'_n(X)\rtarr \tilde{H}'_{n+1}(\SI X)$$
by letting $\SI[f]=[\SI f\com \io_{n+1}]$ for $f:S^n\rtarr X$; that is,  $\SI[f]$ is 
represented by 
$$S^{n+1}\overto{\io_{n+1}} \SI S^n\overto{\SI f}\SI X.$$
This only makes sense for $n\geq 1$. For $n=0$, and a point $x\in X$ that is not in the
component of the basepoint, we define $\SI[x]=[f^{-1}_{*}\cdot f_x]$, where $*\in X$ is 
the basepoint and $f_x$ is the path $t\rtarr x\sma t$ from one cone point to the other
in the unreduced suspension $\SI X$. 
\end{defn}

\begin{lem} If $X$ is a wedge of $n$-spheres, then
$$\SI: \tilde{H}'_n(X)\rtarr \tilde{H}'_{n+1}(\SI X)$$
is an isomorphism.
\end{lem}
\begin{proof}
We claim first that $\tilde{H}'_n(X)$ is the free Abelian group with generators given by the 
inclusions of the wedge summands. Since maps and homotopies of maps $S^n\rtarr X$ have
images in compact subspaces, it suffices to check this on finite wedges, and, when $n\geq 2$, 
we can give the pair $(\times_i S^n,\wed_i S^n)$ the structure of a CW pair with no relative 
$q$-cells for $q< 2n-1$. The claim follows by cellular approximation of maps and homotopies.
Now the conclusion of the lemma follows from the case of a single sphere in view of the
canonical direct sum decompositions. 
\end{proof}

Returning to our CW complex $X$, we take the homotopy classes $[j\com\nu_n^{-1}]$ of the 
composites
$$ S^n\overto{\nu_n^{-1}} D^n/S^{n-1}\overto{j} X^n/X^{n-1}$$
as canonical basis elements of $\tilde{H}'_n(X^n/X^{n-1})$. 

\begin{lem}
The differential $d_n: C_n(X)\rtarr C_{n-1}(X)$ can be identified with the composite
$$d_n': \tilde{H}'_n(X^n/X^{n-1})\overto{(\pa_n)_*} \tilde{H}'_n(\SI (X^{n-1}/X^{n-2}))
\overto{\SI^{-1}} \tilde{H}'_{n-1}(X^{n-1}/X^{n-2}).$$
\end{lem}
\begin{proof}
The identification of the groups is clear: we let the basis element $[j]$ of $C_n(X)$
correspond to the basis element $[j\com\nu_n^{-1}]$ of $\tilde{H}'_n(X^n/X^{n-1})$.
For an $n$-cell $j$ and an $(n-1)$-cell $i$, the following diagram is homotopy commutative 
by the naturality of $\ps$ and $\pi$, the definition of $a_{i,j}$, and our lemma relating
different models of the $n$-sphere:
$$\diagram
S^n \xto[0,3]^{a_{i,j}} \drrto^{\io_n} \dto_{\nu_n^{-1}} & & & S^n \dto^{\io_n} \\
D^n/S^{n-1}\rto^{\ps^{-1}} \dto_{j} & D^n\cup CS^{n-1} \dto^{j\cup Cj} \rto_(0.6){\pi} 
& \SI S^{n-1} \rto^{a_{i,j}} \dto^{\SI j} & \SI S^{n-1} \\
X^n/X^{n-1} \rto^{\ps^{-1}} & X^n\cup CX^{n-1}  \rto^(0.6){\pi} & \SI X^{n-1} \rto^(0.4){\SI\rh}
& \SI(X^{n-1}/X^{n-2}) \uto_{\SI \pi_i}. \\
\enddiagram$$
An inspection of circles shows that this diagram homotopy commutes even when $n=1$. 
The bottom composite is the topological boundary map. Write $d_n'$ in
matrix form,
$$d'_n[j\com\nu_n^{-1}]=\SI^{-1}(\pa_n)_*[j\com\nu_n^{-1}] = 
\textstyle{\sum}_i a_{i,j}'[i\com\nu_{n-1}^{-1}].$$
Then, since the composite
$$\pi_i\com (i\com\nu_{n-1}^{-1}): S^{n-1}\rtarr S^{n-1}$$ 
is the identity map for each $i$, $a'_{i,j}$ is the degree of the map 
$S^n\rtarr S^n$ that we obtain by traversing the diagram counterclockwise from the
top left to the top right. The diagram implies that $a'_{i,j}=a_{i,j}$. 
\end{proof}

\begin{lem} $d_{n-1}\com d_{n}=0$.
\end{lem}
\begin{proof}
The composite $\SI\pa_{n-1}\com \pa_n$ is homotopic to the trivial map since the
following diagram is homotopy commutative and the composite $\SI \pi\com\SI i$ 
is the trivial map:
\begin{small}
$$\diagram
X^n\cup CX^{n-1} \dto_{\ps} \rto^{\pi} & \SI X^{n-1} \dto_{\SI\rh} \rto^(0.4){\SI i} & 
 \SI(X^{n-1}\cup CX^{n-2}) \dto^{\SI\ps} \rto^(0.6){\SI\pi} & \SI^2X^{n-2} \dto^{\SI^2\rh}\\
X^n/X^{n-1} \rto_(0.4){\pa_n} &\SI(X^{n-1}/X^{n-2}) \rdouble &\SI(X^{n-1}/X^{n-2}) \rto_{\SI\pa_{n-1}} &\SI^2X^{n-2}/X^{n-3}.\\
\enddiagram$$
\end{small}
The conclusion follows from our identification of $d_n$ and the naturality of $\SI$.
\end{proof}\index{cellular chain complex|)}

\section{Verification of the axioms}

For a CW complex $X$ with base vertex $*$, define 
$\tilde{C}_*(X)=C_*(X)/C_*(*)$\index{cellular chain complex!reduced}
and define $\tilde{H}_*(X)=H_*(\tilde{C}_*(X))$. This is the reduced homology of $X$.
For a subcomplex $A$ of $X$, define
$$C_*(X,A)=C_*(X)/C_*(A)\iso \tilde{C}_*(X/A)$$
and define 
$$H_*(X,A)=H_*(C_*(X,A))\iso \tilde{H}_*(X/A).$$
The long exact homology sequence of the exact sequence of chain complexes
$$ 0\rtarr C_*(A)\rtarr C_*(X) \rtarr C_*(X,A)\rtarr 0$$
gives the connecting homorphisms $\pa: H_q(X,A)\rtarr H_{q-1}(A)$ and the long
exact sequence called for in the exactness axiom.

If $f: X\rtarr Y$ is a cellular map, it induces maps $X^n/X^{n-1}\rtarr Y^n/Y^{n-1}$
that commute up to homotopy with the topological boundary maps and so induce
homomorphisms $f_n: C_n(X)\rtarr C_n(Y)$ that commute with the differentials. That is,
$f_*$ is a chain map, and it induces a homomorphism $H_*(X)\rtarr H_*(Y)$. 

For any CW complex $X$, $X\times I$ is a CW complex whose cellular chains are isomorphic
to $C_*(X)\otimes C_*(I)$, as we shall verify in the next section. Here $C_*(I)\iso \sI$ 
has basis elements $[0]$ and $[1]$ of degree zero and $[I]$ of degree one 
such that $d([I])=[0]-[1]$. We have observed that, for chain complexes $C$ and $D$, a chain 
map $C\otimes \sI\rtarr D$ can be identified with a chain homotopy between its restrictions 
to $C\otimes \bZ[0]$ and $C\otimes \bZ[1]$. A cellular homotopy $h: X\times I\rtarr Y$ 
induces just such a chain map, hence cellularly homotopic maps induce the same homomorphism 
on homology. The analogous conclusion for pairs follows by consideration of quotient complexes. 

The dimension and additivity axioms are obvious. If $X$ is a point, then $C_*(X)=\bZ$,
concentrated in degree zero. The cellular chain complex of $\amalg X_i$ is the
direct sum of the chain complexes $C_*(X_i)$, and similarly for pairs. Excision is also
obvious. If $X=A\cup B$, then the inclusion $A/A\cap B\rtarr X/B$ is an isomorphism
of CW complexes.

We have dealt so far with the case of integral homology. For more general coefficient 
groups $\pi$, we define
$$ C_*(X,A;\pi)=C_*(X,A)\otimes \pi$$
and proceed in exactly the same fashion to define homology groups and verify the axioms.
Observe that a homomorphism of groups $\pi\rtarr \rh$ induces a natural transformation
$$ H_q(X,A;\pi)\rtarr H_q(X,A;\rh)$$
that commutes with the connecting homomorphisms.

\section{The cellular chains of products}

A nice fact about cellular homology is that the definition leads directly
to an algebraic procedure for the calculation of the homology of Cartesian products. We 
explain the topological point here and return to the algebra later, where we discuss the
K\"{u}nneth theorem for the computation of the homology of tensor products of chain
complexes.

\begin{thm}
If $X$ and $Y$ are CW complexes, then $X\times Y$\index{cellular chain complex!of a product} is 
a CW complex such that
$$C_*(X\times Y)\iso C_*(X)\otimes C_*(Y).$$
\end{thm}
\begin{proof}
We have already seen that $X\times Y$ is a CW complex. Its $n$-skeleton is
$$(X\times Y)^n=\bigcup_{p+q=n}X^p\times Y^q,$$
and it has one $n$-cell, denoted $i\times j$, for each $p$-cell $i$ and $q$-cell $j$.
We define an isomorphism of graded Abelian groups 
$$\ka: C_*(X)\ten C_*(Y)\rtarr C_*(X\times Y)$$
by setting 
$$\ka([i]\ten[j])=(-1)^{pq}[i\times j].$$
It is clear from the definition of the product cell structure that $\ka$ commutes
up to sign with the differentials, and the insertion of the coefficient $(-1)^{pq}$ 
ensures that the signs work out. As we shall see, the coefficient appears
because we write suspension coordinates on the right rather than on the left. To be 
precise about this verification, we fix homeomorphisms
$$(D^n,S^{n-1})\iso (I^n,\pa I^n)$$
by radial contraction to the unit cube centered at $0$ followed by translation to the
unit cube centered at $1/2$. This fixes homeomorphisms
\begin{eqnarray*}
\io_{p,q}: (D^n,S^{n-1}) & \iso &(I^n,\pa I^n) 
= (I^p\times I^q, I^p\times\pa I^q\cup \pa I^p\times I^q) \\
& \iso & (D^p\times D^q, D^p\times S^{q-1}\cup S^{p-1}\times D^q)
\end{eqnarray*}
and thus fixes the product cells $i\times j$. For each $p$ and $q$, the following 
diagrams are homotopy commutative, where $n=p+q$ and where 
$$t: (\SI S^{p-1})\sma S^q = S^{p-1}\sma S^1\sma S^q\rtarr  
S^{p-1}\sma S^q\sma S^1=\SI(S^{p-1}\sma S^q)$$ 
is the transposition map. Note that, by a quick check when $q=1$ and induction, $t$ 
has degree $(-1)^q$.
$$\diagram
S^n \rrto^{\io_n} \dto_{\nu_n^{-1}} & & \SI S^{n-1} \dto^{\SI\nu_{n-1}^{-1}}\\
D^n/S^{n-1} \dto_{\io_{p,q}} & & \SI(D^{n-1}/S^{n-2})\dto^{\SI\io_{p,q-1}}\\
(D^p/S^{p-1})\sma (D^q/S^{q-1})\dto_{\nu_p\sma\nu_q} & &
\SI((D^p/S^{p-1})\sma (D^{q-1}/S^{q-2}))\dto^{\SI(\nu_p\sma\nu_{q-1})} \\
S^p\sma S^q \rto_{\id\sma\io_q} & S^p\sma(\SI S^{q-1})\rdouble & \SI(S^p\sma S^{q-1})\\
\enddiagram$$
and
$$\diagram
S^n \rrto^{\io_n} \dto_{\nu_n^{-1}} & & \SI S^{n-1}\dto^{\SI\nu_{n-1}^{-1}} \\
D^n/S^{n-1} \dto_{\io_{p,q}} & & \SI(D^{n-1}/S^{n-2})\dto^{\SI\io_{p-1,q}} \\
(D^p/S^{p-1})\sma (D^q/S^{q-1})\dto_{\nu_p\sma\nu_q} & & 
\SI((D^{p-1}/S^{p-2})\sma (D^{q}/S^{q-1}))\dto^{\SI(\nu_{p-1}\sma\nu_q)} \\
S^p\sma S^q \rto_{\io_p\sma\id} & (\SI S^{p-1})\sma S^q\rto_{(-1)^q t} & \SI(S^{p-1}\sma S^{q}).\\
\enddiagram$$
The homotopy commutativity would be clear if we worked only with cubes and replaced
the maps $\io_n$ and $\io_{p,q}$ with the evident identifications.  The only point at
issue then would be which copy of $I/\pa I$ is to be interpreted as the suspension
coordinate. The homotopy commutativity of the diagrams as written follows directly.
Now comparison of our description of the cellular differential in terms of the topological 
boundary map and the algebraic description of the differential on tensor products shows 
that $\ka$ is an isomorphism of chain complexes.
\end{proof}

\section{Some examples: $T$, $K$, and $\bR P^n$}

Cellular chains make some computations quite trivial. For example, since $S^n$ is a 
CW complex with one vertex and one $n$-cell, we see immediately that 
$$\tilde{H}_n(S^n;\pi)\iso \pi \ \ \tand \ \ \tilde{H}_q(S^n;\pi)=0 \ \ \ \text{for}\ \ q\neq n.$$

A little less obviously, if we look back at the CW decompositions of the torus $T$,\index{torus} the 
projective plane $\bR P^2$,\index{projective plane} and the Klein bottle\index{Klein bottle} $K$ 
and if we let $j$ denote the unique
$2$-cell in each case, then we find the following descriptions of the cellular chains 
and integral homologies by quick direct inspections. We agree to write $\bZ_n$ for the 
cyclic group $\bZ/n\bZ$.

\begin{exmps} (i) The cell complex $C_*(T)$ has one basis element $[v]$ in degree zero, two
basis elements $[e_1]$ and $[e_2]$ in degree one, and one basis element $[j]$ in degree two.
All basis elements are cycles, hence $H_*(T;Z)=C_*(T)$.

(ii) The cell complex $C_*(\bR P^2)$ has two basis elements $[v_1]$ and $[v_2]$ in degree 
zero, two basis elements $[e_1]$ and $[e_2]$ in degree one, and one basis element $[j]$ in 
degree two. The differentials are given by
$$d([e_1])=[v_1]-[v_2], \ \ d([e_2])=[v_2]-[v_1], \tand d([j])=2[e_1]+2[e_2].$$
Therefore $H_0(\bR P^2;\bZ)=\bZ$ with basis element the homology class of $[v_1]$ (or $[v_2]$),
$H_1(\bR P^2;\bZ) = \bZ_2$ with non-zero element the homology class of $e_1+e_2$, and
$H_q(\bR P^2;\bZ)=0$ for $q\geq 2$.

(iii) The cell complex $C_*(K)$ has one basis element $[v]$ in degree zero, two
basis elements $[e_1]$ and $[e_2]$ in degree one, and one basis element $[j]$ in 
degree two. The only non-zero differential is $d([j])=2[e_2]$. 
Therefore $H_0(K;\bZ)=\bZ$ with basis element the homology class of $[v]$,
$H_1(K;\bZ) = \bZ\oplus \bZ_2$ with $\bZ$ generated by the class of $[e_1]$
and $\bZ_2$ generated by the class of $[e_2]$, and $H_q(K;\bZ)=0$ for $q\geq 2$.
\end{exmps}

However, these examples are misleading. While homology groups are far easier to compute than 
homotopy groups, direct chain level calculation is seldom the method of choice. Rather, one uses
chains as a tool for developing more sophisticated algebraic techniques, notably spectral
sequences. We give an illustration that both shows that chain level calculations are 
sometimes practicable even when there are many non-zero differentials to determine and
indicates why one might not wish to attempt such calculations for really complicated
spaces.

We shall use cellular chains to compute the homology of $\bR P^n$. We think of 
$\bR P^n$\index{projective space!real} as the
quotient of $S^n$ obtained by identifying antipodal points, and we need to know the degree
of the antipodal map.\index{antipodal map}

\begin{lem}
The degree of the antipodal map $a_n: S^n\rtarr S^n$ is $(-1)^{n+1}$.
\end{lem}
\begin{proof}
Since $a_1\htp \id$ via an obvious rotation, the result is clear for $n=1$. The 
homeomorphism $\io_n: S^n\rtarr \SI S^{n-1}$ satisfies
$$\io_n(-x_1,\ldots\!, -x_{n+1})=(-vx_1,\ldots\!,-vx_n)\sma(1-x_{n+1})/2,$$
where $v=1/(\sum x_i^2)^{1/2}$. That is, $\io_n\com a_n=-(\SI a_{n-1})\com \io_n$. 
The conclusion follows by induction on $n$.
\end{proof}

We shall give $\bR P^n$ a CW structure with one $q$-cell for $0\leq q\leq n$ by passage
to quotients from a CW structure on $S^n$ with two $q$-cells for $0\leq q\leq n$. (Note that
this cell structure on $\bR P^2$ will be more economical than the one used in the calculation 
above.) The $q$-skeleton of $S^n$ will be $S^q$, which we identify with the subspace of $S^n$ whose 
points have last $n-q$ coordinates zero. We denote the two $q$-cells of $S^n$ by $j_{\pm}^q$. 
The two vertices are the points $\pm 1$ of $S^0$. Let $E^q_{\pm}$ be the upper and lower 
hemispheres in $S^{q}$, so that 
$$S^{q}=E^q_+\cup E^q_{-} \ \ \tand \ \  S^{q-1}=E^q_+\cap E^q_{-}.$$ 
We shall write
$$\pi_{\pm}: S^{q}\rtarr E^q_{\pm}/S^{q-1}$$
for the quotient maps that identify the lower or upper hemispheres to the basepoint.
Of course, these are homotopy equivalences. We define homeomorphisms
$$j^q_{\pm}: D^q\rtarr E^q_{\pm}\subset S^q$$
by
$$j^q_{\pm}(x_1,\ldots\!,x_q)=(\pm x_1,\ldots\!,\pm x_q,\pm (1-\textstyle{\sum} x_i^2)^{1/2}).$$
This decomposes $S^q$ as the the union of the images of two $q$-cells. The intersection
of these images is $S^{q-1}$ since
$$(x_1,\ldots\!, x_q,(1-\textstyle{\sum} x_i^2)^{1/2})=
(-y_1,\ldots\!,- y_q,-(1-\textstyle{\sum} y_i^2)^{1/2})$$
if and only if $x_i=-y_i$ for each $i$ and $\sum x_i^2=1$. Clearly 
$$j^q_{+}\,|\,S^{q-1}=\id \ \ \ \tand \ \ \ j^q_{-}\,|\,S^{q-1}=a_{q-1}.$$
Inspection of definitions shows that the following diagram commutes:
$$\diagram
S^{q-1} \rto^(0.4){\pi_+}\dto_{a_{q-1}} & E_+^{q-1}/S^{q-2} \dto^{a_{q-1}}& 
D^{q-1}/S^{q-2} \lto_{j_+^{q-1}} \ddouble \\
S^{q-1} \rto_(0.4){\pi_{-}} & E_{-}^{q-1}/S^{q-2}& 
D^{q-1}/S^{q-2} \lto^{j_{-}^{q-1}}.\\
\enddiagram$$
Since $a_{q-1}^2=\id$, it follows that we also obtain a commutative diagram if we 
interchange $+$ and $-$. If we invert the homeomorphisms $j^{q-1}_{\pm}$ and compose 
on the right with the homeomorphism $i_{q-1}: D^{q-1}/S^{q-2} \rtarr S^{q-1}$, then 
the degrees of the four resulting composite homotopy equivalences give the coefficients 
of the differential $d_q$. By composing $i_{q-1}$ with a homeomorphism of degree $-1$ 
if necessary, we can arrange that the degree of $i_{q-1}\com (j_+^{q-1})^{-1}\com\pi_+$ 
is $1$. We then deduce from the lemma and the definition of the differential on the 
cellular chains that
$$d_q[j^{q}_{+}]=(-1)^qd_q[j^q_{-}]=[j^{q-1}_+]+(-1)^q[j^{q-1}_{-}]$$
for all $q\geq 1$. 

Now, identifying antipodal points, we obtain the promised CW decomposition of $\bR P^n$.
If $p: S^n\rtarr \bR P^n$ is the quotient map, then 
$$p\com j^q_+ = p\com j^q_{-}: (D^q,S^{q-1})\rtarr (\bR P^q,\bR P^{q-1}).$$
We call this map $j^q$ and see that these maps give $\bR P^n$ a CW structure. Therefore
$C_q(\bR P^n)=\bZ$ with basis element $[j^q]$ for $q\geq 0$. Moreover, it is immediate 
from the calculation just given that 
$$d[j^q]=(1+(-1)^q)[j^{q-1}]$$
for all $q\geq 1$. This is zero if $q$ is odd and multiplication by $2$ if $q$ is even,
and we read off that
$$H_q(\bR P^n;\bZ)=
\begin{cases}
\bZ  \ \ \ \text{if}\ \ q=0 \\
\bZ_2 \ \ \text{if}\ \ 0<q<n\ \tand\ q\ \text{is odd}\\
\bZ  \ \ \ \text{if}\ \ q=n\ \text{is odd} \\
0 \ \ \ \ \text{otherwise.}
\end{cases}$$
If we work mod $2$, taking $\bZ_2$ as coefficient group, then the answer takes a nicer 
form, namely
$$H_q(\bR P^n;\bZ_2)=
\begin{cases}
\bZ_2  \ \ \text{if}\ 0\leq q\leq n \\
0 \ \ \ \ \text{if} \ q>n.
\end{cases}$$

This calculation well illustrates general facts about the homology of compact connected
closed $n$-manifolds $M$ that we shall prove later. The $n$th integral homology group of 
such a manifold $M$ is $\bZ$ if $M$ is orientable and zero if $M$ is not orientable. The 
$n$th mod $2$ homology group of $M$ is $\bZ_2$ whether or not $M$ is orientable. 

\vspace{.1in}

\begin{center}
PROBLEMS
\end{center}
\begin{enumerate}
\item If $X$ is a finite CW complex, show that $\ch(X)=\ch(H_*(X;k))$ for any field $k$.
\item Let $A$ be a subcomplex of a CW complex $X$, let $Y$ be a  CW complex, and let  
$f: A\rtarr Y$  be a cellular map.  What is the relationship between  $H_*(X,A)$  and  
$H_*(Y\cup_fX,Y)$?  Is there a similar relationship between $\pi_*(X,A)$ and  
$\pi_*(Y\cup_fX,Y)$?  If not, give a counterexample.
\item Fill in the details of the computation of the differentials on the cellular
chains in the examples in \S5.
\item Compute  $H_*(S^m\times S^n)$ for $m\geq 1$  and $n\geq 1$. Convince yourself that 
you can do this by use of CW structures, by direct deduction from the axioms, and by 
the K\"{u}nneth theorem (for which see Chapter 17).
\item Let $p$ be an odd prime number. Regard the cyclic group $\pi$ of order $p$ as the group 
of $p$th  roots of unity contained in $S^1$.  Regard $S^{2n-1}$ as the unit sphere in $\bC^n$,  
$n \geq 2$. Then $\pi\subset S^1$ acts freely on $S^{2n-1}$ via 
$$\ze (z_1,\ldots\!, z_n) = (\ze z_1,\ldots\!,\ze z_n). $$
Let $L^n = S^{2n-1}/\pi$  be the orbit space; it is called a lens space and is an odd primary 
analogue of $\bR P^n$.  The obvious quotient map $S^{2n-1}\rtarr L^n$ is a universal covering. 
\begin{enumerate}
\item[(a)]  Compute the integral homology of $L^n$, $n\geq 2$,  by mimicking the calculation of  
$H_*(\bR P^n)$.
\item[(b)]  Compute $H_*(L^n;\bZ_p)$,  where  $\bZ_p = \bZ/p\bZ$.
\end{enumerate}
\end{enumerate}

\clearpage

\thispagestyle{empty}

\chapter{Derivations of properties from the axioms}

Returning to the axiomatic approach to homology, we assume given a theory on pairs 
of spaces and make some deductions from the axioms. We abbreviate notations by setting 
$E_q(X,A)=H_q(X,A;\pi)$. However, the arguments in this chapter make no use whatever 
of the dimension axiom. A ``generalized homology theory''\index{homology theory!generalized} $E_*$ 
is defined to be a 
system of functors $E_q(X,A)$ and natural transformations $\pa:E_q(X,A)\rtarr E_{q-1}(A)$ 
that satisfy all of our axioms except for the dimension axiom. Similarly, we have the 
notion of a generalized homology theory on CW pairs, and the results of the first section 
of the previous chapter generalize directly to give the following result.

\begin{thm} A homology theory $E_*$ on pairs of spaces determines and is determined
by its restriction to a homology theory $E_*$ on pairs of CW complexes.
\end{thm}

The study of such generalized homology theories pervades modern algebraic topology,
and we shall describe some examples later on. The brave reader may be willing to think 
of $E_*$ in such generality in this chapter. The timorous reader may well prefer to 
think of $E_*(X,A)$ concretely, following our proposal that $E_*(X,A)$ be taken as
an alternative notation for $H_*(X,A;\pi)$.

\section{Reduced homology; based versus unbased spaces}

One of the themes of this chapter is the relationship between homology theories on
pairs of spaces and reduced homology theories on based spaces. The latter are more 
convenient in most advanced work in algebraic topology. For a based space $X$, we 
define the reduced homology\index{reduced homology} of $X$ to be
$$\tilde{E}_q(X)=E_q(X,*).$$
Since the basepoint is a retract of $X$, there results a direct sum 
decomposition
$$ E_*(X) \iso \tilde{E}_*(X)\oplus E_*(*)$$
that is natural with respect to based maps. For $*\in A\subset X$, the summand $E_*(*)$ 
maps isomorphically under the map $E_*(A)\rtarr E_*(X)$, and the exactness axiom implies 
that there is a reduced long exact sequence 
$$\cdots\rtarr \tilde{E}_q(A)\rtarr \tilde{E}_q(X)\rtarr 
E_q(X,A)\overto{\pa} \tilde{E}_{q-1}(A)\rtarr \cdots .$$

We can obtain the unreduced homology groups as special cases of the reduced ones. For an
unbased space $X$, we define a based space $X_+$ by adjoining a disjoint basepoint
to $X$. By the additivity axiom, we see immediately that
$$ E_*(X)=\tilde{E}_*(X_+).$$
Similarly, a map $f: X\rtarr Y$ of unbased spaces induces a map $f_+: X_+\rtarr Y_+$
of based spaces, and the map $f_*$ on unreduced homology coincides with the map
$(f_+)_*$ on reduced homology.

We shall make considerable use of cofiber sequences in this chapter.
To be consistent about this, we should always work with reduced cones and
cofibers. However, it is more convenient to make the convention that we work with 
unreduced cones and cofibers when we apply unreduced homology theories, and we work with 
reduced cones and cofibers when we apply reduced homology theories. In fact, the 
unreduced cone\index{cone!unreduced} on a space $Y$ coincides with the reduced 
cone\index{cone!reduced} on $Y_+$: the line through 
the disjoint basepoint is identified to the cone point when constructing the reduced cone
on $Y_+$. Therefore the unreduced cofiber of an unbased map $f$ coincides with the reduced 
cofiber of the based map $f_+$. Our convention really means that we are always working with 
reduced cofibers, but when we are studying unreduced homology theories we are implicitly 
applying the functor $(-)_+$ to put ourselves in the based context before constructing 
cones and cofibers.

The observant reader will have noticed that the unreduced suspension\index{suspension!unreduced} of 
$X$ is {\em not} the reduced suspension\index{suspension!reduced} on $X_+$. Rather, under either 
interpretation of suspension, $\SI(X_+)$ is homotopy equivalent to the wedge of $\SI(X)$ and a 
circle.

\section{Cofibrations and the homology of pairs}

We use cofibrations to show that the homology of pairs of spaces is in principle a special 
case of the reduced homology of spaces. 

\begin{thm}
For any cofibration\index{cofibration} $i: A\rtarr X$, the quotient map $q: (X,A)\rtarr (X/A,*)$ 
induces an isomorphism
$$E_*(X,A)\rtarr E_*(X/A,*)=\tilde{E}_*(X/A).$$
\end{thm}
\begin{proof}
Consider the (unreduced) cofiber 
$$Ci= X\cup_iCA = X\cup_i A\times I/A\times\sset{1}.$$ 
We have an excisive triad
$$(Ci; X\cup_i A\times [0,2/3], A\times [1/3,1]/A\times \sset{1}).$$
The excision axiom gives that the top inclusion in the following
commutative diagram induces an isomorphism on passage to homology:
$$\diagram
(X\cup_i A\times [0,2/3], A\times [1/3,2/3])\rto \dto_r 
& (Ci,A\times [1/3,1]/A\times \sset{1}) \dto^{\psi}\\
(X,A)\rto_q & (X/A,*)\\
\enddiagram$$
The map $r$ is obtained by restriction of the retraction $Mi\rtarr X$ and
is a homotopy equivalence of pairs. The map $\psi$ collapses $CA$ to a point 
and is also a homotopy equivalence of pairs. The conclusion follows.
\end{proof}

As in our construction of cellular homology, we choose a homotopy inverse 
$\ps^{-1}: X/A\rtarr Ci$ and consider the composite 
$$X/A\overto{\ps^{-1}} Ci\overto{\pi} \SI A$$
to be a topological boundary map
$$\pa: X/A\rtarr \SI A.$$
Observe that we may replace any inclusion $i: A\rtarr X$ by the canonical 
cofibration $A\rtarr Mi$ and then apply the result just given to obtain 
$$E_*(X,A)\iso \tilde{E}_*(Ci).$$

\section{Suspension and the long exact sequence of pairs}

We have a fundamentally important consequence of the results of the previous
section, which should be contrasted with what happened with homotopy groups.
Recall that a basepoint $*\in X$ is nondegenerate if the inclusion 
$\sset{*}\rtarr X$ is a cofibration. This ensures that the inclusion of 
the line through the basepoint in the unreduced suspension of $X$ is a
cofibration, so that the map from the unreduced suspension to the suspension
that collapses out the line through the basepoint is a homotopy equivalence.
We apply reduced homology here, so we use reduced cones and suspensions.

\begin{thm} For a nondegenerately based space $X$, there is a natural isomorphism
$$\SI: \tilde{E}_q(X)\iso \tilde{E}_{q+1}(\SI X).$$
\end{thm}
\begin{proof} Since $CX$ is contractible, its reduced homology is identically
zero. By the reduced long exact sequence, there results an isomorphism
$$\diagram \tilde{E}_{q+1}(\SI X)\iso \tilde{E}_{q+1}(CX/X) 
\overto{\pa} \tilde{E}_q(X). \qed
\enddiagram$$
\renewcommand{\qed}{}\end{proof}

An easy diagram chase gives the following consequence, which describes the axiomatically
given connecting homomorphism of the pair $(X,A)$ in terms of the topological boundary 
map\index{topological boundary map}
$\pa: X/A\rtarr \SI A$ and the suspension isomorphism.

\begin{cor} Let $*\in A\subset X$, where $i: A\rtarr X$ is a cofibration between
nondegenerately based spaces. In the long exact sequence 
$$\cdots\rtarr \tilde{E}_q(A)\rtarr \tilde{E}_q(X)\rtarr 
\tilde{E}_q(X/A)\overto{\pa} \tilde{E}_{q-1}(A)\rtarr \cdots $$
of the pair $(X,A)$, the connecting homomorphism $\pa$ is the composite
$$\tilde{E}_q(X/A)\overto{\pa_*} \tilde{E}_{q}(\SI A)\overto{\SI^{-1}} \tilde{E}_{q-1}(A).$$
\end{cor}

Since $S^0$ consists of two points, $\tilde{E}_*(S^0)=E_*(*)$. Since $S^n$ is the 
suspension of $S^{n-1}$, we have the following special case of the suspension 
isomorphism.

\begin{cor}
For any $n$ and $q$, 
$$\tilde{E}_q(S^n)\iso E_{q-n}(*).$$ 
\end{cor}

Of course, for the theory $H_*(X;\pi)$, this was immediate from our construction in
terms of cellular chains.

\section{Axioms for reduced homology}

In the study of generalized homology theories, it is most convenient to
restrict attention to reduced homology theories\index{homology theory!reduced} defined 
on nondegenerately
based spaces. The results of the previous sections imply that we can do so 
without loss of generality. Again the reader has the choice of bravery or
timorousness in interpreting $E_*$, but we opt for bravery:

\begin{defn} A reduced homology theory $\tilde{E}_*$ consists of functors 
$\tilde{E}_q$ from the homotopy category of nondegenerately based spaces 
to the category of Abelian groups that satisfy the following axioms.
\begin{itemize}
\item EXACTNESS\index{exactness axiom}\ \  If $i: A\rtarr X$ is a cofibration, then the sequence
$$\tilde{E}_q(A)\rtarr \tilde{E}_q(X)\rtarr 
\tilde{E}_q(X/A)$$
is exact.
\item SUSPENSION\index{suspension axiom}\ \ 
For each integer $q$, there is a natural isomorphism
$$\SI: \tilde{E}_q(X)\iso \tilde{E}_{q+1}(\SI X).$$
\item ADDITIVITY\index{additivity axiom}\ \ 
If $X$ is the wedge of a set of nondegenerately based spaces $X_i$, then
the inclusions $X_i\rtarr X$ induce an isomorphism
$$\textstyle{\sum}_i \tilde{E}_*(X_i)\rtarr \tilde{E}_*(X).$$
\item WEAK EQUIVALENCE\index{weak equivalence axiom}\ \ If $f:X\rtarr Y$ is a weak equivalence, then
$$f_*: \tilde{E}_*(X)\rtarr \tilde{E}_*(Y)$$
is an isomorphism.
\end{itemize}
\end{defn}

The reduced form of the dimension axiom would read
$$\tilde{H}_0(S^0)=\pi \ \ \tand \ \ \tilde{H}_q(S^0)=0\ \text{for}\ q\neq 0.$$

\begin{thm} A homology theory $E_*$ on pairs of spaces determines and is 
determined by a reduced homology theory $\tilde{E}_*$ on nondegenerately 
based spaces. 
\end{thm}
\begin{proof}
Given a theory on pairs, we define $\tilde{E}_*(X)=E_*(X,*)$ and deduce
the new axioms. For additivity, the specified wedge is the quotient 
$(\amalg X_i)/(\amalg\sset{*_i})$, where $*_i$ is the basepoint of $X_i$, and our 
result on quotients of cofibrations applies to compute its homology. Conversely, 
assume given a reduced homology theory $\tilde{E}_*$, and define 
$$E_*(X)=\tilde{E}_*(X_+) \ \ \tand \ \ E_*(X,A)=\tilde{E}_*(C(i_+)),$$
where $C(i_+)$ is the cofiber of the based inclusion $i_+: A_+\rtarr X_+$.
Equivalently, $C(i_+)$ is the unreduced cofiber of $i: A\rtarr X$ with its
cone point as basepoint. We must show that the suspension axiom and our restricted 
exactness axiom imply the original, seemingly much stronger, exactness and excision 
axioms. We have the long exact cofiber sequence associated to the based inclusion 
$i_+: A_+\rtarr X_+$, in which each consecutive pair of maps is equivalent to a 
cofibration and the associated quotient map. Noting that $X_+/A_+=X/A$, we define
the connecting homomorphism $\pa_q: E_q(X,A)\rtarr E_{q-1}(A)$ to be the composite 
$$\tilde{E}_q(X_+/A_+)\overto{\pa_*}\tilde{E}_q(\SI A_+)\overto{\SI^{-1}}\tilde{E}_{q-1}(A_+)$$
and find that the exactness and suspension axioms for $\tilde{E}_*$ imply the exactness
axiom for $E_*$. For excision, we could carry out a similarly direct homotopical argument, 
but it is simpler to observe that this follows from the equivalence of theories on pairs of 
spaces with theories on pairs of CW complexes together with the next two theorems. For the 
additivity axiom, we note that the cofiber of a disjoint union of maps is the wedge of the 
cofibers of the given maps.
\end{proof}

\begin{cor} For nondegenerately based spaces $X$, $E_*(X)$ is naturally isomorphic to
$\tilde{E}_*(X)\oplus E_*(*)$.
\end{cor}
\begin{proof}
The long exact sequence in $E_*$ of the pair $(X,*)$ is naturally split in each degree 
by means of the homomorphism induced by the projection $X\rtarr \sset{*}$.
\end{proof}

We require of based CW complexes that the basepoint be a vertex. It is certainly
a nondegenerate basepoint. We give the circle its standard CW structure and so 
deduce a CW structure on the suspension of a based CW complex.

\begin{defn} A reduced homology theory\index{homology theory!reduced} $\tilde{E}_*$ on based 
CW complexes consists 
of functors $\tilde{E}_q$ from the homotopy category of based CW complexes 
to the category of Abelian groups that satisfy the following axioms.
\begin{itemize}
\item EXACTNESS\index{exactness axiom}\ \  If $A$ is a subcomplex of $X$, then the sequence
$$\tilde{E}_q(A)\rtarr \tilde{E}_q(X)\rtarr 
\tilde{E}_q(X/A)$$
is exact.
\item SUSPENSION\index{suspension axiom} \ \ 
For each integer $q$, there is a natural isomorphism
$$\SI: \tilde{E}_q(X)\iso \tilde{E}_{q+1}(\SI X).$$
\item ADDITIVITY\index{additivity axiom}\ \ 
If $X$ is the wedge of a set of based CW complexes $X_i$, then
the inclusions $X_i\rtarr X$ induce an isomorphism
$$\textstyle{\sum}_i \tilde{E}_*(X_i)\rtarr \tilde{E}_*(X).$$
\end{itemize}
\end{defn}

\begin{thm} A reduced homology theory $\tilde{E}_*$ on nondegenerately based spaces
determines and is determined by its restriction to a reduced homology theory on
based CW complexes.
\end{thm}
\begin{proof}
This is immediate by CW approximation of based spaces.
\end{proof}

\begin{thm}
A homology theory $E_*$ on CW pairs determines and is determined by a reduced
homology theory $\tilde{E}_*$ on based CW complexes.
\end{thm}
\begin{proof}
Given a theory on pairs, we define $\tilde{E}_*(X)=E_*(X,*)$ and deduce
the new axioms directly. Conversely, given a reduced theory on based CW 
complexes, we define 
$$ E_*(X)=\tilde{E}_*(X_+)\ \ \tand\ \ E_*(X,A)=\tilde{E}_*(X/A).$$
Of course $X/A$ is homotopy equivalent to $C(i_+)$, where $i_+: A_+\rtarr X_+$ is the 
inclusion. The arguments for exactness and additivity are the same as those given in the 
analogous result for nondegenerately based spaces, but now excision is obvious since if 
$(X;A,B)$ is a CW triad, then the inclusion $A/A\cap B \rtarr X/B$ is an isomorphism of 
based CW complexes.
\end{proof}

\section{Mayer-Vietoris sequences}

The Mayer-Vietoris sequences are long exact sequences associated to excisive triads
that will play a fundamental role in our later proof of the Poincar\'{e} duality theorem. 
We need two preliminaries, both of independent interest. The first is the long exact 
sequence of a triple $(X,A,B)$ of spaces $B\subset A\subset X$, which is just like its
analogue for homotopy groups.

\begin{prop} For a triple $(X,A,B)$, the following sequence\index{triple!exact sequence of} is exact:
$$\cdots \rtarr E_q(A,B)\overto{i_*} E_q(X,B)\overto{j_*} E_q(X,A) \overto{\pa} 
E_{q-1}(A,B)\rtarr \cdots.$$
Here $i:(A,B)\rtarr (X,B)$ and $j:(X,B)\rtarr (X,A)$ are inclusions and $\pa$ is the 
composite
$$E_q(X,A)\overto{\pa}E_{q-1}(A)\rtarr E_{q-1}(A,B).$$
\end{prop}
\begin{proof}
There are two easy arguments. One can either use diagram chasing from the various long
exact sequences of pairs or one can apply CW approximation to replace $(X,A,B)$ by a 
triple of CW complexes. After the replacement, we have that $X/A\iso(X/B)/(A/B)$ as a CW complex,
and the desired sequence is the reduced exact sequence of the pair $(X/B,A/B)$.
\end{proof}

\begin{lem}
Let $(X;A,B)$ be an excisive triad and set $C=A\cap B$. The map 
$$E_*(A,C)\oplus E_*(B,C)\rtarr E_*(X,C)$$
induced by the inclusions of $(A,C)$ and $(B,C)$ in $(X,C)$ is an isomorphism.
\end{lem}
\begin{proof}
Again, there are two easy proofs. One can either pass to homology from the diagram
$$\diagram
(B,C) \ddto_{excision} \drto & & (A,C) \dlto \ddto^{excision} \\
& (X,C) \drto  \dlto & \\
(X,A) & & (X,B) \\
\enddiagram$$
and use algebra or one can approximate $(X;A,B)$ by a CW triad, for which
$$X/C\iso A/C\wed B/C$$
as a CW complex.
\end{proof}

\begin{thm}[Mayer-Vietoris sequence]\index{Mayer-Vietoris sequence} Let 
$(X;A,B)$ be an excisive triad and set $C=A\cap B$.
The following sequence is exact:
$$\cdots \rtarr E_q(C)\overto{\ps} E_q(A)\oplus E_q(B)\overto{\ph} E_q(X)\overto{\DE} 
E_{q-1}(C)\rtarr \cdots.$$
Here, if $i: C\rtarr A$, $j: C\rtarr B$, $k: A\rtarr X$, and $\ell: B\rtarr X$
are the inclusions, then
$$\psi(c)=(i_*(c),j_*(c)), \ \ \ \ \ph(a,b)= k_*(a)-\ell_*(b),$$
and $\DE$ is the composite 
$$E_q(X)\rtarr E_q(X,B)\iso E_q(A,C)\overto{\pa} E_{q-1}(C).$$
\end{thm}
\begin{proof} Note that the definition of $\ph$ requires a sign in order to make $\ph\com\ps = 0$.
The proof of exactness is algebraic diagram chasing and is left as an exercise.  The following 
diagram may help:
$$\diagram
& & E_q(C) \dlto  \ddto^{i_*} \drto & & \\  
& E_q(B) \drto  & & E_q(A) \dlto  & \\   
& & E_q(X) \dlto  \ddto^{j_*} \drto \xto '[0,2] '[4,2]^{\DE} '[4,1] 
\xto '[0,-2] '[4,-2]_{-\DE} '[4,-1] & & \\ 
& E_q(X,A)   & & E_q(X,B)  & \\    
& & E_q(X,C) \ulto  \ddto^{\pa} \urto & & \\ 
& E_q(B,C) \drto_{\pa} \urto \uuto^{\iso} & & E_q(A,C) \dlto^{\pa} \ulto \uuto_{\iso}  & \\ 
& & E_{q-1}(C)  & & \\     
\enddiagram$$
Here the arrow labeled ``$-\DE$'' is in fact $-\DE$ by an algebraic argument from the direct
sum decomposition of $E_q(X,C)$. Alternatively, one can use CW approximation. For a CW triad,
there is a short exact sequence
$$0\rtarr C_*(C)\rtarr C_*(A)\oplus C_*(B)\rtarr C_*(X)\rtarr 0$$
whose associated long exact sequence is the Mayer-Vietoris sequence.
\end{proof}

We shall also need a relative analogue, but the reader may wish to ignore this for now. It 
will become important when we study manifolds with boundary. 

\begin{thm}[Relative Mayer-Vietoris sequence]\index{Mayer-Vietoris sequence!relative}  Let 
$(X;A,B)$ be an excisive \linebreak
triad and set 
$C=A\cap B$. Assume that $X$ is contained in some ambient space $Y$. The following 
sequence is exact:
$$\cdots\rtarr E_q(Y,C)\overto{\ps} E_q(Y,A)\oplus E_q(Y,B)\overto{\ph} E_q(Y,X)\overto{\DE} 
E_{q-1}(Y,C)\rtarr \cdots.$$
Here, if $i: (Y,C)\rtarr (Y,A)$, $j: (Y,C)\rtarr (Y,B)$, $k: (Y,A)\rtarr (Y,X)$, and 
$\ell: (Y,B)\rtarr (Y,X)$ are the inclusions, then
$$\psi(c)=(i_*(c),j_*(c)), \ \ \ \ \ph(a,b)= k_*(a)-\ell_*(b),$$
and $\DE$ is the composite 
$$E_q(Y,X)\overto{\pa} E_{q-1}(X,B)\iso E_{q-1}(A,C)\rtarr E_{q-1}(Y,C).$$
\end{thm}
\begin{proof}
This too is left as an exercise, but it is formally the same exercise.
The relevant diagram is the following one:
$$\diagram
& & E_q(Y,C) \dlto  \ddto \drto & & \\  
& E_q(Y,B) \drto  & & E_q(Y,A) \dlto  & \\   
& & E_q(Y,X) \dlto_{\pa}  \ddto^{\pa} \drto^{\pa} \xto '[0,2] '[4,2]^{\DE} '[4,1] 
\xto '[0,-2] '[4,-2]_{-\DE} '[4,-1] & & \\ 
& E_{q-1}(X,A)  & & E_{q-1}(X,B)  & \\    
& & E_{q-1}(X,C) \ulto  \ddto \urto & & \\ 
& E_{q-1}(B,C) \drto \urto \uuto^{\iso} & & E_{q-1}(A,C) \dlto \ulto 
\uuto_{\iso}  & \\ 
& & E_{q-1}(Y,C)  & & \\     
\enddiagram$$
Alternatively, one can use CW approximation. For a CW triad $(X;A,B)$, with
$X$ a subcomplex of a CW complex $Y$, there is a short exact sequence
$$0\rtarr C_*(Y/C)\rtarr C_*(Y/A)\oplus C_*(Y/B)\rtarr C_*(Y/X)\rtarr 0$$
whose associated long exact sequence is the relative Mayer-Vietoris sequence.
\end{proof}

A comparison of definitions gives a relationship between these sequences.

\begin{cor}
The absolute and relative Mayer-Vietoris sequences are related by the following 
commutative diagram:
$$\diagram
E_q(Y,C)\rto^(0.35){\ps} \dto_{\pa} & E_q(Y,A)\oplus E_q(Y,B) \rto^(0.6){\ph} 
\dto^{\pa+\pa} & E_q(Y,X) \rto^{\DE} \dto^{\pa}
& E_{q-1}(Y,C) \dto^{\pa}\\
E_{q-1}(C)\rto_(0.35){\ps} &
 E_{q-1}(A)\oplus E_{q-1}(B)\rto_(0.6){\ph} & E_{q-1}(X)\rto_{\DE} 
& E_{q-2}(C).\\
\enddiagram$$
\end{cor}

\section{The homology of colimits}

In this section, we let $X$ be the union of an expanding sequence of subspaces $X_i$, 
$i\geq 0$. We have seen that the compactness of spheres $S^n$ and cylinders $S^n\times I$ 
implies that, for any choice of basepoint in $X_0$, the natural map
$$\colim\,\pi_*(X_i)\rtarr \pi_*(X)$$
is an isomorphism.  We shall use the additivity and weak equivalence axioms and the 
Mayer-Vietoris sequence to prove the analogue for homology.

\begin{thm}
The natural map
$$\colim E_*(X_i)\rtarr E_*(X)$$
is an isomorphism.\index{colimit!homology of}
\end{thm}

We record an algebraic description of the colimit of a sequence for use in the proof.

\begin{lem}
Let $f_i: A_i\rtarr A_{i+1}$ be a sequence of homomorphisms of Abelian groups.
Then there is a short exact sequence
$$ 0\rtarr \textstyle{\sum}_i A_i\overto{\al} \textstyle{\sum}_i A_i\overto{\be} \colim\,A_i\rtarr 0,$$
where $\al(a_i)=a_i-f_i(a_i)$ for $a_i\in A_i$ and the restriction of $\be$ to $A_i$
is the canonical map given by the definition of a colimit.
\end{lem}

By the additivity axiom, we may as well assume that $X$ and the $X_i$ are path connected.
The proof makes use of a useful general construction called the ``telescope'' of the $X_i$,
denoted  $\tel\,X_i$. Let $j_i: X_i\rtarr X_{i+1}$ be the given inclusions and consider the 
mapping cylinders
$$M_{i+1}=(X_i\times[i,i+1])\cup X_{i+1}$$
that are obtained by identifying $(x,i+1)$ with $j_i(x)$ for $x\in X_i$.
Inductively, let $Y_0=X_0\times\sset{0}$ and suppose that we have constructed 
$Y_i\supset X_i\times \sset{i}$. Define $Y_{i+1}$ to be the double mapping cylinder
$Y_i\cup M_{i+1}$ obtained by identifying $(x,i)\in Y_i$ with $(x,i)\in M_{i+1}$ for 
$x\in X_i$. Define $\tel\,X_i$\index{telescope} to be the union of the $Y_i$, 
with the colimit topology. Thus 
$$\tel\,X_i= \bigcup_i X_i\times [i,i+1],$$
with the evident identifications at the ends of the cylinders. 

Using the retractions of the mapping cylinders, we obtain composite retractions
$r_i: Y_i\rtarr X_i$ such that the following diagrams commute
$$\diagram
Y_i\rto^{\subset} \dto_{r_i} & Y_{i+1} \dto^{r_{i+1}} \\
X_i \rto_{j_i} & X_{i+1} \\
\enddiagram$$
Since the $r_i$ are homotopy equivalences and since homotopy groups commute with colimits, it
follows that we obtain a weak equivalence
$$r: \tel X_i\rtarr X$$
on passage to colimits. By the weak equivalence axiom, $r$ induces an isomorphism on
homology. It therefore suffices to prove that the natural map
$$\colim\,E_*(X_i)\iso \colim\,E_*(Y_i)\rtarr E_*(\tel\,X_i)$$
is an isomorphism. We define subspaces $A$ and $B$ of $\tel\,X_i$ by choosing $\epz<1$ and
letting 
$$A =X_0\times[0,1]\, \textstyle{\coprod} \,
\textstyle{\coprod}_{i\geq 1}\, X_{2i-1}\times[2i-\epz,2i]
\cup X_{2i}\times [2i,2i+1]$$
and
$$B=\textstyle{\coprod}_{i\geq 0}\,X_{2i}\times[2i+1-\epz,2i+1]
\cup X_{2i+1}\times [2i+1,2i+2].$$
We let  $C=A\cap B$ and find that 
$$C=\textstyle{\coprod}_{i\geq 0}\, X_i\times [i+1-\epz,i+1].$$
This gives an excisive triad, and a quick inspection shows that we have canonical
homotopy equivalences
$$A\htp \textstyle{\coprod}_{i\geq 0} X_{2i},\ \ B\htp\textstyle{\coprod}_{i\geq 0}X_{2i+1},\ \ \tand
C\htp\textstyle{\coprod}_{i\geq 0}X_i.$$
Moreover, under these equivalences the inclusion $C\rtarr A$ has restrictions 
$$\id: X_{2i}\rtarr X_{2i} \ \ \tand\ \ j_{2i+1}: X_{2i+1}\rtarr X_{2i+2},$$ 
while the inclusion $C\rtarr B$ has restrictions
$$j_{2i}: X_{2i}\rtarr X_{2i+1} \ \ \tand\ \ \id: X_{2i+1}\rtarr X_{2i+1}.$$ 
By the additivity axiom,
$$E_*(A)=\textstyle{\sum}_i E_*(X_{2i}),\ \ E_*(B)=\textstyle{\sum}_i E_*(X_{2i+1}),
\ \tand\ E_*(C)=\textstyle{\sum}_i E_*(X_i).$$
We construct the following commutative diagram, whose top row is the Mayer-Vietoris 
sequence of the triad $(\text{tel}\, X_i;A,B)$ and whose bottom row is a short exact sequence
as displayed in our algebraic description of colimits:
$$\diagram
\cdots \rto &  E_q(C) \rto \dto_{\iso} &  E_q(A)\oplus E_q(B) \rto \dto^{\iso}
 & E_q(\tel X_i) \rto \dto^{\iso} & \cdots \\
\cdots \rto & \sum_i E_q(X_i) \rto^{\al'} \dto_{\sum(-1)^{i}}
 & \sum_i E_q(X_i) \rto^{\be'} \dto^{\sum_i(-1)^{i}}
& E_q(X) \rto \ddashed^{\xi}|>\tip & \cdots \\
0 \rto & \sum_i E_q(X_i) \rto^{\al} 
 & \sum_i E_q(X_i) \rto^{\be} 
& \colim E_q(X_i) \rto & 0. \\
\enddiagram$$
By the definition of the maps in the Mayer-Vietoris sequence, $\al'(x_i)=x_i+(j_i)_*(x_i)$
and $\be'_i(x_i)=(-1)^i(k_i)_*(x_i)$ for $x_i\in E_q(X_i)$, where $k_i: X_i\rtarr X$ is 
the inclusion. The commutativity of the lower left square is just the relation 
$$(\textstyle{\sum}_i (-1)^i)\al'(x_i)=(-1)^i(x_i-(j_i)_*(x_i)).$$
The diagram implies the required isomorphism $\xi$.

\begin{rem} There is a general theory of ``homotopy colimits,''\index{homotopy colimit} which are 
up to homotopy versions of
colimits. The telescope is the homotopy colimit of a sequence. The double mapping cylinder
that we used in approximating excisive triads by CW triads is the homotopy pushout of a 
diagram of the shape $\bullet\longleftarrow \bullet\rtarr \bullet$. We implicitly used
homotopy coequalizers in constructing CW approximations of spaces.
\end{rem}

\vspace{.1in}

\begin{center}
PROBLEM
\end{center}
\begin{enumerate}
\item Complete the proof that the Mayer-Vietoris sequence is exact.
\end{enumerate}

\chapter{The Hurewicz and uniqueness theorems}

We now return to the context of ``ordinary homology theories,''\index{homology theory!ordinary} 
namely those that satisfy the dimension axiom. We prove a fundamental relationship, called the Hurewicz theorem,
between homotopy groups and homology groups. We then use it to prove the uniqueness of ordinary 
homology with coefficients in $\pi$.

\section{The Hurewicz theorem}

Although the reader may prefer to think in terms of the cellular homology theory already 
constructed, the proof of the Hurewicz theorem depends only on the axioms. It is this fact
that will allow us to use the result to prove the uniqueness of homology theories in the 
next section. We take $\pi=\bZ$ and delete it from the notation. The dimension axiom 
implicitly fixes a generator $i_0$ of $\tilde{H}_0(S^0)$, and we choose generators $i_n$ of 
$\tilde{H}_n(S^n)$ inductively by setting $\SI i_n=i_{n+1}$. 

\begin{defn}
For based spaces $X$, define the Hurewicz homomorphism\index{Hurewicz homomorphism}
$$h:\pi_n(X)\rtarr \tilde{H}_n(X)$$ 
by 
$$h([f])=f_*(i_n).$$
\end{defn}

\begin{lem} If $n\geq 1$, then $h$ is a homomorphism for all $X$. 
\end{lem}
\begin{proof}
For maps $f,g: S^n\rtarr X$, $[f+g]$ is represented by the composite
$$S^n\overto{p} S^n\wed S^n\overto{f\wed g} X\wed X\overto{\triangledown} X,$$
where $p$ is the pinch map and $\triangledown$ is the codiagonal map; that is,
$\triangledown$ restricts to the identity on each wedge summand. Since
$p_*(i_n)=i_n+i_n$ and $\triangledown$ induces addition on $\tilde{H}_*(X)$,
the conclusion follows.
\end{proof}

\begin{lem} The Hurewicz homomorphism is natural and the following diagram 
commutes for $n\geq 0$:
$$\diagram
\pi_n(X) \dto_{\SI} \rto & \tilde{H}_n(X) \dto^{\SI} \\
\pi_{n+1}(\SI X) \rto_{h} & \tilde{H}_{n+1}(\SI X).\\
\enddiagram$$
\end{lem}
\begin{proof}
The naturality of $h$ is clear, and the naturality of $\SI$ on homology
implies the commutativity of the diagram:
$$(h\com \SI)([f])=(\SI f)_*(\SI i_n)=\SI(f_*(i_n))=\SI(h([f])). \qed $$
\renewcommand{\qed}{}\end{proof}

\begin{lem} Let $X$ be a wedge of $n$-spheres. Then
$$h:\pi_n(X)\rtarr \tilde{H}_n(X)$$ 
is the Abelianization homomorphism if $n=1$ and is an isomorphism if $n>1$. 
\end{lem}
\begin{proof}
When $X$ is a single sphere, $h[\id]=i_n$ and the conclusion is obvious.
In general, $\pi_n(X)$ is the free group if $n=1$ or the free Abelian group
if $n\geq 2$ with generators given by the inclusions of the wedge summands.
Since $h$ maps these generators to the canonical generators of 
the free Abelian group $\tilde{H}_n(X)$, the conclusion follows.
\end{proof}

That is all that we shall need in the next section, but we can generalize
the lemma to arbitrary $(n-1)$-connected based spaces $X$.

\begin{thm}[Hurewicz]\index{Hurewicz theorem}
Let $X$ be any $(n-1)$-connected based space. Then
$$h:\pi_n(X)\rtarr \tilde{H}_n(X)$$ 
is the Abelianization homomorphism if $n=1$ and is an isomorphism if $n>1$. 
\end{thm}
\begin{proof}
We can assume without loss of generality that $X$ is a CW complex with a single
vertex, based attaching maps, and no $q$-cells for $1\leq q<n$. The inclusion 
of the $(n+1)$-skeleton in $X$ induces an isomorphism on $\pi_n$ by the cellular 
approximation theorem and induces an isomorphism on $\tilde{H}_n$ by our 
cellular construction of homology or by a deduction from the axioms that will be 
given in the next section. Thus we may assume without loss of generality that $X=X^{n+1}$. 
Then $X$ is the cofiber of a map $f: K\rtarr L$, where $K$ and $L$ are both wedges of 
$n$-spheres. We have the following commutative diagram:
$$\diagram
\pi_n(K) \rto \dto & \pi_n(L)\rto \dto & \pi_n(X)\rto \dto & 0 \\
\tilde{H}_n(K) \rto & \tilde{H}_n(L) \rto & \tilde{H}_n(X) \rto & 0.\\
\enddiagram$$
The lemma gives the conclusion for the two left vertical arrows. 
Since $X/L$ is a wedge of $(n+1)$-spheres, the bottom row is exact by our long 
exact homology sequences and the known homology of wedges of spheres. When $n=1$,
a corollary of the van Kampen theorem gives that $\pi_1(X)$ is the quotient of
$\pi_1(L)$ by the normal subgroup generated by the image of $\pi_1(K)$. An easy
algebraic exercise shows that the sequence obtained from the top row by passage
to Abelianizations is therefore exact. If $n>1$, the homotopy excision theorem 
implies that the top row is exact. To see this, factor $f$ as the composite of
the inclusion $K\rtarr Mf$ and the deformation retraction $r:Mf\rtarr L$. Since
$X=Cf$, we have the following commutative diagram, in which the top row is exact: 
$$\diagram
\pi_n(K) \rto \ddouble & \pi_n(Mf)\rto \dto^{r_*} & \pi_n(Mf,K)\rto \dto & 0 \\
\pi_n(K) \rto & \pi_n(L) \rto & \pi_n(X) \rto & 0.\\
\enddiagram$$
Since $K$ and $L$ are $(n-1)$-connected and $n>1$, a corollary of the homotopy 
excision theorem gives that $X$ is $(n-1)$-connected and 
$\pi_n(Mf,K)\rtarr \pi_n(X)$ is an isomorphism.
\end{proof}

\section{The uniqueness of the homology of CW complexes}

We assume given an ordinary homology theory on CW pairs and describe how it must 
be computed. We focus on integral homology, taking $\pi=\bZ$ and deleting it from 
the notation. With a moment's reflection on the case $n=0$, we see that the 
Hurewicz theorem gives a natural isomorphism\index{reduced homology!provisional definition}
$$\tilde{H}'_n(X)\rtarr \tilde{H}_n(X)$$
for $(n-1)$-connected based spaces $X$. Here the groups on the left are defined
in terms of homotopy groups and were used in our construction of cellular chains,
while the groups on the right are those of our given homology theory. We use the
groups on the right to construct cellular chains in our given theory, and we find
that the isomorphism is compatible with differentials. From here, to prove uniqueness,
we only need to check from the axioms that our given theory is computable from the 
homology groups of these cellular chain complexes. 

Thus let $X$ be a CW complex. For each integer $n$, define
$$C_n(X)=H_n(X^n,X^{n-1})\iso\tilde{H}_n(X^n/X^{n-1}).$$
Define 
$$d: C_n(X)\rtarr C_{n-1}(X)$$
to be the composite
$$H_n(X^n,X^{n-1})\overto{\pa}H_{n-1}(X^{n-1})
\rtarr H_{n-1}(X^{n-1},X^{n-2}).$$
It is not hard to check that $d\com d=0$.
\begin{thm}
$C_*(X)$ is isomorphic to the cellular chain complex of $X$.\index{cellular chain complex}
\end{thm}
\begin{proof}
Since $X^n/X^{n-1}$ is the wedge of an $n$-sphere for each $n$-cell of $X$, we
see by the additivity axiom that $C_n(X)$ is the free Abelian group with one 
generator $[j]$ for each $n$-cell $j$. We must compare the differential with
the one that we defined earlier. Let $i:X^{n-1}\rtarr X^n$ be the inclusion.
We see from our proof of the suspension isomorphism that $d$ coincides with the 
composite
$$\tilde{H}_n(X^n/X^{n-1})\iso \tilde{H}_n(Ci)
\to \tilde{H}_n(\SI X^{n-1})
\overto{\SI^{-1}}\tilde{H}_{n-1}(X^{n-1})\to \tilde{H}_{n-1}(X^{n-1}/X^{n-2}).$$
By the naturality of the Hurewicz homomorphism and its commutation with suspension,
this coincides with the differential that we defined originally. 
\end{proof}

Similarly, if we start with a homology theory $H_*(-;\pi)$, we can use the axioms to
construct a chain complex $C_*(X;\pi)$, and a comparison of definitions then gives an 
isomorphism of chain complexes
$$C_*(X;\pi)\iso C_*(X)\otimes \pi.$$
We have identified our axiomatically derived chain complex of $X$ with the cellular 
chain complex of $X$, and we again adopt the notation $C_*(X,A)=\tilde{C}_*(X/A)$. 

\begin{thm}\index{homology theory!ordinary}
There is a natural isomorphism
$$H_*(X,A)\iso H_*(C_*(X,A))$$
under which the natural transformation $\pa$ agrees with the natural transformation
induced by the connecting homomorphisms associated to the short exact sequences
$$0\rtarr C_*(A)\rtarr C_*(X)\rtarr C_*(X,A)\rtarr 0.$$
\end{thm}
\begin{proof}
In view of our comparison of theories on pairs of spaces and theories on pairs of
CW complexes and our comparison of theories on pairs with reduced theories, it 
suffices to obtain a natural isomorphism of reduced theories on based CW complexes $X$.
By the additivity axiom, we may as well assume that $X$ is connected. More precisely,
we must obtain a system of natural isomorphisms 
$$\tilde{H}_n(X)\iso {H}_*(\tilde{C}_n(X))$$
that commute with the suspension isomorphisms.

By the dimension and additivity axioms, we know the homology of wedges of spheres. 
Since $X^n/X^{n-1}$ is a wedge of $n$-spheres,
the long exact homology sequence associated to the cofiber sequence 
$$X^{n-1}\rtarr X^n\rtarr X^n/X^{n-1}$$
and an induction on $n$ imply that
$$\tilde{H}_q(X^{n-1})\rtarr \tilde{H}_q(X^{n})$$
is an isomorphism for $q<n-1$ and
$$\tilde{H_q}(X^n)=0$$
for $q>n$. Of course, the analogues for cellular homology are obvious. Note in particular
that $\tilde{H}_n(X^{n+1})\iso \tilde{H}_n(X^{n+i})$ for all $i>1$. Since homology
commutes with colimits on sequences of inclusions, this implies that the inclusion 
$X^{n+1}\rtarr X$ induces an isomorphism
$$ \tilde{H}_n(X^{n+1})\rtarr \tilde{H}_n(X).$$
Using these facts, we easily check from the exactness axiom that the rows and columns are 
exact in the following commutative diagram:
$$\diagram
& \tilde{H}_{n+1}(X^{n+1}/X^n)  \drto^{d_{n+1}} \dto_{\pa}  &  & 0 \dto \\
0  \rto & \tilde{H}_n(X^n) \rto_{\rh_*} \dto_{i_*} & \tilde{H}_n(X^n/X^{n-1})
\rto^{\pa} \drto_{d_n} & \tilde{H}_{n-1}(X^{n-1}) \dto \\
& \tilde{H}_n(X)\iso\tilde{H}_n(X^{n+1}) \dto & & \tilde{H}_{n-1}(X^{n-1}/X^{n-2}). \\
& 0 & & \\
\enddiagram$$
Define $\al: \tilde{H}_n(X)\rtarr H_n(\tilde{C}_*(X))$ by letting $\al(x)$ be the
homology class of $\rh_*(y)$ for any $y$ such that $i_*(y)=x$. It is an exercise in
diagram chasing and the definition of the homology of a chain complex to check that 
$\al$ is a well defined isomorphism. 

The reduced chain complex of $\SI X$ can be 
identified with the suspension of the reduced chain complex of $X$. That is,
$$\tilde{C}_{n+1}(\SI X)\iso \tilde{C}_n(X),$$
compatibly with the differential. All maps in the diagram commute with suspension,
and this implies that the isomorphisms $\al$ commute with the suspension isomorphisms.
\end{proof}

\vspace{.1in}

\begin{center}
PROBLEMS
\end{center}
\begin{enumerate}
\item Let $\pi$ be any group. Construct a connected CW complex $K(\pi,1)$ such that
$\pi_1(K(\pi,1))=\pi$ and $\pi_q(K(\pi,1))=0$ for $q\neq 1$. 
\item* In Problem 1, it is rarely the case that $K(\pi,1)$ can be constructed as a compact
manifold. What is a necessary condition on $\pi$ for this to happen?
\item Let $n\geq1$ and let $\pi$ be an Abelian group. Construct a connected CW complex 
$M(\pi,n)$ such that $\tilde{H}_n(X;\bZ)=\pi$ and $\tilde{H}_q(X;\bZ)=0$ for $q\neq n$.
(Hint: construct $M(\pi,n)$ as the cofiber of a map between wedges of spheres.)
The spaces $M(\pi,n)$ are called Moore spaces.\index{Moore space}
\item Let $n\geq1$ and let $\pi$ be an Abelian group. Construct a connected CW complex 
$K(\pi,n)$ such that $\pi_n(X)=\pi$ and $\pi_q(X)=0$ for $q\neq n$. (Hint:
start with $M(\pi,n)$, using the Hurewicz theorem, and kill its higher homotopy groups.)
The spaces $K(\pi,n)$ are called Eilenberg-Mac\,Lane spaces.\index{Eilenberg-Mac\,Lane space}
\item There are familiar spaces that give $K(\bZ,1)$, $K(\bZ_2,1)$, and $K(\bZ,2)$. Name them.
\item Let $X$ be any connected CW complex whose only non-vanishing homotopy
group is $\pi_n(X)\iso \pi$. Construct a homotopy equivalence $K(\pi,n)\rtarr X$,
where $K(\pi,n)$ is the Eilenberg-Mac\,Lane space that you have constructed.
\item* For groups $\pi$ and $\rh$,  compute  $[K(\pi,n),K(\rh,n)]$;  here $[-,-]$ 
means based homotopy classes of based maps.
\end{enumerate}

\clearpage

\thispagestyle{empty}

\chapter{Singular homology theory}

We explain, without giving full details, how the standard approach to singular homology 
theory fits into our framework. We also introduce simplicial sets and spaces and their
geometric realization. These notions play a fundamental role in modern algebraic topology.

\section{The singular chain complex}

The standard topological $n$-simplex\index{nsimplex@$n$-simplex!topological} is the subspace
$$\DE_n=\sset{(t_0,\ldots\!,t_n)|0\leq t_i\leq 1,\ \textstyle{\sum} t_i=1}$$
of $\bR^{n+1}$. There are ``face maps''\index{face map} 
$$\de_i: \DE_{n-1}\rtarr \DE_n,\ \ 0\leq i\leq n,$$
specified by
$$\de_i(t_0,\ldots\!,t_{n-1})=(t_0,\ldots\!,t_{i-1},0,t_i,\ldots\!,t_{n-1})$$
and ``degeneracy maps''\index{degeneracy map}
$$\si_i: \DE_{n+1}\rtarr \DE_{n},\ \ 0\leq i\leq n,$$
specified by
$$\si_i(t_0,\ldots\!,t_{n+1})=(t_0,\ldots\!,t_{i-1},t_i+t_{i+1},\ldots\!, t_{n+1}).$$

For a space $X$, define $S_nX$ to be the set of continuous maps $f:\DE_n\rtarr X$.
In particular, regarding a point of $X$ as the map that sends $1$ to $x$, we may 
identify the underlying set of $X$ with $S_0X$. Define the $i$th face operator
$$d_i: S_nX\rtarr S_{n-1}X,\ \ 0\leq i\leq n,$$
by 
$$d_i(f)(u)=f(\de_i(u)),$$
where $u\in \DE_{n-1}$, and define the $i$th degeneracy operator
$$s_i: S_nX\rtarr S_{n+1}X, \ \ 0\leq i\leq n,$$
by 
$$s_i(f)(v)=f(\si_i(v)),$$
where $v\in \DE_{n+1}$. The following identities are easily checked:
$$d_i\com d_j=d_{j-1}\com d_i \ \ \ \text{if}\ \ i<j $$
$$d_i\com s_j=\begin{cases}
s_{j-1}\com d_i  \ \  \ \ \ \text{if}\ \  i<j \\
\id \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if} \ \ i=j\ \ \text{or}\ \ i=j+1\\
s_j\com d_{i-1} \ \ \ \ \ \text{if} \ \ i>j+1.
\end{cases}$$
$$s_i\com s_j=s_{j+1}\com s_i \ \ \text{if}\ \ i\leq j.$$
A map $f: \DE_n\rtarr X$ is called a singular $n$-simplex.\index{nsimplex@$n$-simplex!singular} It 
is said to be nondegenerate
if it is not of the form $s_i(g)$ for any $i$ and $g$. Let $C_n(X)$\index{singular chain
complex} be the free Abelian
group generated by the nondegenerate $n$-simplexes, and think of $C_n(X)$ as the quotient
of the free Abelian group generated by all singular $n$-simplexes by the subgroup generated
by the degenerate $n$-simplexes. Define
$$d = \sum_{i=0}^{n}(-1)^i d_i: C_n(X)\rtarr C_{n-1}(X).$$
The identities ensure that $C_*(X)$ is then a well defined chain complex. In fact,
$$d\com d =\sum_{i=0}^{n-1}\sum _{j=0}^{n}(-1)^{i+j}d_i\com d_j,$$
and, for $i<j$, the $(i,j)$th and $(j-1,i)$th summands add to zero. This gives that $d\com d=0$
before quotienting out the degenerate simplexes, and the degenerate simplexes span a subcomplex.

The singular homology\index{singular homology} of $X$ is usually defined in terms of this chain complex:
$$H_*(X;\pi)=H_*(C_*(X)\otimes\pi).$$

\section{Geometric realization}

One can prove the compatibility of this definition with our definition by checking the axioms and 
quoting the uniqueness of homology. We instead describe how the new definition fits into our 
original definition in terms of CW approximation and cellular chain complexes.  We define a
space $\GA X$, called the ``geometric realization of the total singular complex 
of $X$,''\index{geometric realization} as 
follows. As a set
$$\GA X=\textstyle{\coprod}_{n\geq 0} (S_nX\times \DE_n)/(\sim),$$
where the equivalence relation $\sim$ is generated by
$$(f,\de_iu)\sim(d_i(f),u) \ \ \text{for}\ \ f:\DE_n\rtarr X \ \ \tand\ \ u\in \DE_{n-1}$$
and
$$(f,\si_iv)\sim(s_i(f),v) \ \ \text{for}\ \ f:\DE_n\rtarr X \ \ \tand\ \ v\in \DE_{n+1}.$$
Topologize $\GA X$ by giving 
$$\textstyle{\coprod}_{0\leq n\leq q} (S_nX\times \DE_n)/(\sim)$$ 
the quotient
topology and then giving $\GA X$ the topology of the union. Define $\ga:\GA X\rtarr X$ by
$$\ga|f,u|=f(u)\ \ \text{for}\ \ f:\DE_n\rtarr X \ \ \tand\ \ u\in \DE_{n},$$
where $|f,u|$ denotes the equivalence class of $(f,u)$. Now the following two theorems imply
that that this construction provides a canonical way of realizing our original construction 
of homology.

\begin{thm}
For any space $X$, $\GA X$ is a CW complex with one $n$-cell for each nondegenerate 
singular $n$-simplex, and the cellular chain complex $C_*(\GA X)$ is naturally isomorphic
to the singular chain complex $C_*(X)$.
\end{thm}

\begin{thm}
For any space $X$, the map $\ga: \GA X\rtarr X$ is a weak equivalence.
\end{thm}

Thus the singular chain complex of $X$ is the cellular chain complex of a functorial CW
approximation of $X$, and this shows that our original construction of homology coincides
with the classical construction in terms of singular chains. Each approach has its 
mathematical and pedagogical advantages. 

\section{Proofs of the theorems}

We give a detailed outline of how the required CW decomposition of $\GA X$ is obtained
and sketch the proof that $\ga$ is a weak equivalence.

Let $\bar{X}=\coprod_{n\geq 0}S_nX\times \DE_n$. Define functions 
$$\la: \bar{X}\rtarr \bar{X} \ \ \tand\ \ \rh: \bar{X}\rtarr \bar{X}$$
by 
$$\la(f,u)=(g,\si_{j_1}\cdots\si_{j_p}u)$$
if $f=s_{j_p}\cdots s_{j_1}g$ where $g$ is nondegenerate and $0\leq j_1<\cdots <j_p$ and
$$\rh(f,u)=(d_{i_1}\cdots d_{i_q}f,v)$$
if $u=\de_{i_q}\cdots\de_{i_1}v$ where $v$ is an interior point and $0\leq i_1<\cdots<i_q$.
Note that the unique point of $\DE_0$ is interior. Say that a point $(f,u)$ is nondegenerate
if $f$ is nondegenerate and $u$ is interior. A combinatorial argument from the 
definitions gives the following observation.

\begin{lem} The composite $\la\com\rh$ carries each point of $\bar{X}$ to the unique
nondegenerate point to which it is equivalent.
\end{lem}

Let $(\GA X)^n$ be the image in $\GA X$ of $\coprod_{m\leq n}S_mX\times \DE_m$. Then
$$(\GA X)^n-(\GA X)^{n-1} = 
\sset{\text{nondegenerate}\ n\text{-simplexes}}\times\sset{\DE_n-\pa\DE_n}.$$
This implies that $\GA X$ is a CW complex whose $n$-cells are the maps
$$\tilde{f}:(\DE_n,\pa\DE_n)\rtarr ((\GA X)^n,(\GA X)^{n-1})$$
specified by $\tilde{f}(u)=|f,u|$ for a nondegenerate $n$-simplex $f$. Here we think of 
$(\DE_n,\pa\DE_n)$ as the domains of cells via oriented homeomorphisms 
with $(D^n,S^{n-1})$.

To compute $d$ on the cellular chains $C_*(\GA X)$, we must compute the degrees of the composites
$$S^{n-1}\iso \pa \DE_n\overto{\tilde{f}} (\GA X)^{n-1}/(\GA X)^{n-2}
\overto{\pi_{\tilde{g}}}\DE_{n-1}/\pa\DE_{n-1}\iso S^{n-1}$$
for nondegenerate $n$-simplexes $f$ and $(n-1)$-simplexes $g$. The only relevant $g$ are the
$d_if$ since $f$ traverses these $g$ on the various faces of $\pa \DE_n$. Let 
$\bar{\de}_i:\pa \DE_n\rtarr \DE_{n-1}/\pa\DE_{n-1}$ be the map that collapses all faces 
of $\pa \DE_n$ other than $\de_i\DE_{n-1}$ to the basepoint and is $\de_i^{-1}$ on 
$\de_i\DE_{n-1}$. Then, with $g=d_if$, the composite above reduces to the map
$$S^{n-1}\iso \pa \DE_n\overto{\bar{\de}_i}\DE_{n-1}/\pa\DE_{n-1}\iso S^{n-1}.$$
It is not hard to check that the degree of this map is $(-1)^i$ (provided that we choose
our homeomorphisms sensibly). If $n=2$, the three maps $\de_i$ are given by
$$\de_0(1-t,t)=(0,1-t,t)$$
$$\de_1(1-t,t)=(1-t,0,t)$$
$$\de_2(1-t,t)=(1-t,t,0)$$
and we can visualize the maps $\bar{\de_i}$ as follows:
$$\diagram
 & & & & & & & & & \\
 & & \ddlline_{\de_0}|\tip & & & & & & & \\
 & & \uuto \rrto \ddllto & \ulline_{\de_2}|\tip \dllline^{\de_1}|\tip &  & 
\rto^{\bar{\de}_i} & & \uuto \rrto & \ulline|\tip & \\
 & & & & & & & & & \\
 & & & & & & & & & \\
\enddiagram$$
The alternation of orientations and thus of signs should be clear. This shows that
$C_*(\GA X)=C_*(X)$, as claimed.

We must still explain why $\ga:\GA X\rtarr X$ is a weak equivalence. In fact, it is 
tautologically obvious that $\ga$ induces an epimorphism on all homotopy groups: a map 
of pairs
$$f:(\DE_n,\pa\DE_n)\rtarr (X,x)$$
determines the map of pairs 
$$\tilde{f}: (\DE_n,\pa\DE_n)\rtarr (\GA X,|x,1|)$$
specified by $\tilde{f}(u)=|f,u|$, and $\ga\com\tilde{f}=f$. Injectivity is more delicate,
and we shall only give a sketch. Given a map $g:(\DE_n,\pa\DE_n)\rtarr (\GA X,|x,1|)$, we
may first apply cellular approximation to obtain a homotopy of $g$ with a cellular map 
and we may then subdivide the domain and apply a further homotopy so as to obtain
a map $g'\htp g$ such that $g'$ is simplicial, in the sense that $g'\com e$ is a cell of 
$\GA X$ for every cell $e$ of the subdivision of $\DE_n$. Suppose that $\ga\com g$ and 
thus $\ga\com g'$ is homotopic to the constant map $c_x$ at the point $x$. We may view a
homotopy $h:\ga\com g'\htp c_x$ as a map 
$$h: (\DE_n\times I, \pa\DE_n\times I\cup \DE_n\times\sset{1})\rtarr (X,x).$$
We can simplicially subdivide $\DE_n\times I$ so finely that our subdivided 
$\DE_n=\DE_n\times\sset{0}$ is a subcomplex. We can then lift $h$ simplex by simplex
to a simplicial map 
$$\tilde{h}: (\DE_n\times I, \pa\DE_n\times I\cup \DE_n\times\sset{1})\rtarr (\GA X,|x,1|)$$
such that $\tilde{h}$ restricts to $\tilde{g'}$ on $\DE_n\times\sset{0}$ and $\ga\com\tilde{h}=h$.

\section{Simplicial objects in algebraic topology}

A simplicial set\index{simplicial set} $K_*$ is a sequence of sets $K_n$, $n\geq 0$, connected 
by face and degeneracy
operators $d_i: K_n\rtarr K_{n-1}$ and $s_i:K_n\rtarr K_{n+1}$, $0\leq i\leq n$, that satisfy
the commutation relations that we displayed for the total singular complex $S_*X=\sset{S_nX}$ 
of a space $X$. Thus $S_*$ is a functor from spaces to simplicial sets. 

We may define the geometric realization\index{geometric realization} $|K_*|$ of general simplicial 
sets exactly as we defined
the geometric realization $\GA X=|S_*X|$ of the total singular complex of a topological space.
In fact, the total singular complex and geometric realization functors are adjoint. If $\sS\sS$
is the category of simplicial sets and $\sU$ the category of spaces, then
$$\sU(|K_*|,X)\iso \sS\sS(K_*,S_*X).$$
The identity map of $S_*X$ on the right corresponds to $\ga: |S_*X|\rtarr X$ on the left. In
general, for a map $f_*: K_*\rtarr S_*X$ of simplicial sets, the corresponding map of spaces
is the composite
$$ |K_*|\overto{|f_*|} |S_*X| \overto{\ga} X.$$

In fact, one can develop homotopy theory and homology in the category of simplicial sets in a
fashion parallel to and, in a suitable sense, equivalent to the development that we have here
given for topological spaces. For example, we have the chain complex $C_*(K_*)$ defined exactly
as we defined the singular chain complex, using the alternating sum of the face maps, and there
result homology groups 
$$H_*(K_*;\pi)=H_*(C_*(K_*)\otimes\pi).$$
Exactly as in the case of $S_*X$, $|K_*|$ is a CW complex and $C_*(K_*)$ is naturally isomorphic
to the cellular chain complex $C_*(|K_*|)$. Singular homology is the special case obtained by 
taking $K_*=S_*X$ for spaces $X$. The passage back and forth between simplicial sets and 
topological spaces plays a major role in many applications.

The ideas generalize. One can define a simplicial object\index{simplicial object} in any 
category $\sC$ as a sequence of
objects $K_n$ of $\sC$ connected by face and degeneracy maps\index{face map}\index{degeneracy
map} in $\sC$ that satisfy the commutation
relations that we have displayed. Thus we have simplicial groups, simplicial Abelian groups,
simplicial spaces, and so forth. We can think of simplicial sets as discrete simplicial spaces,
and we then see that geometric realization generalizes directly to a functor $|-|$ from the
category $\sS\sU$ of simplicial spaces to the category $\sU$ of spaces. This provides a very
useful way of constructing spaces with desirable properties.

We note one of the principal features of geometric realization. Define the product $X_*\times Y_*$
of simplicial spaces $X_*$ and $Y_*$ to be the simplicial space whose space of $n$-simplexes is
$X_n\times Y_n$, with faces and degeneracies $d_i\times d_i$ and $s_i\times s_i$. The projections
induce maps of simplicial spaces from $X_*\times Y_*$ to $X_*$ and $Y_*$. On passage to geometric
realization, these give the coordinates of a map
$$|X_*\times Y_*|\rtarr |X_*|\times |Y_*|.$$
It turns out that this map is always a homeomorphism. 

Now restrict attention to simplicial sets $K_*$ and $L_*$.  Then the homeomorphism just 
specified is a map between CW complexes. However, it is not a cellular map; rather, it 
takes the $n$-skeleton of $|K_*\times L_*|$ to the $2n$-skeleton of $|K_*|\times |L_*|$. 
It is homotopic to a cellular map, no longer a homeomorphism, and
there results a chain homotopy equivalence
$$C_*(|K_*\times L_*|)\rtarr C_*(|K_*|)\otimes C_*(|L_*|).$$
In particular, for spaces $X$ and $Y$, there is a natural chain homotopy equivalence
from the singular chain complex $C_*(X\times Y)$ to the tensor product $C_*(X)\otimes C_*(Y)$.
One can be explicit about this but, pedagogically, one technical advantage of approaching
homology via CW complexes is that it leaves us free to work directly with the natural cell
structures on Cartesian products of CW complexes and to postpone the introduction of 
chain homotopy equivalences such as these to a later stage of the development.

\section{Classifying spaces and $K(\pi,n)$s}

We illustrate these ideas by defining the ``classifying spaces'' and ``universal bundles'' associated to
topological groups $G$ and describing how this leads to a beautiful conceptual construction of the
Eilenberg-Mac\,Lane spaces $K(\pi,n)$ associated to discrete Abelian groups $\pi$. Recall that
these are spaces such that\index{Eilenberg-Mac\,Lane space}
$$\pi_q(K(\pi,n))=\begin{cases}
\pi \ \ \text{if} \ q=n \\
0 \ \ \text{if}\ q\neq n.
\end{cases}$$

We define a map $p_*: E_*(G)\rtarr B_*(G)$ of simplicial topological
spaces. Let $E_n(G)=G^{n+1}$ and $B_n(G)=G^n$, and let $p_n: G^{n+1}\rtarr G^n$ be the
projection on the first $n$ coordinates. The faces and degeneracies are defined on $E_n(G)$ by
$$d_i(g_1,\ldots\!,g_{n+1})=\begin{cases}
(g_2,\ldots\!,g_{n+1}) \ \ \ \ \ \text{if}\ \ i=0 \\
(g_1,\ldots\!,g_{i-1},g_ig_{i+1},g_{i+2},\ldots\!,g_{n+1}) \ \ \text{if} \ \ 1\leq i\leq n 
\end{cases}$$
and
$$s_i(g_1,\ldots\!,g_{n+1})=(g_1,\ldots\!,g_{i-1},e,g_i,\ldots\!,g_{n+1}) \ \ \text{if}\ \ 0\leq i\leq n.$$
The faces and degeneracies on $B_n(G)$ are defined in the same way, except that the last coordinate
$g_{n+1}$ is omitted and the last face operation $d_n$ takes the form
$$d_n(g_1,\ldots\!,g_{n})=(g_1,\ldots\!,g_{n-1}).$$
Certainly $p_*$ is a map of simplicial spaces. If we let $G$ act from the right on $E_n(G)$ by 
multiplication on the last coordinate,
$$(g_1,\ldots\!,g_n, g_{n+1})g=(g_1,\ldots\!,g_n,g_{n+1}g),$$
then $E_*(G)$ is a simplicial $G$-space. That is, the action of $G$ commutes with the face and
degeneracy maps. We may view $B_n(G)$ as the orbit space $E_n(G)/G$. We define
$$E(G)=|E_*(G)|, \ \ B(G) = |B_*(G)|, \ \ \tand\ \ p=|p_*(G)|: E(G)\rtarr B(G).$$
Then $E(G)$ inherits a free right action by $G$, and $B(G)$ is the orbit space $E(G)/G$. The
space $BG$ is called the classifying space of $G$.\index{classifying space}

The space $E(G)$ is the union of the images $E(G)^n$ of the spaces
$\coprod_{m\leq n}G^{m+1}\times\DE_m$, and
$$E(G)^{n}-E(G)^{n-1} = (G^n-W)\times G\times (\DE_n-\pa\DE_n),$$
where $W\subset G^n$ is the ``fat wedge'' consisting of those points at least one of whose
coordinates is the identity element $e$. Similarly, we have subspaces $B(G)^n$ such that
$$B(G)^{n}-B(G)^{n-1} = (G^n-W)\times (\DE_n-\pa\DE_n).$$
The map $p$ restricts to the projection between these subspaces. Intuitively, it looks as 
if $p$ should be a bundle with fiber $G$, and this is indeed the case if the identity element
of $G$ is a nondegenerate basepoint. This condition is enough to ensure local triviality as
we glue together over the  filtration $\sset{B(G)^n}$. It is less intuitive, but true, that the
space $E(G)$ is contractible. By the long exact homotopy sequence, these facts imply that
$$\pi_{q+1}(BG)\iso \pi_q(G)$$
for all $q\geq 0$. 

For topological groups $G$ and $H$, the obvious shuffle homeomorphisms 
$$(G\times H)^n\iso G^n\times H^n$$
specify isomorphisms of simplicial spaces 
$$E_*(G\times H)\iso E_*(G)\times E_*(H) \ \ \tand \ \ B_*(G\times H)\iso B_*(G)\times B_*(H)$$
that are compatible with the projections. Since geometric realization commutes with products,
we conclude that $B(G\times H)$ is homeomorphic to $B(G)\times B(H)$. Thus $B$ is a 
product-preserving functor on the category of topological groups. 

Now suppose that $G$ is a commutative topological group.\index{topological group!commutative} Then 
its multiplication $G\times G\rtarr G$
and inverse map $G\rtarr G$ are homomorphisms. We conclude that $B(G)$ and $E(G)$ are again 
commutative topological groups. The multiplication on $B(G)$ is determined by the multiplication 
on $G$ as the composite
$$B(G)\times B(G)\iso B(G\times G)\rtarr B(G).$$
Moreover, the map $p: E(G)\rtarr B(G)$ and the inclusion of $G$ in $E(G)$ as the fiber over the 
basepoint (the unique point in $B_0(G)$) are homomorphisms. This allows us to iterate the 
construction, setting $B^0(G)=G$ and $B^n(G)=B(B^{n-1}(G))$ for $n\geq 1$. Specializing to a 
discrete Abelian group $\pi$, we define 
$$K(\pi,n)=B^n(\pi).$$
As promised, we have 
$$\pi_q(K(\pi,n))=\pi_{q-1}(K(\pi,n-1))=\cdots=\pi_{q-n}(K(\pi,0))=\begin{cases}
\pi \ \ \text{if} \ q=n \\
0 \ \ \text{if}\ q\neq n.
\end{cases}$$

\begin{center}
PROBLEMS
\end{center}
\begin{enumerate}
\item  Let $X$ be a space that satisfies the hypotheses used to construct a universal 
cover $\tilde{X}$. Let $\pi=\pi_1(X)$ and consider the action of the group $\pi$
on the space $\tilde{X}$ given by the isomorphism of $\pi$ with Aut$(\tilde{X})$.
Let $A$ be an Abelian group and let $\bZ[\pi]$ act trivially on $A$, 
$a\cdot\si = a$ for $\si\in\pi$ and $a\in A$. Do one or both of the following, 
and convince yourself that the other choice also works. 
\begin{enumerate}
\item[(a)] [Cellular chains] Assume that $X$ is a CW complex. Show that $\tilde{X}$ is a CW complex 
such that the action of $\pi$ on $\tilde{X}$ induces an action of the group ring $\bZ[\pi]$ 
on the cellular chain complex $C_*(\tilde{X})$ such that each $C_q(\tilde{X})$ is a free 
$\bZ[\pi]$-module and $$C_*(X;A)\iso A\ten_{\bZ[\pi]}C_*(\tilde X).$$
\item[(b)] [Singular chains] Show that the action of $\pi$ on $\tilde{X}$ induces an action of 
$\bZ[\pi]$ on the singular chain complex $C_*(\tilde{X})$ such that each $C_q(\tilde{X})$ is
a free $\bZ[\pi]$-module and $$C_*(X;A)\iso A\ten_{\bZ[\pi]}C_*(\tilde X).$$
\end{enumerate}
\item Let $\pi$ be a group and let $K(\pi,1)$ be a connected CW complex such that
$\pi_1(K(\pi,1))=\pi$ and $\pi_q(K(\pi,1))=0$ for $q\neq 1$. Use Problem 1 to show that 
there is an isomorphism 
$$H_*(K(\pi,1);A)\iso \Tor_*^{\bZ[\pi]}(A,\bZ).$$
\item Let $p: Y\rtarr X$ be a covering space with finite fibers, say of cardinality $n$.
Using singular chains, construct a homomorphism $t: H_*(X;A)\rtarr H_*(Y;A)$ such
that the composite $p_*\com t: H_*(X;A)\rtarr H_*(X;A)$ is multiplication by $n$;
$t$ is called a ``transfer\index{transfer homomorphism} homomorphism.''
\end{enumerate}

\clearpage

\thispagestyle{empty}

\chapter{Some more homological algebra}

The reader will by now appreciate that the calculation of homology groups, although far
simpler than the calculation of homotopy groups, can still be a difficult task. In
practice, one seldom uses chains explicitly; rather, one uses them to prove algebraic
theorems that simplify topological calculations. Indeed, if one focuses on singular 
chains, then one eschews chain level computations in principle as well as in practice.

We here recall some classical results in homological algebra that explain how to calculate
$H_*(X;\pi)$ from $H_*(X)\equiv H_*(X;\bZ)$ and how to calculate $H_*(X\times Y)$ 
from $H_*(X)\ten H_*(Y)$. We then say a little about cochain complexes in
preparation for the definition of cohomology groups.

We again work over a general commutative ring $R$, although the main example will be $R=\bZ$. 
Tensor products are understood to be taken over $R$.

\section{Universal coefficients in homology}

Let $X$ and $Y$ be chain complexes over $R$. We think of $H_*(X)\otimes H_*(Y)$ as a 
graded $R$-module which, in degree $n$, is $\sum_{p+q=n}H_p(X)\otimes H_q(Y)$. We
define
$$\al: H_*(X)\otimes H_*(Y)\rtarr H_*(X\otimes Y)$$
by $\al([x]\otimes[y])=[x\otimes y]$ for cycles $x$ and $y$ that represent homology
classes $[x]$ and $[y]$. As a special case, for an $R$-module $M$ we have
$$\al: H_*(X)\otimes M\rtarr H_*(X\otimes M).$$

We omit the proof of the following standard result, but we shall shortly give the
quite similar proof of a cohomological analogue. Recall that an $R$-module $M$ is
said to be flat if the functor $M\otimes N$ is exact (that is, preserves exact sequences
in the variable $N$). We say that a graded $R$-module is flat if each of its terms is flat.

We assume that the reader has seen torsion products, which measure the failure of tensor 
products to be exact functors. For a principal ideal domain (PID) $R$, the only torsion 
product\index{torsion product}\index{Tor}
is the first one, denoted $\Tor_1^R(M,N)$. It can be computed by constructing a short exact
sequence 
$$0\rtarr F_1\rtarr F_0 \rtarr M \rtarr 0$$
and tensoring with $N$ to obtain an exact seqence
$$0 \rtarr \Tor^R_1(M,N)\rtarr F_1\ten N\rtarr F_0\ten N\rtarr M\ten N\rtarr 0,$$
where $F_1$ and $F_0$ are free $R$-modules. That is, we choose an epimorphism 
$F_0\rtarr M$ and note that, since $R$ is a PID, its kernel $F_1$ is also free.

\begin{thm}[Universal coefficient]\index{universal coefficient theorem} Let $R$ be a 
$PID$ and let $X$ be a flat chain complex over $R$. Then, for each $n$, 
there is a natural short exact sequence
$$ 0\rtarr H_n(X)\otimes M\overto{\al} H_n(X\otimes M)\overto{\be} \Tor^R_1(H_{n-1}(X),M)\rtarr 0.$$
The sequence splits, so that 
$$H_n(X\otimes M)\iso (H_n(X)\otimes M)\oplus \Tor^R_1(H_{n-1}(X),M),$$
but the splitting is not natural. 
\end{thm}

In Chapter 20 \S3, we shall see an important class of examples in which the splitting is very far from
being natural. 

\begin{cor}
If $R$ is a field, then
$$\al: H_*(X)\otimes M\rtarr H_*(X;M)$$
is a natural isomorphism. 
\end{cor}

\section{The K\"{u}nneth theorem}

The universal coefficient theorem in homology is a special case of the  K\"{u}nneth theorem.

\begin{thm}[K\"{u}nneth]\index{Kunneth theorem@K\"unneth theorem}
Let $R$ be a $PID$ and let $X$ be a flat chain complex and $Y$ be any
chain complex. Then, for each $n$, there is a natural short exact sequence
$$0\rtarr \sum_{p+q=n}H_p(X)\ten H_q(Y)\overto{\al} H_n(X\otimes Y)\overto{\be}
\sum_{p+q=n-1}\Tor^R_1(H_p(X),H_q(Y))\rtarr 0.$$
The sequence splits, so that 
$$H_n(X\otimes Y)\iso (\sum_{p+q=n}H_p(X)\ten H_q(Y))
\oplus (\sum_{p+q=n-1}\Tor^R_1(H_p(X),H_q(Y))),$$
but the splitting is not natural. 
\end{thm}

Returning to topology for a moment, observe that this applies directly to the 
computation of the homology of the Cartesian product of CW complexes $X$ and $Y$ 
in view of the isomorphism

$$C_*(X\times Y) \iso C_*(X)\ten C_*(Y).$$

\begin{cor}
If $R$ is a field, then
$$\al: H_*(X)\otimes H_*(Y) \rtarr H_*(X\otimes Y)$$
is a natural isomorphism. 
\end{cor}

We prove the corollary to give the idea. The general case is proved by an elaboration
of the argument. In fact, in practice, algebraic topologists carry out the vast majority 
of their calculations using a field of coefficients, and
it is then the corollary that is relevant to the study of the homology of Cartesian
products. There is a simple but important technical point to make about this. Let us 
for the moment remember to indicate the ring over which we are taking tensor products.
For chain complexes $X$ and $Y$ over $\bZ$, we have
$$(X\ten_{\bZ}R)\ten_R (Y\ten_{\bZ}R)\iso (X\ten_{\bZ}Y)\ten_{\bZ}R.$$
We can therefore use the corollary to compute $H_*(X\otimes_{\bZ}Y;R)$ from $H_*(X;R)$ and 
$H_*(Y;R)$. 

\begin{proof}[Proof of the corollary] Assume first that $X_i=0$ for $i\neq p$, so that 
$X=X_p$ is just an $R$-module with no differential. The square commutes and the row 
and column are exact in the diagram
$$\diagram
& & 0 \dto & & \\
0 \rto & B_q(Y) \rto & Z_q(Y) \dto \rto & H_q(Y) \rto & 0. \\
 & Y_{q+1} \rto_{d_{q+1}} \uto^{d_{q+1}} & Y_q \dto^{d_q} & &  \\
 & & Y_{q-1} & & \\
\enddiagram$$
Since all modules over a field are free and thus flat, this remains true when we tensor 
the diagram with $X_p$. This proves that if $n=p+q$, then 
$$Z_n(X_p\ten Y) = X_p\ten Z_q(Y), \ \ B_n(X_p\ten Y)=X_p\ten B_q(Y),$$
and therefore
$$H_n(X\ten Y) = X_p\ten H_q(Y).$$
In the general case, regard the graded modules $Z(X)$ and $B(X)$ as chain complexes with 
zero differential. The exact sequences 
$$ 0\rtarr Z_p(X)\rtarr X_p\overto{d_p} B_{p-1}(X) \rtarr 0 $$
of $R$-modules define a short exact seqence of chain complexes since $d_{p-1}\com d_p=0$.
Define the suspension of a graded $R$-module $N$ by $(\SI N)_{n+1}= N_n$.
Tensoring with $Y$, we obtain a short exact sequence of chain complexes
$$ 0\rtarr Z(X)\ten Y \rtarr X \ten Y \rtarr \SI B(X)\ten Y \rtarr 0. $$
It follows from the first part and additivity that 
$$H_*(Z(X)\ten Y) =Z(X)\ten H_*(Y) \ \ \tand \ \ H_*(\SI B(X)\ten Y) =\SI B(X)\ten H_*(Y).$$
Moreover, by inspection of definitions, the connecting homomorphism of the long exact
sequence of homology modules associated to our short exact sequence of chain
complexes is just the inclusion $B\ten H_*(Y)\rtarr Z\ten H_*(Y)$. In particular,
the long exact sequence breaks up into short exact sequences
$$ 0 \rtarr B(X)\ten H_*(Y) \rtarr Z(X)\ten H_*(Y) \rtarr H_*(X\ten Y)\rtarr 0.$$
However, since tensoring with $H_*(Y)$ is an exact functor, the cokernel of the inclusion   
$B\ten H_*(Y)\rtarr Z\ten H_*(Y)$ is $H_*(X)\ten H_*(Y)$. The conclusion follows.
\end{proof}

\section{Hom functors and universal coefficients in cohomology}

For a chain complex $X=X_*$, we define the dual cochain complex\index{cochain complex!dual} $X^*$ by 
setting
$$ X^q=\Hom (X_q,R) \ \ \tand \ \ d^q=(-1)^q \Hom (d_{q+1},\id).$$
As with tensor products, we understand Hom\index{Hom} to mean $\Hom_R$ when $R$ is clear from 
the context. On elements, for an $R$-map $f: X_q\rtarr R$ and an element $x\in X_{q+1}$,
$$(d^qf)(x) = (-1)^qf(d_q(x)).$$
More generally, for an $R$-module $M$, we define a cochain complex $\Hom(X,M)$ in the same 
way. The sign is conventional and is designed to facilitate the definition of $\Hom(X,Y)$ for 
a chain complex $X$ and cochain complex $Y$; however, we shall not have occasion to use 
the latter definition. 

In analogy with the notation $H_*(X;M)= H_*(X\ten M)$, we write 
$$H^*(X;M)=H^*(\Hom(X,M)).$$
We have a cohomological version of the universal coefficient theorem.
We assume that the reader has seen Ext modules,\index{Ext} which measure the failure of Hom
to be an exact functor. For a PID $R$, the only Ext module
is the first one, denoted $\Ext^1_R(M,N)$. It can be computed by constructing a short exact
sequence 
$$0\rtarr F_1\rtarr F_0 \rtarr M \rtarr 0$$
and applying Hom to obtain an exact seqence
$$0 \rtarr \Hom(M,N)\rtarr \Hom(F_0,N)\rtarr \Hom(F_1,N)\rtarr \Ext^1_R(M,N) \rtarr 0,$$
where $F_1$ and $F_0$ are free $R$-modules.

For each $n$, define
$$\al: H^n(\Hom(X,M))\rtarr \Hom(H_n(X),M)$$
by letting $\al[f]([x])= f(x)$
for a cohomology class $[f]$ represented by a ``cocycle'' $f: X_n\rtarr M$ and a homology class
$[x]$ represented by a cycle $x$. It is easy to check that $f(x)$ is independent of the 
choices of $f$ and $x$ since $x$ is a cycle and $f$ is a cocycle.

\begin{thm}[Universal coefficient]\index{universal coefficient theorem}
Let $R$ be a $PID$ and let $X$ be a free chain complex over $R$. Then, for each $n$, 
there is a natural short exact sequence
$$ 0\rtarr \Ext_R^1(H_{n-1}(X),M) \overto{\be} H^n(X;M)\overto{\al} \Hom(H_n(X),M)\rtarr 0.$$
The sequence splits, so that 
$$H^n(X; M)\iso \Hom(H_n(X),M) \oplus \Ext_R^1(H_{n-1}(X),M),$$
but the splitting is not natural. 
\end{thm}

\begin{cor}
If $R$ is a field, then
$$\al: H^*(X;M)\rtarr \Hom(H_*(X),M)$$
is a natural isomorphism. 
\end{cor}

Again, there is a technical point to be made here. If $X$ is a complex of free Abelian groups
and $M$ is an $R$-module, such as $R$ itself, then
$$ \Hom_{\bZ}(X,M)\iso \Hom_R(X\ten_{\bZ}R,M).$$
One way to see this is to observe that, if $B$ is a basis for a free Abelian group $F$, then
$\Hom_{\bZ}(F,M)$ and $\Hom_R(F\ten_{\bZ}R,M)$ are both in canonical bijective correspondence
with maps of sets $B\rtarr M$. More algebraically, a homomorphism $f: F\rtarr M$ of Abelian
groups determines the corresponding map of $R$-modules as the composite of $f\ten\id$ and the
action of $R$ on $M$:
$$ F\ten_{\bZ} R\rtarr M\ten_{\bZ} R\rtarr M.$$

\section{Proof of the universal coefficient theorem}

We need two properties of $\Ext$ in the proof. First, $\Ext^1_R(F,M)=0$ for a free $R$-module $F$.
Second, when $R$ is a PID, a short exact sequence 
$$0\rtarr L'\rtarr L\rtarr L''\rtarr 0$$
of $R$-modules gives rise to a six-term exact sequence
\begin{eqnarray*}
\lefteqn{0\rtarr \Hom(L'',M)\rtarr \Hom(L,M)\rtarr \Hom(L',M)} \hspace{.8in} \\
& \overto{\de} \Ext^1_R(L'',M)\rtarr \Ext^1_R(L,M)\rtarr \Ext^1_R(L',M)\rtarr 0.
\end{eqnarray*}

\begin{proof}[Proof of the universal coefficient theorem] We write
$B_n=B_n(X)$,
$Z_n=Z_n(X)$, and $H_n=H_n(X)$ to abbreviate notation. Since each 
$X_n$ is a free $R$-module and $R$ is a PID, 
each $B_n$ and $Z_n$ is also free. We have short exact sequences
$$\diagram
0\rto & B_n \rto^{i_n} & Z_n\rto^{\pi_n} & H_n\rto & 0 \\
\enddiagram$$
and 
$$\diagram
0\rto & Z_n \rto^{j_n} & X_n \rto<.5ex>^{d_n} & B_{n-1} \ldashed<.5ex>^{\si_n}|>\tip \rto & 0; \\
\enddiagram$$
we choose a splitting $\si_n$ of the second. Writing $f^*=\Hom(f,M)$ consistently, we obtain a 
commutative diagram with exact rows and columns
$$\diagram
 & & 0 & 0 \dto & \\
0 \rto & \Hom(H_n,M) \rto^{\pi_n^*} & \Hom(Z_n,M) \rto^{i_n^*} \uto 
& \Hom(B_n,M) \dto^{d_{n+1}^*} & \\
\cdots \rto & \Hom(X_{n-1},M) \rto^{d_n^*} \dto_{j_{n-1}^*} & \Hom(X_n,M) \uto_{j_n^*}
\rto^{d_{n+1}^*} \ddashed<-.5ex>_{\si_n^*}|>\tip & \Hom(X_{n+1},M) \rto & \cdots \\
 & \Hom(Z_{n-1},M) \rto_{i_{n-1}^*} \dto
& \Hom(B_{n-1},M) \rto^{\de} \uto<-.5ex>_{d_n^*} \urdashed_{0}|>\tip & \Ext^1_R(H_{n-1},M)\rto & 0 \\
 & 0 & 0 \uto & & \\
\enddiagram$$
By inspection of the diagram, we see that the canonical map $\al$ coincides with the composite
$$H^n(X;M)=\ker\,d^*_{n+1}/\im\,d^*_n = \ker\,i^*_nj^*_n/\im\,d_n^*i^*_{n-1} \overto{j_n^*}
 \im \pi^*_n\,\overto{(\pi_n^*)^{-1}} \Hom(H_n,M).$$
Since $j_n^*$ is an epimorphism, so is $\al$. The kernel of $\al$ is
$\im\,d_n^*/\im\,d_n^*i^*_{n-1}$, and $\de(d_n^*)^{-1}$ maps this group isomorphically onto
$\Ext^1_R(H_{n-1},M)$. The composite $\de\si_n^*$ induces the required splitting.
\end{proof}

\section{Relations between $\otimes$ and Hom}

We shall need some observations about cochain complexes and tensor products, and
we first recall some general facts about the category of $R$-modules. For $R$-modules $L$, $M$, 
and $N$, we have an adjunction
$$\Hom(L\ten M,N)\iso \Hom(L,\Hom(M,N)).$$
We also have a natural homomorphism
$$\Hom(L,M)\ten N\rtarr \Hom(L,M\ten N),$$
and this is an isomorphism if either $L$ or $N$ is a finitely generated projective $R$-module.
Again, we have a natural map
$$\Hom(L,M)\ten \Hom(L',M')\rtarr \Hom(L\ten L',M\ten M'),$$
which is an isomorphism if $L$ and $L'$ are finitely generated and projective or if $L$ is
finitely generated and projective and $M=R$. 

We can replace $L$ and $L'$ by chain complexes and obtain similar maps, inserting signs
where needed. For example, a chain homotopy $X\ten \sI\rtarr X'$ between chain maps 
$f,g: X\rtarr X'$ induces a chain map
$$\Hom(X',M)\rtarr \Hom (X\ten \sI,M)\iso \Hom(\sI,\Hom(X,M))\iso \Hom(X,M)\ten\sI^*,$$
where $\sI^*=\Hom(\sI,R)$. It should be clear that this implies that our original chain
homotopy induces a homotopy of cochain maps 
$$f^*\htp g^*: \Hom(X',M)\rtarr \Hom(X,M).$$

If $Y$ and $Y'$ are cochain complexes, then we have the natural homomorphism
$$\al: H^*(Y)\ten H^*(Y')\rtarr H^*(Y\ten Y')$$
given by $\al([y]\ten [y'])=[y\ten y']$, exactly as for chain complexes. (In fact, by
regrading, we may view this as a special case of the map for chain complexes.) The K\"{u}nneth
theorem applies to this map. For its flatness hypothesis, it is useful to remember that, for
any Noetherian ring $R$, the dual $\Hom(F,R)$ of a free $R$-module is a flat $R$-module. 

As indicated above, if $Y=\Hom(X,M)$ and $Y'=\Hom(X',M')$ for chain complexes $X$ and $X'$ 
and $R$-modules $M$ and $M'$, then we also have the map of cochain complexes
$$\om: \Hom(X,M)\ten \Hom(X',M')\rtarr \Hom(X\ten X',M\ten M')$$
specified by the formula
$$\om(f\ten f')(x\ten x') = (-1)^{(\text{deg}\,f')(\text{deg}\, x)}f(x)\ten f'(x').$$
We continue to write $\om$ for the map it induces on cohomology, and we then have the composite
$$\om\com\al: H^*(X;M)\ten H^*(X';M')\rtarr H^*(X\ten X';M\ten M').$$
When $M=M'=A$ is a commutative $R$-algebra, we may compose with the map 
$$ H^*(X\ten X';A\ten A) \rtarr H^*(X\ten X';A)$$
induced by the multiplication of $A$ to obtain a map
$$H^*(X;A)\ten H^*(X';A)\rtarr H^*(X\ten X';A).$$
We are especially interested in the case when $R=\bZ$ and $A$ is either $\bZ$ or a field.

\chapter{Axiomatic and cellular cohomology theory}

We give a treatment of cohomology that is precisely parallel to our treatment of
homology. The essential new feature is the cup product structure that makes the
cohomology of $X$ with coefficients in a commutative ring $R$ a commutative graded 
$R$-algebra. This additional structure ties together the cohomology groups in 
different degrees and is fundamentally important to most of the applications.  

\section{Axioms for cohomology}

Fix an Abelian group $\pi$ and consider pairs of spaces $(X,A)$. We
shall see that $\pi$ determines a ``cohomology theory on pairs $(X,A)$.''\index{cohomology
theory}

\begin{thm} For integers $q$, there exist {\em contravariant} functors $H^q(X,A;\pi)$ 
from the homotopy category of pairs of spaces to the category of Abelian groups together 
with natural transformations $\de: H^q(A;\pi)\rtarr H^{q+1}(X,A;\pi)$, where
$H^q(X;\pi)$ is defined to be $H^q(X,\emptyset;\pi)$. These functors and natural
transformations satisfy and are characterized by the following axioms.
\begin{itemize}
\item DIMENSION\index{dimension axiom}\ \  If $X$ is a point, then $H^0(X;\pi) = \pi$ and $H^q(X;\pi)=0$
for all other integers.
\item EXACTNESS\index{exactness axiom}\ \ The following sequence is exact, where the unlabeled arrows
are induced by the inclusions $A\rtarr X$ and $(X,\emptyset)\rtarr (X,A)$:
$$\cdots\rtarr H^q(X,A;\pi)\rtarr H^q(X;\pi)\rtarr 
H^q(A;\pi)\overto{\de}  H^{q+1}(X,A;\pi)\rtarr \cdots .$$
\item EXCISION\index{excision axiom}\ \ 
If $(X;A,B)$ is an excisive triad, so that $X$ is the union of the interiors
of $A$ and $B$, then the inclusion $(A,A\cap B)\rtarr (X,B)$ induces an
isomorphism
$$H^*(X,B;\pi)\rtarr H^*(A,A\cap B;\pi).$$
\item ADDITIVITY\index{additivity axiom}\ \ 
If $(X,A)$ is the disjoint union of a set of pairs $(X_i,A_i)$, then
the inclusions $(X_i,A_i)\rtarr (X,A)$ induce an isomorphism
$$ H^*(X,A;\pi)\rtarr \textstyle{\prod}_i\, H^*(X_i,A_i;\pi).$$
\item WEAK EQUIVALENCE\index{weak equivalence axiom}\ \  If $f:(X,A)\rtarr (Y,B)$ is a weak 
equivalence, then
$$f^*: H^*(Y,B;\pi)\rtarr H^*(X,A;\pi)$$
is an isomorphism.
\end{itemize}
\end{thm}

We write $f^*$ instead of $H^*(f)$ or $H^q(f)$. As in homology, our approximation theorems 
for spaces, pairs, maps, homotopies, and excisive triads directly imply that such a theory 
determines and is determined by an appropriate theory defined on CW pairs, as spelled out 
in the following CW version of the theorem.\index{cohomology theory}

\begin{thm} For integers $q$, there exist functors $H^q(X,A;\pi)$ from the 
homotopy category of pairs of CW complexes to the category of 
Abelian groups together with natural transformations $\de: H^q(A)\rtarr H^{q+1}(X,A;\pi)$, 
where $H^q(X;\pi)$ is defined to be $H^q(X,\emptyset;\pi)$. These functors and natural
transformations satisfy and are characterized by the following axioms.
\begin{itemize}
\item DIMENSION\index{dimension axiom}\ \  If $X$ is a point, then $H^0(X;\pi) = \pi$ and $H^q(X;\pi)=0$
for all other integers.
\item EXACTNESS\index{exactness axiom}\ \  The following sequence is exact, where the unlabeled arrows
are induced by the inclusions $A\rtarr X$ and $(X,\emptyset)\rtarr (X,A)$:
$$\cdots\rtarr H^q(X,A;\pi)\rtarr H^q(X;\pi)\rtarr 
H^q(A;\pi)\overto{\de}  H^{q+1}(X,A;\pi)\rtarr \cdots .$$
\item EXCISION\index{excision axiom}\ \ 
If $X$ is the union of subcomplexes $A$ and $B$, then the inclusion 
$(A,A\cap B)\rtarr (X,B)$ induces an isomorphism
$$ H^*(X,B;\pi) \rtarr H^*(A,A\cap B;\pi).$$
\item ADDITIVITY\index{additivity axiom}\ \ 
If $(X,A)$ is the disjoint union of a set of pairs $(X_i,A_i)$, then
the inclusions $(X_i,A_i)\rtarr (X,A)$ induce an isomorphism
$$H^*(X,A;\pi)\rtarr \textstyle{\prod}_i\, H^*(X_i,A_i;\pi).$$
\end{itemize}
Such a theory determines and is determined by a theory as in the previous 
theorem.
\end{thm}

\section{Cellular and singular cohomology}

We define the cellular cochains\index{cellular cochains} of a CW pair $(X,A)$ with 
coefficients in an Abelian group $\pi$ to be
$$C^*(X,A;\pi)=\Hom(C_*(X,A),\pi).$$ 
We then define the cellular cohomology groups to be
$$H^*(X,A;\pi)=H^*(C^*(X,A;\pi)).$$
If $M$ is a module over a commutative ring $R$, we have a natural identification
$$C^*(X,A;M)\iso \Hom_R(C_*(X,A)\ten R,M)$$
which allows us to do homological algebra over $R$ rather than over $\bZ$ when
convenient. In particular, if $R$ is a field, then
$$ H^*(X,A;M)\iso \Hom_R(H_*(X,A;R),M).$$
In general, with $R=\bZ$, we have a natural and splittable short exact sequence
$$0\rtarr \Ext^1_{\bZ}(H_{n-1}(X,A),\pi)\rtarr  H^n(X,A;\pi)\rtarr \Hom(H_n(X,A),\pi)\rtarr 0.$$

The verification of the axioms listed in the previous section is immediate, as in homology. The
fact that cellularly homotopic maps induce the same map on cohomology uses our observations 
relating homotopies of chain complexes with homotopies of cochain complexes. For exactness, the 
fact that our chain complexes are free over $\bZ$ implies that we have a short exact sequence of 
cochain complexes
$$ 0\rtarr C^*(X,A;\pi)\rtarr C^*(X;\pi)\rtarr C^*(A;\pi) \rtarr 0.$$
The required natural long exact sequence follows. The rest is the same as in homology.

For general spaces $X$, we can use $\GA X=|S_*X|$ as a canonical CW approximation functor.
We define the singular cochains\index{singular cochains} of $X$ to be the cellular cochains 
of $\GA X$. Then our
passage from the cohomology of CW complexes to the cohomology of general spaces can be
realized by taking the cohomology of singular cochain complexes.

\section{Cup products in cohomology}

If $X$ and $Y$ are CW complexes, we have an isomorphism 
$$C_*(X\times Y)\iso C_*(X)\ten C_*(Y)$$
of chain complexes and therefore, for any Abelian groups $\pi$ and $\pi'$, an
isomorphism of cochain complexes
$$C^*(X\times Y;\pi\ten \pi')\iso \Hom(C_*(X)\ten C_*(Y),\pi\ten \pi').$$
By our observations about cochain complexes, there results a natural homomorphism
$$H^*(X;\pi)\ten H^*(Y;\pi')\rtarr H^*(X\times Y;\pi\ten\pi').$$
If $X=Y$ and if $\pi=\pi'=R$ is a commutative ring, we can use the 
diagonal map $\DE: X\rtarr X\times X$ and the product $R\ten R\rtarr R$ to 
obtain a ``cup product''\index{cup product}
$$\cup: H^*(X;R)\ten_R H^*(X;R)\rtarr H^*(X;R).$$
More precisely, for $p\geq 0$ and $q\geq 0$, we have a product
$$\cup: H^p(X;R)\ten_R H^q(X;R)\rtarr H^{p+q}(X;R).$$
We have noted that we can use $C_*(X;R)$ instead of $C_*(X)$ and so justify 
tensoring over $R$ rather than $\bZ$. This product makes $H^*(X;R)$ into 
a graded unital, associative, and ``commutative'' $R$-algebra. Here commutativity
is understood in the appropriate graded sense,\index{commutativity!graded} namely
$$xy=(-1)^{pq} yx \ \ \text{if}\ \ \text{deg}\,x=p\ \tand\ \text{deg}\,y=q.$$
The image of $1\in R=H^0(*;R)$ under the map $\pi^*:H^0(*;R)\rtarr H^0(X;R)$ 
induced by the unique map $\pi: X\rtarr \sset{*}$ is the unit (= identity element) for the product. 
In fact, the diagrams that say that $H^*(X;R)$ is unital, associative, and commutative 
result by passing to cohomology from the evident commutative diagrams
$$\diagram
& X \dto^{\DE} \drdouble \dldouble & \\
X\times * & X\times X \lto^{\id\times \pi} \rto_{\pi\times\id} &  *\times X, \\
\enddiagram$$
$$\diagram
X \rto^{\DE} \dto_{\DE} & X\times X \dto^{\DE\times \id} \\
X\times X \rto_(0.4){\id\times\DE} & X\times X\times X,\\
\enddiagram$$
and
$$\diagram
& X \dlto_{\DE} \drto^{\DE} & \\
X\times X\rrto_t & & X\times X.\\
\enddiagram$$
Here $t: X\times Y\rtarr Y\times X$ is the transposition, $t(x,y)=(y,x)$. The following
diagrams commute in homology and cohomology with cofficients in $R$:
$$\diagram
H_*(X)\ten_{R} H_*(Y)\dto_{\ta} \rto^(0.55){\al} & H_*(X\times Y) \dto^{t_*}\\
H_*(Y)\ten_{R} H_*(X) \rto^(0.55){\al} & H_*(Y\times X)\\
\enddiagram$$
and
$$\diagram
H^*(X)\ten_{R} H^*(Y)\dto_{\ta} \rto^(0.55){\al} & H^*(X\times Y) \dto^{t^*}\\
H^*(Y)\ten_{R} H^*(X) \rto^(0.55){\al} & H^*(Y\times X). 
\enddiagram$$
In both diagrams, 
$$\ta(x\ten y)=(-1)^{pq} y\ten x \ \ \text{if}\ \ \text{deg}\,x=p\ \tand\ \text{deg}\,y=q.$$
The reason is that, on the topological level, $t$ permutes $p$-cells past $q$-cells and, on
the level of cellular chains, this involves the transposition
$$ S^{p+q} = S^p\sma S^q \rtarr S^q\sma S^p = S^{p+q}.$$
We leave it as an exercise that this map has degree $(-1)^{pq}$. It is this fact that forces
the cup product to be commutative in the graded sense.

In principle, the way to compute cup products is to pass to cellular chains from a cellular
approximation to the diagonal map $\DE$. The point is that $\DE$ fails to be cellular since
it carries the $n$-skeleton of $X$ to the $2n$-skeleton of $X\times X$. In practice, this
does not work very well and more indirect means of computation must be used.

\section{An example: $\bR P^n$ and the Borsuk-Ulam theorem}

Remember that $\bR P^n$\index{projective space!real} is a CW complex with one $q$-cell for 
each $q\leq n$. The differential
on $C_q(\bR P^n)\iso \bZ$ is zero if $q$ is odd and multiplication by $2$ if $q$ is even.
When we dualize to $C^*(\bR P^n)$, we find that the differential on $C^q(\bR P^n)$ is 
multiplication by $2$ if $q$ is odd and zero if $q$ is even. We read off that
$$H^q(\bR P^n;\bZ)=
\begin{cases}
\bZ  \ \ \ \text{if}\ \ q=0 \\
\bZ_2 \ \ \text{if}\ \ 0<q\leq n\ \tand\ q\ \text{is even}\\
\bZ  \ \ \ \text{if}\ \ q=n\ \text{is odd} \\
0 \ \ \ \ \text{otherwise.}
\end{cases}$$
If we work mod $2$, taking $\bZ_2$ as coefficient group, then the answer takes a nicer 
form, namely
$$H^q(\bR P^n;\bZ_2)=
\begin{cases}
\bZ_2  \ \ \text{if}\ 0\leq q\leq n \\
0 \ \ \ \ \text{if} \ q>n.
\end{cases}$$
The reader may find it instructive to compare with the calculations in homology, checking
the correctness of the calculation by comparison with the universal coefficient theorem.

We shall later use Poincar\'{e} duality to give a quick proof that the cohomology algebra
$H^*(\bR P^n;\bZ_2)$ is a truncated polynomial algebra $\bZ_2[x]/(x^{n+1})$, where $\deg\,x=1$.
That is, for $1\leq q\leq n$, the unique non-zero element of $H^q(\bR P^n;\bZ_2)$ is the
$q$th power of $x$. This means that the elements are so tightly bound together that knowledge 
of the cohomological behavior of a map $f: \bR P^m\rtarr \bR P^n$ on cohomology in degree one
determines its behavior on cohomology in all higher degrees. We assume that $m\geq 1$ and
$n\geq 1$ to avoid triviality.

\begin{prop} Let $f: \bR P^m\rtarr \bR P^n$ be a map such that
$f_*: \pi_1(\bR P^m)\rtarr \pi_1(\bR P^n)$ is non-zero. Then $m\leq n$.
\end{prop}
\begin{proof}
Since $\pi_1(\bR P^1)=\bZ$ and  $\pi_1(\bR P^m)=\bZ_2$ if $m\geq 2$, the result is certainly
true if $n=1$. Thus assume that $n>1$ and assume for a contradiction that $m>n$. By the 
naturality of the Hurewicz isomorphism, $f_*: H_1(\bR P^m;\bZ)\rtarr H_1(\bR P^n;\bZ)$ is 
non-zero. By our universal coefficient theorems, the same is true for mod $2$ homology and 
for mod $2$ cohomology. That is, if $x$ is the non-zero element of $H^1(\bR P^n;\bZ_2)$, then 
$f^*(x)$ is the non-zero element of $H^1(\bR P^m;\bZ_2)$. 
By the naturality of cup products
$$ (f^*(x))^m = f^*(x^m).$$
However, the left side is non-zero in $H^m(\bR P^m;\bZ_2)$ and the right side is zero since
$x^m=0$ by our assumption that $m>n$. The contradiction establishes the conclusion.
\end{proof}

We use this fact together with covering space theory to prove a celebrated result
known as the Borsuk-Ulam theorem. A map $g: S^m\rtarr S^n$ is said to be antipodal\index{antipodal
map} if it takes pairs of antipodal points to pairs of antipodal points. It then induces a map 
$f: \bR P^m\rtarr \bR P^n$ such that the following diagram commutes:
$$\diagram
S^m\rto^g \dto_{p_m} & S^n  \dto^{p_n}\\
\bR P^m \rto_f & \bR P^n,\\
\enddiagram$$
where $p_m$ and $p_n$ are the canonical coverings.

\begin{thm} If $m>n\geq 1$, then there exist no antipodal maps $S^m\rtarr S^n$.
\end{thm} 
\begin{proof}
Suppose given an antipodal map $g:S^m\rtarr S^n$. According to
the proposition, $f_*:\pi_1(\bR P^m)\rtarr \pi_1(\bR P^n)$ is zero. According to
the fundamental theorem of covering space theory, there is a map 
$\tilde{f}: \bR P^m\rtarr S^n$ such that $p_n\com\tilde{f}= f$. Let $s\in S^m$.
Then $\tilde{f}(p_m(s))=\tilde{f}(p_m(-s))$ must be either $g(s)$ or $g(-s)$,
since these are the only two points in $p_n^{-1}(f(p_m(s)))$. Thus either $t=s$
or $t=-s$ satisfies $\tilde{f}(p_m(t))=g(t)$. Therefore, by the fundamental
theorem of covering space theory, the maps $\tilde{f}\com p_m$ and $g$ must be 
equal since they agree on a point. This is absurd: $\tilde{f}\com p_m$ takes 
antipodal points to the same point, while $g$ was assumed to be antipodal.
\end{proof}

\begin{thm}[Borsuk-Ulam]\index{Borsuk-Ulam theorem}
For any continuous map $f: S^n\rtarr \bR^n$, there exists $x\in S^n$ such that
$f(x)=f(-x)$.
\end{thm}
\begin{proof}
Suppose for a contradiction that $f(x)\neq f(-x)$ for all $x$. We could then define a
continuous antipodal map $g: S^n\rtarr S^{n-1}$ by letting $g(x)$ be the point at 
which the vector from $0$ through $f(x)-f(-x)$ intersects $S^{n-1}$.
\end{proof}

\section{Obstruction theory}

We give an outline of one of the most striking features of cohomology: the 
cohomology groups of a space $X$ with coefficients in the homotopy groups of a
space $Y$ control the construction of homotopy classes of maps $X\rtarr Y$.
As a matter of motivation, this helps explain why one is interested in general
coefficient groups. It also explains why the letter $\pi$ is so often used to 
denote coefficient groups.

\begin{defn} Fix  $n\geq 1$. A connected space  $X$  is said to be  $n$-simple\index{nsimple 
space@$n$-simple space} if  
$\pi _{1}(X)$ is Abelian and acts trivially on the homotopy groups $\pi _{q}(X)$  
for  $q\leq n$; $X$ is said to be simple\index{simple space} if it is $n$-simple for all $n$.
\end{defn}

Let  $(X,A)$  be a relative CW complex with relative skeleta $X^n$ and let  $Y$  be  an 
$n$-simple space. The
assumption on $Y$ has the effect that we need not worry about basepoints. Let  
$f: X^{n}\rtarr Y$ be a map.  We ask when  $f$ can be extended to  
a map $X^{n+1}\rtarr Y$ that restricts to the given map on $A$. 

If we compose the attaching maps $S^{n} \rightarrow  X$ of cells of  $X\setminus A$  
with  $f$, we obtain elements of  $\pi_{n}(Y)$.  These elements specify a well defined 
``obstruction cocycle''\index{obstruction cocycle}
\[ c_{f}\in C^{n+1}(X,A;{\pi}_{n}(Y)). \]
Clearly, by considering extensions cell by cell,  $f$  extends to  $X^{n+1}$  if and only 
if  $c_{f} = 0$.  This is not a computable
criterion. However, if we allow ourselves to modify $f$ a little, then we can refine the
criterion to a cohomological one that often is computable. If  $f$  and  $f'$  
are maps  $X^{n} \rightarrow  Y$  and  $h$  is a homotopy rel $A$  of the 
restrictions of  $f$  and  $f'$  to   $X^{n-1}$,  
then  $f$,  $f'$,  and  $h$  together define a map 
\[ h(f,f'): (X\times I)^{n} \rtarr  Y. \]
Applying  $c_{h(f,f')}$ to cells  $j\times I$,  we obtain a ``deformation 
cochain''\index{deformation cochain}
\[ d_{f,f',h}\in C^{n}(X,A;{\pi}_{n}(Y)) \]
such that  $\delta d_{f,f',h} = c_{f}-c_{f'}$.  Moreover, given  $f$  and  $d$,  
there exists  $f'$  that coincides with  $f$  on  $X^{n-1}$  and satisfies  
$d_{f,f'} = d$,  where the constant homotopy  $h$  is understood.  This gives the 
following result.

\begin{thm}
For  $f: X^{n}\rtarr Y$,  the restriction of  $f$  to  
$X^{n-1}$  extends to a map  $X^{n+1}\rightarrow  Y$ if and only if  $[c_{f}]=0$
in $H^{n+1}(X,A;{\pi}_{n}(Y))$.  
\end{thm}

It is natural to ask further when such extensions are unique up to homotopy,
and a similar argument gives the answer.

\begin{thm}
Given maps  $f, f': X^{n}\rightarrow  Y$  and a homotopy rel $A$ of their 
restrictions to  $X^{n-1}$,  there is an obstruction class in  
$H^{n}(X,A;{\pi}_{n}(Y))$  that vanishes 
if and only if the restriction of the given homotopy to  $X^{n-2}$  
extends to a homotopy  $f\simeq f'$ rel $A$. 
\end{thm}

\vspace{.1in}

\begin{center}
PROBLEMS
\end{center}

The first few problems here are parallel to those at the end of Chapter 16.
\begin{enumerate}
\item Let $X$ be a space that satisfies the hypotheses used to construct a universal 
cover $\tilde{X}$ and let $A$ be an Abelian group. Using cellular or singular chains, 
show that 
$$C^*(X;A)\iso \Hom_{\bZ[\pi]}(C_*(\tilde{X}),A).$$
\item Show that there is an isomorphism
$$H^*(K(\pi,1);A) \iso \Ext^*_{\bZ[\pi]}(\bZ,A).$$
When $A$ is a commutative ring, the Ext groups have algebraically defined products,
constructed as follows. The evident isomorphism $\bZ\iso\bZ\ten\bZ$ is covered by a 
map of free $\bZ[\pi]$-resolutions  
$P \rtarr P\ten P$,  where  $\bZ[\pi]$  acts diagonally on tensor products, 
$\al(x\ten y) = \al x\ten \al y$.  This chain map is unique 
up to chain homotopy.  It induces a map of chain complexes
$$\Hom_{\bZ[\pi]}(P,A)\ten \Hom_{\bZ[\pi]}(P,A) \rtarr \Hom_{\bZ[\pi]}(P,A)$$
and therefore an induced product on  Ext$^*_{\bZ[\pi]}(\bZ,A)$.  
Convince yourself that the isomorphism above preserves 
products and explain the intuition (don't worry about technical exactitude).
\item* Now use homological algebra to determine $H^*(\bR P^{\infty};\bZ_2)$ as a ring.
\item Use the previous problem to deduce the ring structure on $H^*(\bR P^n;\bZ_2)$
for each $n\geq 1$.
\item Let $p: Y\rtarr X$ be a covering space with finite fibers, say of cardinality $n$. 
Construct a ``transfer homomorphism''\index{transfer homomorphism} 
$t: H^*(Y;A)\rtarr H^*(X;A)$ and show that $t\com p^*: H^*(X;A)\rtarr H^*(X;A)$
is multiplication by $n$.
\item Let $X$ and $Y$ be CW complexes. Show that the interchange map 
$$t: X\times Y\rtarr Y\times X$$
satisfies $t_*([i]\ten[j])=(-1)^{pq}[j]\ten[i]$ for a $p$-cell of $X$ and a $q$-cell of $Y$.
Deduce that the cohomology ring $H^*(X)$ is commutative in the graded 
sense:\index{commutativity!graded}
$$ x\cup y = (-1)^{pq}y\cup x \ \ \text{if}\ \ \text{deg}\,x=p\ \tand\ \text{deg}\,y=q.$$
\end{enumerate}

An ``$H$-space''\index{Hspace@$H$-space} is a space $X$ with a basepoint $e$ and a product 
$\ph: X\times X\rtarr X$ such that the maps $\la: X\rtarr X$ and $\rh: X\rtarr X$
given by left and right multiplication by $e$ are each homotopic to the identity map. 
Note that $\la$ and $\rh$ specify a map $X\wed X\rtarr X$ that is homotopic to the
codiagonal or folding map $\bigtriangledown$, which restricts to the identity on each 
wedge summand. The following two problems are optional review exercises. 

\begin{enumerate}
\item[7.] If $e$ is a nondegenerate basepoint for $X$, then $\ph$ is homotopic to a product 
$\ph'$ such that left and right multiplication by $e$ under the product $\ph'$ are 
both identity maps.
\item[8.] Show that the product on $\pi_1(X,e)$ induced by the based map 
$\ph': X\times X\rtarr X$ agrees with the multiplication given by composition
of paths and that both products are commutative.
\item[9.] For an $H$-space $X$, the following diagram is commutative:
$$\diagram
X\times X \dto_{\ph} \rrto^{\DE\times\DE} & & X\times X\times X\times X 
\rrto^{\id\times t\times \id} & & X\times X\times X\times X \dto^{\ph\times \ph} \\
X \xto[0,4]_{\DE} & & & & X\times X
\enddiagram$$
(Check it: it is too trivial to write down.) Let $X$ be $(n-1)$-connected, $n\geq 2$, and
let $x\in H^n(X)$. 
\begin{enumerate}
\item[(a)] Show that $\ph^*(x) = x\ten 1 + 1\ten x$.
\item[(b)] Show that 
$$(\DE\times\DE)^*(\id\times\, t\times \id)^*(\ph\times \ph)^*(x\ten x)
=x^2\ten 1 +(1+(-1)^n)(x\ten x)+1\ten x^2.$$
\item[(c)] Prove that, if $n$ is even, then either $2(x\ten x)=0$ in $H^*(X\times X)$ or
$x^2\neq 0$. Deduce that $S^{n}$ cannot be an $H$-space if $n$ is even.
\end{enumerate}
\end{enumerate}

\chapter{Derivations of properties from the axioms}

Returning to the axiomatic approach to cohomology, we assume given a theory on pairs 
of spaces and give some deductions from the axioms. This may be viewed as a dualized
review of what we did in homology, and we generally omit the proofs. The only significant
difference that we will encounter is in the computation of the cohomology of colimits.
In a final section, we show the uniqueness of (ordinary) cohomology with coefficients
in $\pi$. 

Prior to that section, we make no use of the dimension axiom in this chapter.
A ``generalized cohomology theory''
\index{cohomology theory!generalized} $E^*$ is defined to be a 
system of functors $E^q(X,A)$ 
and natural transformations $\de:E^q(A)\rtarr E^{q+1}(X,A)$ that satisfy all of our axioms 
except for the dimension axiom. Similarly, we have the notion of a generalized cohomology 
theory on CW pairs, and the following result holds.

\begin{thm} A cohomology theory $E^*$ on pairs of spaces determines and is determined
by its restriction to a cohomology theory $E^*$ on pairs of CW complexes.
\end{thm}

\section{Reduced cohomology groups and their properties}

For a based space $X$, we define the reduced cohomology\index{reduced cohomology} of $X$ to be
$$\tilde{E}^q(X)=E^q(X,*).$$
There results a direct sum decomposition
$$ E^*(X) \iso \tilde{E}^*(X)\oplus E^*(*)$$
that is natural with respect to based maps. For $*\in A\subset X$, the summand $E^*(*)$ 
maps isomorphically under the map $E^*(X)\rtarr E^*(A)$, and the exactness axiom implies 
that there is a reduced long exact sequence 
$$\cdots\rtarr \tilde{E}^{q-1}(A)\overto{\de} E^q(X,A) \rtarr \tilde{E}^q(X)\rtarr \tilde{E}^q(A)
\rtarr \cdots.$$

The unreduced cohomology groups are recovered as the special cases 
$$ E^*(X)=\tilde{E}^*(X_+)$$
of reduced ones, and similarly for maps. Relative cohomology groups are also special
cases of reduced ones.

\begin{thm}
For any cofibration\index{cofibration} $i: A\rtarr X$, the quotient map $q: (X,A)\rtarr (X/A,*)$ 
induces an isomorphism
$$\tilde{E}^*(X/A)=E^*(X/A,*)\iso E^*(X,A).$$
\end{thm}

We may replace any inclusion $i: A\rtarr X$ by the canonical cofibration $A\rtarr Mi$ 
and then apply the result just given to obtain an isomorphism
$$ E^*(X,A)\iso \tilde{E}^*(Ci).$$

\begin{thm} For a nondegenerately based space $X$, there is a natural isomorphism
$$\SI: \tilde{E}^q(X)\iso \tilde{E}^{q+1}(\SI X).$$
\end{thm}

\begin{cor} Let $*\in A\subset X$, where $i: A\rtarr X$ is a cofibration between
nondegenerately based spaces. In the long exact sequence 
$$\cdots\rtarr \tilde{E}^{q-1}(A)\overto{\de} \tilde{E}^q(X/A)\rtarr \tilde{E}^q(X)\rtarr 
\tilde{E}^q(A)\rtarr \cdots $$
of the pair $(X,A)$, the connecting homomorphism $\de$ is the composite
$$\tilde{E}^{q-1}(A)\overto{\SI} \tilde{E}^{q}(\SI A) \overto{\pa^*}\tilde{E}^q(X/A).$$
\end{cor}

\begin{cor}
For any $n$ and $q$, 
$$\tilde{E}^q(S^n)\iso \tilde{E}^{q-n}(*).$$ 
\end{cor}

\section{Axioms for reduced cohomology}

\begin{defn} A reduced cohomology theory\index{cohomology theory!reduced} $\tilde{E}^*$ consists of functors 
$\tilde{E}^q$ from the homotopy category of nondegenerately based spaces 
to the category of Abelian groups that satisfy the following axioms.
\begin{itemize}
\item EXACTNESS\index{exactness axiom}\ \  If $i: A\rtarr X$ is a cofibration, then the sequence
$$\tilde{E}^q(X/A)\rtarr \tilde{E}^q(X)\rtarr 
\tilde{E}^q(A)$$
is exact.
\item SUSPENSION\index{suspension axiom}\ \ 
For each integer $q$, there is a natural isomorphism
$$\SI: \tilde{E}^q(X)\iso \tilde{E}^{q+1}(\SI X).$$
\item ADDITIVITY\index{additivity axiom}\ \ 
If $X$ is the wedge of a set of nondegenerately based spaces $X_i$, then
the inclusions $X_i\rtarr X$ induce an isomorphism
$$\tilde{E}^*(X) \rtarr \textstyle{\prod}_i\, \tilde{E}^*(X_i).$$
\item WEAK EQUIVALENCE\index{weak equivalence axiom}\ \ If $f:X\rtarr Y$ is a 
weak equivalence, then
$$f^*: \tilde{E}^*(Y)\rtarr \tilde{E}^*(X)$$
is an isomorphism.
\end{itemize}
\end{defn}

The reduced form of the dimension axiom would read
$$\tilde{H}^0(S^0)=\pi \ \ \tand \ \ \tilde{H}^q(S^0)=0\ \text{for}\ q\neq 0.$$

\begin{thm} A cohomology theory $E^*$ on pairs of spaces determines and is 
determined by a reduced cohomology theory $\tilde{E}^*$ on nondegenerately 
based spaces.
\end{thm}

\begin{defn} A reduced cohomology theory\index{cohomology theory!reduced} $\tilde{E}^*$ on 
based CW complexes consists 
of functors $\tilde{E}^q$ from the homotopy category of based CW complexes 
to the category of Abelian groups that satisfy the following axioms.
\begin{itemize}
\item EXACTNESS\index{exactness axiom}\ \  If $A$ is a subcomplex of $X$, then the sequence
$$\tilde{E}^q(X/A)\rtarr \tilde{E}^q(X)\rtarr 
\tilde{E}^q(A)$$
is exact.
\item SUSPENSION\index{suspension axiom}\ \ 
For each integer $q$, there is a natural isomorphism
$$\SI: \tilde{E}^q(X)\iso \tilde{E}^{q+1}(\SI X).$$
\item ADDITIVITY\index{additivity axiom}\ \ 
If $X$ is the wedge of a set of based CW complexes $X_i$, then
the inclusions $X_i\rtarr X$ induce an isomorphism
$$\tilde{E}^*(X) \rtarr \textstyle{\prod}_i\, \tilde{E}^*(X_i).$$
\end{itemize}
\end{defn}

\begin{thm} A reduced cohomology theory $\tilde{E}^*$ on nondegenerately based spaces
determines and is determined by its restriction to a reduced cohomology theory on
based CW complexes.
\end{thm}

\begin{thm}
A cohomology theory $E^*$ on CW pairs determines and is determined by a reduced
cohomology theory $\tilde{E}^*$ on based CW complexes.
\end{thm}

\section{Mayer-Vietoris sequences in cohomology}

We have Mayer-Vietoris sequences in cohomology just like those in homology. The proofs are
the same. Poincar\'{e} duality between the homology and cohomology of manifolds will be proved 
by an inductive comparison of homology and cohomology Mayer-Vietoris sequences. We record two
preliminaries.

\begin{prop} For a triple $(X,A,B)$, the following sequence is exact:\index{triple!exact sequence
of}
$$\cdots E^{q-1}(A,B) \overto{\de} E^q(X,A)\overto{j^*} E^q(X,B)\overto{i^*} 
E^{q}(A,B)\rtarr \cdots.$$
Here $i:(A,B)\rtarr (X,B)$ and $j:(X,B)\rtarr (X,A)$ are inclusions and $\de$ is the composite
$$E^{q-1}(A,B)\rtarr E^{q-1}(A)\overto{\de} E^q(X,A).$$
\end{prop}

Now let $(X;A,B)$ be an excisive triad and set $C=A\cap B$.

\begin{lem}
The map 
$$E^*(X,C) \rtarr E^*(A,C)\oplus E^*(B,C) $$
induced by the inclusions of $(A,C)$ and $(B,C)$ in $(X,C)$ is an isomorphism.
\end{lem}

\begin{thm}[Mayer-Vietoris sequence]\index{Mayer-Vietoris sequence} Let $(X;A,B)$ be an excisive 
triad and set $C=A\cap B$.
The following sequence is exact:
$$\cdots \rtarr E^{q-1}(C)\overto{\DE^*} E^q(X) \overto{\ph^*} E^q(A)\oplus E^q(B)\overto{\ps^*} 
E^{q}(C)\rtarr \cdots.$$
Here, if $i: C\rtarr A$, $j: C\rtarr B$, $k: A\rtarr X$, and $\ell: B\rtarr X$
are the inclusions, then
$$\ph^*(\ch)= (k^*(\ch),\ell^*(\ch))\ \ \tand\ \ \psi^*(\al,\be)=i^*(\al)-j^*(\be) $$
and $\DE^*$ is the composite 
$$E^{q-1}(C)\overto{\de} E^q(A,C)\iso E^q(X,B)\rtarr E^q(X).$$
\end{thm}

For the relative version, let $X$ be contained in some ambient space $Y$.

\begin{thm}[Relative Mayer-Vietoris sequence]\index{Mayer-Vietoris sequence!relative} The 
following sequence is exact:
$$\cdots\rtarr E^{q-1}(Y,C)\overto{\DE^*} E^q(Y,X)\overto{\ph^*} E^q(Y,A)\oplus E^q(Y,B)
\overto{\ps^*} E^{q}(Y,C)\rtarr \cdots.$$
Here, if $i: (Y,C)\rtarr (Y,A)$, $j: (Y,C)\rtarr (Y,B)$, $k: (Y,A)\rtarr (Y,X)$, and 
$\ell: (Y,B)\rtarr (Y,X)$ are the inclusions, then
$$ \ph^*(\ch)= (k^*(\ch),\ell^*(\ch)) \ \ \tand\ \ \psi^*(\al,\be)=i^*(\al)-j^*(\be)$$
and $\DE^*$ is the composite 
$$E^{q-1}(Y,C)\rtarr E^{q-1}(A,C)\iso E^{q-1}(X,B)\overto{\de} E^{q}(Y,X).$$
\end{thm}

\begin{cor}
The absolute and relative Mayer-Vietoris sequences are related by the following 
commutative diagram:
$$\diagram
E^{q-1}(C)\rto^{\DE^*} \dto_{\de}  & E^{q}(X) \rto^(0.4){\ph^*} \dto^{\de} &
 E^{q}(A)\oplus E^{q}(B)\rto^(0.6){\ps^*} \dto^{\de+\de} &  E^{q}(C) \dto^{\de} \\
E^q(Y,C)\rto_(0.46){\DE^*} &  E^{q+1}(Y,X) \rto_(0.35){\ph^*} & 
E^{q+1}(Y,A)\oplus E^{q+1}(Y,B) \rto_(0.65){\ps^*}  & E^{q+1}(Y,C).\\
\enddiagram$$
\end{cor}

\section{Lim$^1$ and the cohomology of colimits}

In this section, we let $X$ be the union of an expanding sequence of subspaces $X_i$, 
$i\geq 0$. We shall use the additivity and weak equivalence axioms and the 
Mayer-Vietoris sequence to explain how to compute $E^*(X)$. The answer is more subtle
than in homology because, algebraically, limits are less well behaved than colimits:
they are not exact functors from diagrams of Abelian groups to Abelian groups. Rather
than go into the general theory, we simply display how the ``first right derived functor''
$\lim^{1}$\index{lima@$\lim^{1}$} of an inverse sequence of Abelian groups can be computed.

\begin{lem}
Let $f_i: A_{i+1}\rtarr A_{i}$, $i\geq 1$, be a sequence of homomorphisms of Abelian groups.
Then there is an exact sequence
$$ 0\rtarr \lim\,A_i\overto{\be} \textstyle{\prod}_i A_i\overto{\al} 
\textstyle{\prod}_i A_i\rtarr \lim^{1}A_i\rtarr 0,$$
where $\al$ is the difference of the identity map and the map with coordinates $f_i$ and $\be$
is the map whose projection to $A_i$ is the canonical map given by the definition of a limit.
\end{lem}

That is, we may as well define $\lim^{1}A_i$ to be the displayed cokernel. We then have the
following result.

\begin{thm}\index{colimit!cohomology of}
For each $q$, there is a natural short exact sequence
$$0 \rtarr {\lim}^{1}\,E^{q-1}(X_i)\rtarr E^q(X)\overto{\pi} \lim\,E^q(X_i) \rtarr 0,$$
where $\pi$ is induced by the inclusions $X_i\rtarr X$.
\end{thm}
\begin{proof}
We use the notations and constructions in the proof that homology commutes with
colimits and consider the excisive triad $(\tel\,X_i;A,B)$ with $C=A\cap B$ constructed
there. By the additivity axiom,
$$E^*(A)=\textstyle{\prod}_i\, E^*(X_{2i}),\ \ E^*(B)
=\textstyle{\prod}_i\, E^*(X_{2i+1}),\ \tand\ E^*(C)=\textstyle{\prod}_i\, E^*(X_i).$$
We construct the following commutative diagram, whose top row is the cohomology Mayer-Vietoris 
sequence of the triad $(\tel\, X_i;A,B)$ and whose bottom row is an exact sequence
of the sort displayed in the previous lemma.
\begin{footnotesize}
$$\diagram
\cdots \rto &  E^q(\tel X_i) \rto \dto_{\iso} &  E^q(A)\oplus E^q(B) \rto \dto_{\iso}
 &  E^q(C) \rto \dto^{\iso} & E^{q+1}(\tel\,X_i) \dto^{\iso} \rto & \cdots \\
\cdots \rto & E^q(X) \rto^{\be'} \dto_{\pi'}
& \prod E^q(X_i) \rto^{\al'} \dto_{\prod(-1)^{i}}
 & \prod_i E^q(X_i)  \dto^{\prod_i(-1)^{i}} \rto  & E^{q+1}(X) \rto & \cdots \\
0 \rto & \lim\,E^q(X_i) \rto^{\be} & \prod_i E^q(X_i) \rto^{\al} 
 & \prod_i E^q(X_i) \rto & \lim^1 E^q(X_i) \rto & 0. \\
\enddiagram$$
\end{footnotesize}
The commutativity of the bottom middle square is a comparison based on the sign used in
the Mayer-Vietoris sequence. Here the map $\pi'$ differs by alternating signs from the 
canonical map $\pi$, but this does not affect the conclusion. A chase of the diagram 
implies the result.
\end{proof}

The $\lim^1$ ``error terms'' are a nuisance, and it is important to know when they vanish.
We say that an inverse sequence $f_i: A_{i+1}\rtarr A_i$ satisfies the Mittag-Leffler
condition\index{Mittag-Leffler condition} if, for each fixed $i$, there exists $j\geq i$ such 
that, for every $k>j$, the
image of the composite $A_k\rtarr A_i$ is equal to the image of the composite $A_j\rtarr A_i$.
For example, this holds if all but finitely many of the $f_i$ are epimorphisms or if
the $A_i$ are all finite. As a matter of algebra, we have the following vanishing result.

\begin{lem}
If the inverse sequence  $f_i: A_{i+1}\rtarr A_i$ satisfies the Mittag-Leffler condition,
then $\lim^1\,A_i=0$.
\end{lem}

For example, for $q<n$, the inclusion $X^n\rtarr X^{n+1}$ of skeleta in a CW complex induces
an isomorphism $H^q(X^{n+1};\pi)\rtarr H^q(X^n;\pi)$ and we conclude that the canonical map
$$H^q(X;\pi)\rtarr H^q(X^n;\pi)$$ is an isomorphism for $q<n$. This is needed in the proof 
of the uniqueness of ordinary cohomology.

\section{The uniqueness of the cohomology of CW complexes}

As with homology, one reason for defining ordinary cohomology with coefficients in an
Abelian group $\pi$\index{cohomology theory!ordinary} in terms of cellular cochains 
is the inevitability of the definition. 
If we assume given a theory that satisfies the axioms, 
we see that the cochains with coefficients in $\pi$ of a CW 
complex $X$ can be redefined by
$$C^n(X;\pi)=H^n(X^n,X^{n-1};\pi),$$
with differential
$$d: C^n(X;\pi)\rtarr C^{n+1}(X;\pi)$$
the composite
$$H^n(X^n,X^{n-1};\pi)\rtarr H^{n}(X^{n})
\overto{\de} H^{n+1}(X^{n+1},X^{n}).$$
That is, the following result holds.

\begin{thm}\index{cellular cochains}
$C^*(X;\pi)$ as just defined is isomorphic to $\Hom(C_*(X),\pi)$.
\end{thm}

We define the reduced cochains $\tilde{C}^*(X;\pi)$\index{reduced cochains} of a based 
space $X$ to be the kernel 
of the map $C^*(X;\pi)\rtarr C^*(*;\pi)$ induced by the inclusion $\sset{*}\rtarr X$. For a 
CW pair $(X,A)$, we define $C^*(X,A;\pi)$ to be the kernel of the epimorphism
$$C^*(X;\pi)\rtarr C^*(A;\pi)$$
induced by the inclusion $A\rtarr X$. The analogue of the previous result for reduced and
relative cochains follows directly. This leads to a uniqueness theorem for cohomology just
like that for homology.

\begin{thm}\index{cohomology theory!ordinary}
There is a natural isomorphism
$$H^*(X,A;\pi)\iso H^*(C^*(X,A;\pi))$$
under which the natural transformation $\de$ agrees with the natural transformation
induced by the connecting homomorphisms associated to the short exact sequences
$$0\rtarr C^*(X,A;\pi)\rtarr C^*(X;\pi)\rtarr C^*(A;\pi)\rtarr 0.$$
\end{thm}
\begin{proof}
It suffices to obtain a natural isomorphism of reduced theories on based CW complexes $X$.
We have seen that $\tilde{H}^q(X;\pi)\iso \tilde{H}^q(X^n;\pi)$ for $q<n$, and we obtain a
diagram dual to that used in the proof of the analogue in homology by arrow reversal.
We leave further details as an exercise for the reader.
\end{proof}

\vspace{.1in}

\begin{center}
PROBLEMS
\end{center}
\begin{enumerate}
\item Complete the proof of the uniqueness theorem for cohomology.
\end{enumerate}

In the following sequence of problems, we take cohomology with coefficients
in a commutative ring $R$ and we write $\ten$ for $\ten_R$.

\begin{enumerate}
\item[2.]  Let $A$ and $B$ be subspaces of a space $X$. Construct a relative 
cup product\index{cup product!relative}
$$H^p(X,A)\otimes H^q(X,B)\rtarr H^{p+q}(X,A\cup B)$$
and show that the following diagram is commutative:
$$\diagram
H^p(X,A)\otimes H^q(X,B) \rto  \dto & H^{p+q}(X,A\cup B) \dto \\
H^p(X)\otimes H^q(X)\rto & H^{p+q}(X).\\
\enddiagram$$
The horizontal arrows are cup products; the vertical arrows are induced from 
$X\rtarr (X,A)$, and so forth.
\item[3.] Let  $X$ have a basepoint $*\in A\cap B$. Deduce a commutative diagram
$$\diagram
H^p(X,A)\otimes H^q(X,B) \rto \dto & H^{p+q}(X,A\cup B) \dto \\
\tilde{H}^p(X)\otimes \tilde{H}^q(X)\rto & \tilde{H}^{p+q}(X).\\
\enddiagram$$
\item[4.]  Let  $X = A\cup B$,  where $A $ and  $B$  are contractible and  $A\cap B \neq\emptyset$.  
Deduce that the cup product 
$$\tilde{H}^p(X)\otimes \tilde{H}^q(X)\rtarr \tilde{H}^{p+q}(X)$$ 
is the zero homomorphism.
\item[5.]  Let  $X = \SI Y = Y\sma S^1$.  Deduce that the cup product  
$$\tilde{H}^p(X)\otimes \tilde{H}^q(X)\rtarr \tilde{H}^{p+q}(X)$$ 
is the zero homomorphism.
\end{enumerate}

Commentary: Additively, cohomology groups are ``stable,''\index{stable} in the sense that
$$\tilde{H}^p(Y) \iso \tilde{H}^{p+1}(\SI Y).$$
Cup products are ``unstable,'' in the sense that they vanish on suspensions.
This is an indication of how much more information they carry than the mere
additive groups. The proof given by this sequence of exercises actually
applies to any ``multiplicative'' cohomology theory,\index{cohomology theory!multiplicative}
that is, any theory that has suitable cup products.

\chapter{The Poincar\'e duality theorem}

The crucial starting point for applications of algebraic topology to geometric topology
is the Poincar\'e duality theorem. It gives a tight algebraic constraint on the homology
and cohomology groups of compact manifolds.

\section{Statement of the theorem}

It is apparent that there is a kind of duality relating the construction of homology
and cohomology. In its simplest form, this is reflected by the fact that evaluation
of cochains on chains gives a natural homomorphism
$$C^p(X;\pi)\ten C_p(X;\rh) \rtarr \pi\ten \rh.$$
This passes to homology and cohomology to give an evaluation pairing\index{evaluation pairing}
$$H^p(X;\pi)\ten H_p(X;\rh) \rtarr \pi\ten \rh.$$
Taking $\pi=\rh$ to be a commutative ring $R$ and using its product, there results a pairing
$$H^p(X;R)\ten_R H_p(X;R) \rtarr R.$$
It is usually written $\langle\al,x\rangle$ for $\al\in H^p(X;R)$ and $x\in H_p(X;R)$. 
When $R$ is a field and the $H_p(X;R)$ are finite dimensional vector spaces, the adjoint
of this pairing is an isomorphism
$$H^p(X;R)\iso \Hom_R( H_p(X;R), R).$$
That is, the cohomology groups of $X$ are the vector space duals of the homology groups
of $X$.

Now let $M$ be a compact manifold of dimension $n$. We shall study manifolds without
boundary in this chapter, turning to manifolds with boundary in the next. We do not 
assume that $M$ is differentiable. It is known that $M$ can be given the structure of a 
finite CW complex, and its homology and cohomology groups are therefore finitely generated. 
When $M$ is differentiable, it is not hard to prove this using Morse theory, but it is a 
deep theorem in the general topological case. We shall not go into the proof but shall
take the result as known. 

We have the cup product
$$ H^p(M;R)\ten H^{n-p}(M;R) \rtarr H^n(M;R).$$
If $R$ is a field and $M$ is ``$R$-orientable,''\index{Rorientable@$R$-orientable} then there 
is an ``$R$-fundamental class''\index{Rfundamental class@$R$-fundamental class}
$z\in H_n(M;R)$. The composite of the cup product and evaluation on $z$ gives a 
cup product pairing\index{cup product pairing}
$$ H^p(M;R)\ten H^{n-p}(M;R) \rtarr R. $$
One version of the Poincar\'e duality theorem asserts that this pairing is nonsingular,
so that its adjoint is an isomorphism
$$ H^p(M;R)\iso \Hom_R(H^{n-p}(M;R),R)\iso H_{n-p}(M;R).$$
In fact, Poincar\'e duality does not require the commutative ring $R$ to be a field, 
and it is useful to allow $R$-modules $\pi$ as coefficients in our homology and cohomology 
groups. We shall gradually make sense of and prove the following theorem.

\begin{thm}[Poincar\'e duality]\index{Poincare duality theorem@Poincar\'e duality theorem} Let 
$M$ be a compact $R$-oriented\index{Roriented@$R$-oriented manifold} $n$-manifold. \linebreak
Then, for an $R$-module $\pi$, there is an isomorphism 
$$D: H^p(M;\pi)\rtarr H_{n-p}(M;\pi).$$
\end{thm}

We shall define the notion of an $R$-orientation\index{Rorientation@$R$-orientation} and an 
$R$-fundamental class of a manifold in \S3, and we shall later prove the following result.

\begin{prop} If $M$ is a compact $n$-manifold, then an $R$-orientation of $M$ determines and is 
determined by an $R$-fundamental class $z\in H_n(M;R)$.
\end{prop}

The isomorphism $D$ is given by the adjoint of the cup product pairing determined 
by $z$, but it is more convenient to describe it in terms of the ``cap product.'' For any space 
$X$ and $R$-module $\pi$, there is a cap product\index{cap product}
$$ \cap: H^p(X;\pi)\ten_R H_n(X;R) \rtarr H_{n-p}(X;\pi).$$
We shall define it in the next section. The isomorphism $D$ is specified by
$$D(\al)=\al\cap z.$$
When $\pi=R$, we shall prove that the cap product, cup product, and evaluation pairing are 
related by the fundamental identity
$$\langle\al\cup\be,x\rangle = \langle\be,\al\cap x\rangle.$$
Taking $x=z$, this shows that in this case $D$ is adjoint to the cup product pairing 
determined by $z$. 

We explain a few consequences before beginning to fill in the details and proofs. Let $M$
be a connected compact oriented (= $\bZ$-oriented) $n$-manifold. Taking integer 
coefficients,
we have $D:H^p(M)\iso H_{n-p}(M)$. With $p=0$, this shows that $H_n(M)\iso \bZ$ with generator
the fundamental class\index{fundamental class} $z$. With $p=n$, it shows that $H^n(M)\iso\bZ$ 
with generator $\ze$ dual
to $z$, $\langle\ze,z\rangle=1$. The relation between $\cup$ and $\cap$ has the following 
consequence.

\begin{cor} Let $T_p\subset H^p(M)$ be the torsion subgroup. The cup product pairing
$\al\ten\be \rtarr \langle\al\be,z\rangle$ induces a nonsingular pairing
$$H^p(M)/T_p \ten H^{n-p}(M)/T_{n-p}\rtarr \bZ.$$
\end{cor}
\begin{proof}
If $\al\in T_p$, say $r\al=0$, and $\be\in H^{n-p}(M)$, then $r(\al\cup\be)=0$ and therefore 
$\al\cup\be=0$ since $H^n(M)=\bZ$. Thus the pairing vanishes on torsion elements. Since
$\Ext^1_{\bZ}(\bZ_r,\bZ)=\bZ_r$ and each $H_p(M)$ is finitely generated, 
$\Ext^1_{\bZ}(H_*(M),\bZ)$ is a torsion group. By the universal coefficient theorem, this
implies that
$$H^p(M)/T_p = \Hom(H_p(M),\bZ).$$
Thus, if $\al\in H^p(M)$ projects to a generator of the free Abelian group $H^p(M)/T_p$,
then there exists $a\in H_p(M)$ such that $\langle\al,a\rangle=1$. By Poincar\'e duality,
there exists $\be\in H^{n-p}(M)$ such that $\be\cap z=a$. Then
$$\langle\be\cup\al,z\rangle = \langle\al,\be\cap z\rangle = 1.\qed$$
\renewcommand{\qed}{}\end{proof}

We shall see that any simply connected manifold, such as $\bC P^n$, is orientable. The
previous result allows us to compute the cup products of $\bC P^n$.

\begin{cor} As a graded ring, $H^*(\bC P^n)$\index{projective space!complex} is the 
truncated polynomial algebra 
$\bZ[\al]/(\al^{n+1})$, where {\em deg}$\,\al=2$. That is, $H^{2q}(\bC P^n)$ is the free
Abelian group with generator $\al^q$ for $1\leq q\leq n$. 
\end{cor}
\begin{proof}
We know that $\bC P^n$ is a CW complex with one $2q$-cell for each $q$, $0\leq q\leq n$. 
Therefore the conclusion is correct additively: $H^{2q}(\bC P^n)$ is a free Abelian
group on one generator for $0\leq q\leq n$. Moreover $\bC P^{n-1}$ is the $(2n-1)$-skeleton
of $\bC P^n$, and the inclusion $\bC P^{n-1}\rtarr \bC P^n$ therefore induces an isomorphism
$H^{2q}(\bC P^n)\rtarr H^{2q}(\bC P^{n-1})$ for $q<n$. We proceed by induction on $n$, the
conclusion being obvious for $\bC P^1\iso S^2$. The induction hypothesis implies that if 
$\al$ generates $H^2(\bC P^n)$, then $\al^q$ generates $H^{2q}(\bC P^n)$ for $q<n$. By
the previous result, there exists $\be\in H^{2n-2}(\bC P^n)$ such that 
$\langle \al\cup\be, z\rangle = 1$. Clearly $\be$ must be a generator, so that 
$\be=\pm \al^{n-1}$, and therefore $\al^n$ must generate $H^{2n}(\bC P^n)$.
\end{proof}

In the presence of torsion in the cohomology of $M$, it is convenient to work with 
coefficients in a field. We shall see that an oriented manifold is $R$-oriented for
any commutative ring $R$. The same argument as for integer coefficients gives the
following more convenient nonsingular pairing result.

\begin{cor}
Let $M$ be a connected compact $R$-oriented $n$-manifold, where $R$ is a field.
Then $\al\ten\be\rtarr \langle \al\cup\be,z\rangle$ defines a nonsingular pairing
$$H^p(M;R)\ten_R H^{n-p}(M;R)\rtarr R.$$
\end{cor}

We shall see that every manifold is $\bZ_2$-oriented, and an argument exactly like
that for $\bC P^n$ allows us to compute the cup products in $H^*(\bR P^n;\bZ_2)$.
We used this information in our proof of the Borsuk-Ulam theorem.

\begin{cor} As a graded ring, $H^*(\bR P^n;\bZ_2)$\index{projective space!real} is the 
truncated polynomial algebra 
$\bZ_2[\al]/(\al^{n+1})$, where {\em deg}$\,\al=1$. That is, $\al^q$ is the non-zero element
of $H^{q}(\bR P^n;\bZ_2)$ for $1\leq q\leq n$. 
\end{cor}

\section{The definition of the cap product}

To define the cap product, we may as well assume that $X$ is a CW complex, by CW 
approximation. The diagonal map $\DE: X\rtarr X\times X$ is not cellular, but it
is homotopic to a cellular map $\DE'$. Thus we have a chain map
$$\DE'_*: C_*(X)\rtarr C_*(X\times X)\iso C_*(X)\ten C_*(X).$$
It carries $C_n(X)$ to $\sum C_p(X)\ten C_{n-p}(X)$. Tensoring over a commutative
ring $R$ and using that $R\iso R\ten_R R$, we obtain
$$\DE'_*: C_*(X;R)\rtarr C_*(X\times X;R)\iso C_*(X;R)\ten_R C_*(X;R).$$
For an $R$-module $\pi$, we define\index{cap product}
$$\cap: C^*(X;\pi)\ten_R C_*(X;R) \rtarr C_*(X;\pi)$$
to be the composite
$$\diagram
C^*(X;\pi)\ten_R C_*(X;R) \dto^{\id\ten\DE'_*}\\
C^*(X;\pi)\ten_R C_*(X;R)\ten_R C_*(X;R) \dto^{\epz\ten\id}\\
\pi\ten_R C_*(X;R)\iso C_*(X;\pi).\\
\enddiagram$$
Here $\epz$ evaluates cochains on chains. Precisely, it must be interpreted as zero on
$C^p(X;\pi)\ten_R C_q(X;R)$ if $p\neq q$ and the evident evaluation map
$$\Hom_R(C_p(X;R),\pi)\ten_R C_p(X;R)\rtarr \pi$$
if $p=q$. Therefore the cap product is given degreewise by maps
$$\cap: C^p(X;\pi)\ten_R C_n(X;R) \rtarr C_{n-p}(X;\pi).$$
To understand this, it makes sense to think in terms of $C^*(X;\pi)$ regraded by
negative degrees and so thought of as a chain complex rather than a cochain complex.
Our convention on cochains that $(d\al)(x)=(-1)^{p+1}\al(dx)$ for $\al\in C^p(X;\pi)$
and $x\in C_{p+1}(X;R)$ means that $\epz\com d=0$, where $d$ is the tensor product
differential on the chain complex $C^*(X;\pi)\ten_R C_*(X;R)$. That is, $\epz$ is a
map of chain complexes, where $\pi$ is thought of as a chain complex concentrated in
degree zero, with zero differential. It follows that $\cap$ is a chain map. That is,
$$d(\al\cap x) = (d\al)\cap x + (-1)^{\text{deg}\,\al}\al\cap dx.$$
Using the evident natural map from the tensor product of homologies to the homology
of a tensor product, we see that $\cap$ passes to homology to induce a pairing
$$\cap: H^*(X;\pi)\ten_R H_*(X;R) \rtarr H_*(X;\pi).$$

To relate the cap and cup products, recall that the latter is induced by
$${\DE'}^*: C^*(X;R)\ten_RC^*(X;R)\iso C^*(X\times X;R)\rtarr C^*(X;R).$$
It is trivial that the following diagram commutes:
$$\diagram
C^*(X\times X;R)\ten_R C_*(X;R) \rto^(0.47){\id\ten\DE'_*} \dto_{{\DE'}^*\ten\id}
& C^*(X\times X;R)\ten_R C_*(X\times X;R) \dto^{\epz} \\
C^*(X;R)\ten_R C_*(X;R) \rto_{\epz} & R.\\
\enddiagram$$
We may identify the chains and cochains of $X\times X$ on the top row with tensor
products of chains and cochains of $X$. After this identification, the right-hand
map $\epz$ becomes the composite
$$\diagram 
C^*(X;R)\ten_R C^*(X;R)\ten_R C_*(X;R)\ten_R C_*(X;R) \dto^{\id\ten t\ten \id}\\
C^*(X;R)\ten_R C_*(X;R)\ten_R C^*(X;R)\ten_R C_*(X;R) \dto^{\epz\ten\epz} \\
R\ten_R R\iso R.\\
\enddiagram$$
Noting the agreement of signs introduced by the two maps $t$, we see that this
composite is the same as the composite
$$\diagram
C^*(X;R)\ten_R C^*(X;R)\ten_R C_*(X;R)\ten_R C_*(X;R) \dto^{t\ten \id\ten \id}\\
C^*(X;R)\ten_R C^*(X;R)\ten_R C_*(X;R)\ten_R C_*(X;R) \dto^{\id\ten\epz\ten\id} \\
C^*(X;R)\ten_R C_*(X;R)\dto^{\epz}\\
R.\\
\enddiagram$$
Inspecting definitions, we see that, on elements, these observations prove the 
fundamental identity
$$\langle\al\cup\be,x\rangle = \langle\be,\al\cap x\rangle.$$

For use in the proof of the Poincar\'e duality theorem, we observe that the cap product
generalizes to relative cap products\index{cap product!relative}
$$\cap: H^p(X,A;\pi)\ten_R H_n(X,A;R) \rtarr H_{n-p}(X;\pi)$$
and 
$$\cap: H^p(X;\pi)\ten_R H_n(X,A;R) \rtarr H_{n-p}(X,A;\pi)$$
for pairs $(X,A)$. Indeed, we may assume that $(X,A)$ is a CW pair and that $\DE'$ 
restricts to a map $A\rtarr A\times A$ that is homotopic to the diagonal of $A$.
Via the quotient map $X\rtarr X/A$, $\DE'$ induces relative diagonal approximations
$$\DE'_*: C_*(X,A;R) \rtarr C_*(X,A;R)\ten C_*(X;R)$$
and 
$$\DE'_*: C_*(X,A;R) \rtarr C_*(X;R)\ten C_*(X,A;R).$$
These combine with the evident evaluation maps to give the required relative cap products.

\section{Orientations and fundamental classes}

Let $M$ be an $n$-manifold, not necessarily compact; the extra generality will be crucial
to our proof of the Poincar\'e duality theorem. For $x\in M$, we can choose a
coordinate chart $U\iso \bR^n$ with $x\in U$. By excision, exactness, and
homotopy invariance, we have isomorphisms
$$H_i(M,M-x)\iso H_i(U,U-x)\iso \tilde{H}_{i-1}(U-x)\iso \tilde{H}_{i-1}(S^{n-1}).$$
This holds with any coefficient group, but we agree to take coefficients in a given
commutative ring $R$. Thus $H_i(M,M-x)=0$ if $i\neq n$ and $H_n(M,M-x)\iso R$. We
think of $H_n(M,M-x)$ as a free $R$-module on one generator, but the generator
(which corresponds to a unit of the ring $R$) is unspecified. Intuitively, an
$R$-orientation of $M$ is a consistent choice of generators.

\begin{defn}
An $R$-fundamental class\index{Rfundamental class@$R$-fundamental class} of $M$ at a 
subspace $X$ is an element $z\in H_n(M,M-X)$ such that,
for each $x\in X$, the image of $z$ under the map
$$H_n(M,M-X)\rtarr H_n(M,M-x)$$
induced by the inclusion $(M,M-X)\rtarr (M,M-x)$ is a generator. If $X=M$, we refer to 
$z\in H_n(M)$ as a fundamental class of $M$. An $R$-orientation\index{Rorientation@$R$-orientation} 
of $M$ is an open cover $\sset{U_i}$ and $R$-fundamental classes $z_i$ of $M$ at $U_i$ such 
that if $U_i\cap U_j$ is non-empty, then $z_i$ and $z_j$ map to the same element of 
$H_n(M,M-U_i\cap U_j)$.
\end{defn}

We say that $M$ is $R$-orientable\index{Rorientable@$R$-orientable} if it admits an 
$R$-orientation. When $R=\bZ$, we refer
to orientations\index{orientation} and orientability\index{orientable manifold}. There are various 
equivalent ways of formulating these 
notions. We leave it as an exercise for the reader to reconcile the present definition of 
orientability with any other definition he or she may have seen.

Clearly an $R$-fundamental class $z$ determines an $R$-orientation: given any open cover
$\sset{U_i}$, we take $z_i$ to be the image of $z$ in $H_n(M,M-U_i)$. The converse holds
when $M$ is compact. To show this, we need the following vanishing theorem, which we shall 
prove in the next section.

\begin{thm}[Vanishing]\index{vanishing theorem} Let $M$ be an $n$-manifold. For any coefficient 
group $\pi$, $H_i(M;\pi)=0$ if $i>n$, and $\tilde{H}_n(M;\pi)=0$ 
if $M$ is connected and is not compact.
\end{thm}

We can use this together with Mayer-Vietoris sequences to construct $R$-fun\-da\-men\-tal classes
at compact subspaces from $R$-orientations. To avoid trivialities, we tacitly assume that $n>0$.
(The trivial case $n=0$ forced the use of reduced  homology in the statement; where arguments use
reduced homology below, it is only to ensure that what we write is correct in dimension zero.)

\begin{thm}
Let $K$ be a compact subset of $M$. Then, for any coefficient group $\pi$, $H_i(M,M-K;\pi)=0$ 
if $i>n$, and an $R$-orientation of $M$ determines an $R$-fundamental class of $M$ at $K$. 
In particular, if $M$ is compact, then an $R$-orientation of $M$ determines an $R$-fundamental 
class of $M$.
\end{thm}
\begin{proof}
First assume that $K$ is contained in a coordinate chart $U\iso \bR^n$. By excision and 
exactness, we then have
$$H_i(M,M-K;\pi)\iso H_i(U,U-K;\pi)\iso \tilde{H}_{i-1}(U-K;\pi).$$
Since $U-K$ is open in $U$, the vanishing theorem implies that $\tilde{H}_{i-1}(U-K;\pi)=0$
for $i>n$. In fact, a lemma used in the proof of the vanishing theorem will prove this
directly. In this case, an $R$-fundamental class in $H_n(M,M-U)$ maps to an $R$-fundamental 
class in $H_n(M,M-K)$. A general compact subset $K$ of $M$ can be written as the union of 
finitely many compact subsets, each of which is contained in a coordinate chart. By induction,
it suffices to prove the result for $K\cup L$ under the assumption that it holds for $K$, $L$,
and $K\cap L$. With any coefficients, we have the Mayer-Vietoris sequence
\begin{multline*}
\cdots \rtarr H_{i+1}(M,M-K\cap L)\overto{\DE}H_i(M,M-K\cup L)\\
\overto{\ps}H_i(M,M-K)\oplus H_i(M,M-L)\overto{\ph} H_i(M,M-K\cap L) \rtarr \cdots.
\end{multline*}
The vanishing of $H_i(M,M-K\cup L;\pi)$ for $i>n$ follows directly. Now take $i=n$ and
take coefficients in $R$. Then $\ps$ is a monomorphism. The $R$-fun\-da\-men\-tal classes 
$z_K\in H_n(M,M-K)$ and $z_L\in H_n(M,M-L)$ determined by a given $R$-orientation both map
to the $R$-fundamental class $z_{K\cap L}\in H_n(M,M-K\cap L)$ determined by the given
$R$-orientation. Therefore
$$\ph(z_K,z_L) = z_{K\cap L}- z_{K\cap L} = 0$$ 
and there exists a unique $z_{K\cup L}\in H_n(M,M-K\cup L)$ such that 
$$ \ps(z_{K\cup L}) = (z_K,z_L).$$
Clearly $z_{K\cup L}$ is an $R$-fundamental class of $M$ at $K\cup L$. 
\end{proof}

The vanishing theorem also implies the following dichotomy, which we have already noticed
in our examples of explicit calculations.

\begin{cor}
Let $M$ be a connected compact $n$-manifold, $n>0$. Then either $M$ is not orientable and 
$H_n(M;\bZ)=0$ or $M$ is orientable\index{orientable} and the map
$$H_n(M;\bZ) \rtarr H_n(M,M-x;\bZ)\iso \bZ$$
is an isomorphism for every $x\in M$.
\end{cor}
\begin{proof}
Since $M-x$ is connected and not compact, $H_n(M-x;\pi)=0$ and thus
$$H_n(M;\pi) \rtarr H_n(M,M-x;\pi)\iso \pi$$
is a monomorphism for all coefficient groups $\pi$. In particular, by the universal
coefficient theorem,
$$H_n(M;\bZ)\ten \bZ_q \rtarr H_n(M,M-x;\bZ)\ten \bZ_q\iso \bZ_q$$
is a monomorphism for all positive integers $q$. If $H_n(M;\bZ)\neq 0$, then 
$H_n(M;\bZ)\iso \bZ$ with generator mapped to some multiple of a generator of 
$H_n(M,M-x;\bZ)$. By the mod $q$ monomorphism, the coefficient must be $\pm 1$. 
\end{proof}

As an aside, the corollary leads to a striking example of the failure of the naturality
of the splitting in the universal coefficient theorem. Consider a connected, compact,
non-orientable $n$-manifold $M$. Let $x\in M$ and write $M_x$ for the pair $(M,M-x)$. 
Since $M$ is $\bZ_2$-orientable, the middle vertical arrow in the following diagram is 
an isomorphism between copies of $\bZ_2$:
$$\diagram
0 \rto & H_n(M)\ten \bZ_2 \rto \dto_0 & H_n(M;\bZ_2) \rto \dto^{\iso} 
& \Tor_1^{\bZ}(H_{n-1}(M),\bZ_2) \rto \dto^0 & 0 \\
0 \rto & H_n(M_x)\ten \bZ_2 \rto & H_n(M_x;\bZ_2) \rto 
& \Tor_1^{\bZ}(H_{n-1}(M_x),\bZ_2) \rto & 0. \\
\enddiagram$$
Clearly $H_{n-1}(M,M-x)=0$, and the corollary gives that $H_n(M)=0$. Thus the left and
right vertical arrows are zero. If the splittings of the rows were natural, this would 
imply that the middle vertical arrow is also zero. 

\section{The proof of the vanishing theorem}

Let $M$ be an $n$-manifold, $n>0$. Take all homology groups with coefficients in a given
Abelian group $\pi$ in this section. We must prove the intuitively obvious statement that 
$H_i(M)=0$ for $i>n$ and the much more subtle statement that $H_n(M)=0$ if $M$ is
connected and is not compact. The last statement is perhaps the technical heart of our
proof of the Poincar\'e duality theorem. 

We begin with the general observation that homology is ``compactly 
supported''\index{compactly supported homology} in the
sense of the following result.

\begin{lem} For any space $X$ and element $x\in H_q(X)$, there is a compact subspace
$K$ of $X$ and an element $k\in H_q(K)$ that maps to $x$.
\end{lem}
\begin{proof}
Let $\ga: Y\rtarr X$ be a CW approximation of $X$ and let $x=\ga_*(y)$. If $y$ is represented 
by a cycle $z\in C_q(Y)$, then $z$, as a finite linear combination of $q$-cells, is an
element of $C_q(L)$ for some finite subcomplex $L$ of $Y$. Let $K=\ga(L)$ and let $k$ be 
the image of the homology class represented by $z$. Then $K$ is compact and $k$ maps to $x$. 
\end{proof}

We need two lemmas about open subsets of $\bR^n$ to prove the vanishing theorem, the first
of which is just a special case.

\begin{lem} If $U$ is open in $\bR^n$, then $H_i(U)=0$ for $i\geq n$.
\end{lem}
\begin{proof}
Let $s\in H_i(U)$, $i\geq n$. There is a compact subspace $K$ of $U$ and an element 
$k\in H_i(K)$ that maps to $s$. We may decompose $\bR^n$ as a CW complex whose $n$-cells 
are small $n$-cubes in such a way that there is a finite subcomplex $L$ of $\bR^n$ with 
$K\subset L\subset U$. (To be precise, use a cubical grid with small enough mesh.)  For $i>0$, 
the connecting homomorphisms $\pa$ are isomorphisms in the commutative diagram
$$\diagram
H_{i+1}(\bR^n,L) \rto \dto_{\pa} & H_{i+1}(\bR^n,U) \dto^{\pa} \\
H_i(L) \rto & H_i(U).\\
\enddiagram$$
Since $(\bR^n,L)$ has no relative $q$-cells for $q > n$, the groups on the left are zero for 
$i\geq n$. Since $s$ is in the image of $H_i(L)$, $s=0$. 
\end{proof}
\begin{lem}
Let $U$ be open in $\bR^n$. Suppose that $t\in H_n(\bR^n,U)$ maps to zero in $H_n(\bR^n,\bR^n-x)$
for all $x\in \bR^n-U$. Then $t=0$. 
\end{lem}
\begin{proof}
We prove the equivalent statement that if $s\in \tilde{H}_{n-1}(U)$ maps to zero in
$\tilde{H}_{n-1}(\bR^n-x)$ for all $x\in \bR^n-U$, then $s=0$. Choose a compact subspace
$K$ of $U$ such that $s$ is in the image of $\tilde{H}_{n-1}(K)$. Then $K$ is contained
in an open subset $V$ whose closure $\bar{V}$ is compact and contained in $U$, hence $s$
is the image of an element $r\in\tilde{H}_{n-1}(V)$. We claim that $r$ maps to zero in
$\tilde{H}_{n-1}(U)$, so that $s=0$. Of course, $r$ maps to zero in
$\tilde{H}_{n-1}(\bR^n-x)$ if $x\not\!\!{\in}\,U$. Let $T$ be an open contractible subset of $\bR^n$
such that $\bar{V}\subset T$ and $\bar{T}$ is compact. For example, $T$ could be a large
enough open cube. Let $L=T-(T\cap U)$. For each $x\in \bar{L}$, choose a closed cube $D$ that 
contains $x$ and is disjoint from $V$. A finite set $\sset{D_1,\ldots\!,D_q}$ of these 
cubes covers $\bar{L}$. Let $C_i=D_i\cap T$ and observe that $(\bR^n-D_i)\cap T = T-C_i$. 
We see by induction on $p$ that $r$ maps to zero in $\tilde{H}_{n-1}(T-(C_1\cup\cdots\cup C_p))$ 
for $0\leq p\leq q$. This is clear if $p=0$. For the inductive step, observe that
$$T-(C_1\cup\cdots\cup C_p) = (T-(C_1\cup\cdots\cup C_{p-1}))\cap (\bR^n-D_p)$$
and that $H_n((T-(C_1\cup\cdots\cup C_{p-1}))\cup (\bR^n-D_p)) = 0$ by the previous lemma.
Therefore the map
$$\tilde{H}_{n-1}(T-(C_1\cup\cdots\cup C_p))\rtarr 
\tilde{H}_{n-1}(T-(C_1\cup\cdots\cup C_{p-1}))\oplus \tilde{H}_{n-1}(\bR^n-D_p)$$
in the Mayer-Vietoris sequence is a monomorphism. Since $r\in\tilde{H}_{n-1}(V)$ maps to
zero in the two right-hand terms, by the induction hypothesis and the contractibility of
$D_p$ to a point $x\not\!\!{\in}\,U$, it maps to zero in the left-hand term. Since
$$V\subset T-(C_1\cup\cdots\cup C_q)\subset T\cap U\subset U,$$
this implies our claim that $r$ maps to zero in $\tilde{H}_{n-1}(U)$.
\end{proof}

\begin{proof}[Proof of the vanishing theorem]
Let $s\in H_i(M)$. We must prove that $s=0$ if $i>n$ and if $i=n$ when $M$ is connected
and not compact. Choose a compact subspace $K$ of $M$ such that $s$ is in the image of
$H_i(K)$. Then $K$ is contained in some finite union $U_1\cup\cdots\cup U_q$ of
coordinate charts, and it suffices to prove that $H_i(U_1\cup\cdots\cup U_q)=0$ for the
specified values of $i$. Inductively, using that $H_i(U)=0$ for $i\geq n$ when $U$ is
an open subset of a coordinate chart, it suffices to prove that $H_i(U\cup V)=0$ for
the specified values of $i$ when $U$ is a coordinate chart and $V$ is an open subspace
of $M$ such that $H_i(V)=0$ for the specified values of $i$. We have the Mayer-Vietoris 
sequence
$$H_i(U)\oplus H_i(V) \rtarr H_i(U\cup V) \rtarr \tilde{H}_{i-1}(U\cap V) \rtarr 
\tilde{H}_{i-1}(U)\oplus \tilde{H}_{i-1}(V).$$
If $i>n$, the vanishing of $H_i(U\cup V)$ follows immediately. Thus assume that $M$ is
connected and not compact and consider the case $i=n$. We have $H_n(U)=0$, $H_n(V)=0$,
and $\tilde{H}_{n-1}(U)=0$. It follows that $H_n(U\cup V)=0$ if and only if
$i_*: \tilde{H}_{n-1}(U\cap V)\rtarr \tilde{H}_{n-1}(V)$ is a monomorphism, where 
$i: U\cap V\rtarr V$ is the inclusion. 

We claim first that $H_n(M)\rtarr H_n(M,M-y)$ is the zero homomorphism for any $y\in M$. 
If $x\in M$ and $L$ is a path in $M$ connecting $x$ to $y$, then the diagram
$$\diagram
& & H_n(M,M-x) \\
H_n(M) \rto & H_n(M,M-L) \urto^{\iso} \drto_{\iso} & \\
& & H_n(M,M-y) \\
\enddiagram$$
shows that if $s\in H_n(M)$ maps to zero in $H_n(M,M-x)$, then it maps to zero in $H_n(M,M-y)$.
If $s$ is in the image of $H_n(K)$ where $K$ is compact, we may choose a point $x\in M-K$.
Then the map $K\rtarr M\rtarr (M,M-x)$ factors through $(M-x,M-x)$ and therefore $s$ maps to
zero in $H_n(M,M-x)$. This proves our claim.

Now consider the following diagram, where $y\in U- U\cap V$:
$$\diagram
& & H_n(U\cup V) \rto \dlto & H_n(M) \dto^{0}\\
H_n(V,U\cap V)\dto_{\pa} \rto & H_n(U\cup V, U\cap V)  \dlto^{\pa} \rrto & & H_n(M,M-y) \\
\tilde{H}_{n-1}(U\cap V) \dto_{i_*}  & H_n(U,U\cap V) \lto^{\pa} \rrto \uto
& & H_n(U,U-y) \uto_{\iso}\\
\tilde{H}_{n-1}(V). &  & & \\
\enddiagram$$
Let $r\in \ker\,i_*$. Since $\tilde{H}_{n-1}(U)=0$, the bottom map $\pa$ is an epimorphism
and there exists $s\in H_n(U,U\cap V)$ such that $\pa(s)=r$. We claim that $s$ maps to zero
in $H_n(U,U-y)$ for every $y\in U-(U\cap V)$. By the previous lemma, this will imply that
$s=0$ and thus $r=0$, so that $i_*$ is indeed a monomorphism. Since $i_*(r)=0$, there exists
$t\in H_n(V,U\cap V)$ such that $\pa(t)=r$. Let $s'$ and $t'$ be the images of $s$
and $t$ in $H_n(U\cup V,U\cap V)$. Then $\pa(s'-t')=0$, hence there exists $w\in H_n(U\cup V)$
that maps to $s'-t'$. Since $w$ maps to zero in $H_n(M,M-y)$, so does $s'-t'$. Since the map
$(V,U\cap V)\rtarr (M,M-y)$ factors through $(M-y,M-y)$, $t$ and thus also $t'$ maps to zero
in $H_n(M,M-y)$. Therefore $s'$ maps to zero in $H_n(M,M-y)$ and thus $s$ maps to zero in 
$H_n(U,U-y)$, as claimed.
\end{proof}

\section{The proof of the Poincar\'e duality theorem}

Let $M$ be an $R$-oriented $n$-manifold, not necessarily compact. Unless otherwise specified,
we take homology and cohomology with coefficients in a given $R$-module $\pi$ in this section.
Remember that homology
is a covariant functor with compact supports. Cohomology is a contravariant functor, and
it does not have compact supports. We would like to prove the Poincar\'e duality theorem
by inductive comparisons of Mayer-Vietoris sequences, and the opposite variance of 
homology and cohomology makes it unclear how to proceed. To get around this, we introduce
a variant of cohomology that does have compact supports and has enough covariant functoriality
to allow us to proceed by comparisons of Mayer-Vietoris sequences. 

Consider the set $\sK$ of compact subspaces $K$ of $M$. This set is directed under inclusion; to
conform with our earlier discussion of colimits, we may view $\sK$ as a category whose objects
are the compact subspaces $K$ and whose maps are the inclusions between them. We define
$$H^q_c(M) = \colim H^q(M,M-K),$$
where the colimit is taken with respect to the homomorphisms
$$H^q(M,M-K) \rtarr H^q(M,M-L)$$
induced by the inclusions $(M,M-L)\subset (M,M-K)$ for $K\subset L$. This is the cohomology of
$M$ with compact supports.\index{cohomology with compact 
supports}\index{compactly supported cohomology} Intuitively, thinking in 
terms of singular cohomology, its elements
are represented by cocycles that vanish off some compact subspace. 

A map $f:M\rtarr N$ is said to be proper if $f^{-1}(L)$ is compact in $M$ when $L$ is compact
in $N$. This holds, for example, if $f$ is the inclusion of a closed subspace.  For such $f$,
we obtain an induced homomorphism $f^*: H^*_c(N)\rtarr H^*_c(M)$ in an evident way. However,
we shall make no use of this contravariant functoriality.

What we shall use is a kind of covariant functoriality that will allow us to compare long
exact sequences in homology and cohomology. Explicitly, for an open subspace $U$ of $M$, we 
obtain a homomorphism $H^q_c(U)\rtarr H^q_c(M)$ by passage to colimits from the excision
isomorphisms 
$$ H^q(U,U-K) \rtarr H^q(M,M-K)$$
for compact subspaces $K$ of $U$. 

For each compact subspace $K$ of $M$, the $R$-orientation of $M$ determines a fundamental 
class $z_K\in H_n(M,M-K;R)$. Taking the relative cap product with $z_K$, we obtain a duality
homomorphism 
$$D_K: H^p(M,M-K) \rtarr H_{n-p}(M).$$
If $K\subset L$, the following diagram commutes:
$$\diagram
H^p(M,M-K) \rrto \drto_{D_K} & & H^p(M,M-L) \dlto^{D_L}\\
& H_{n-p}(M). & \\
\enddiagram$$
We may therefore pass to colimits to obtain a duality homomorphism 
$$D: H^p_c(M) \rtarr H_{n-p}(M).$$
If $U$ is open in $M$ and is given the induced $R$-orientation, then the following naturality
diagram commutes:
$$\diagram
 H^p_c(U) \dto  \rto^(0.43){D} & H_{n-p}(U) \dto \\
 H^p_c(M) \rto_(0.4){D} & H_{n-p}(M).\\
\enddiagram$$
If $M$ itself is compact, then $M$ is cofinal among the compact subspaces of $M$. Therefore
$H^p_c(M) = H^p(M)$, and the present duality map $D$ coincides with that of the Poincar\'e
duality theorem as originally stated. We shall prove a generalization to not necessarily 
compact manifolds.

\begin{thm}[Poincar\'e duality]\index{Poincare duality theorem@Poincar\'e duality theorem}
Let $M$ be an $R$-oriented $n$-manifold. Then \linebreak
$D: H^p_c(M)\rtarr H_{n-p}(M)$ is an isomorphism.
\end{thm}
\begin{proof}
We shall prove that $D: H^p_c(U)\rtarr H_{n-p}(U)$ is an isomorphism for every open
subspace $U$ of $M$. The proof proceeds in five steps.
\begin{proof}[Step 1] {\em The result holds for any coordinate chart $U$.}\\
We may take $U=M=\bR^n$. The compact cubes $K$ are cofinal among the compact subspaces of $\bR^n$. 
For such $K$ and for $x\in K$,
$$H^p(\bR^n,\bR^n-K)\iso H^p(\bR^n,\bR^n-x)\iso \tilde{H}^{p-1}(S^{n-1})\iso \tilde{H}^p(S^n).$$
The maps of the colimit system defining $H^p_c(\bR^n)$ are clearly isomorphisms. By the definition 
of the cap product, we see that $D: H^n(\bR^n,\bR^n-x)\rtarr H_0(\bR^n)$ is an isomorphism. 
Therefore $D_K$ is an isomorphism for every compact cube $K$ and so 
$D: H^n_c(\bR^n)\rtarr H_0(\bR^n)$ is an isomorphism.  
\end{proof}
\begin{proof}[Step 2] {\em If the result holds for open subspaces $U$ and $V$ and their 
intersection, then it holds for their union.}\\
Let $W=U\cap V$ and $Z=U\cup V$. The compact subspaces of $Z$ that are unions of a compact 
subspace $K$ of $U$ and a compact subspace $L$ of $V$ are cofinal among all of the compact
subspaces of $Z$. For such $K$ and $L$, we have the following commutative diagram with
exact rows. We let $J=K\cap L$ and $N=K\cup L$, and we write $U_K = (U,U-K)$, and so on, 
to abbreviate notation.
\begin{small}
$$\diagram
\rto & H^p(Z_J) \rto \dto_{\iso}& H^p(Z_K)\oplus H^p(Z_L) \rto \dto^{\iso}
&  H^p(Z_N) \rto \ddouble & H^{p+1}(Z_J) \rto \dto^{\iso} & \\
\rto & H^p(W_J) \rto \dto_D & H^p(U_K)\oplus H^p(V_L) \rto \dto^{D\oplus D}
&  H^p(Z_N) \rto \dto^D & H^{p+1}(W_J) \rto \dto^D & \\
\rto & H_{n-p}(W) \rto & H_{n-p}(U)\oplus H_{n-p}(V) \rto & H_{n-p}(Z) \rto & H_{n-p-1}(W)\rto &\\
\enddiagram$$
\end{small}
The top row is the relative Mayer-Vietoris sequence of the triad $(Z;Z-K,Z-L)$.
The middle row results from the top row by excision isomorphisms. The bottom row is the
absolute Mayer-Vietoris sequence of the triad $(Z;U,V)$. The left two squares commute by
naturality. The right square commutes by a diagram chase from the definition of the
cap product. The entire diagram is natural with respect to pairs $(K,L)$. We obtain a 
commutative diagram with exact rows on passage to colimits, and the conclusion follows
by the five lemma.
\end{proof}
\begin{proof}[Step 3] {\em If the result holds for each $U_i$ in a totally ordered set of
open subspaces $\sset{U_i}$, then it holds for the union $U$ of the $U_i$.}\\
Any compact subspace $K$ of $U$ is contained in a finite union of the $U_i$ and therefore
in one of the $U_i$. Since homology is compactly supported, it follows that 
$\colim H_{n-p}(U_i)\iso H_{n-p}(U)$. On the cohomology side, we have
\begin{eqnarray*}
\colim_i\,H^p_c(U_i) & = & \colim_i\colim_{\sset{K| K\subset U_i}} H^p(U_i,U_i-K) \\
& \iso & \colim_{\sset{K\subset U}}\colim_{\sset{i|K\subset U_i}} H^p(U_i,U_i-K) \\
& \iso & \colim_{\sset{K\subset U}} H^p(U,U-K) = H^p_c(U).
\end{eqnarray*}
Here the first isomorphism is an (algebraic) interchange of colimits isomorphism: both
composite colimits are isomorphic to $\colim H^p_c(U_i,U_i-K)$, where the colimit
runs over the pairs $(K,i)$ such that $K\subset U_i$. The second isomorphism holds 
since $\colim_{\sset{i|K\subset U_i}} H^p(U_i,U_i-K)\iso H^p(U,U-K)$ because the 
colimit is taken over a system of inverses of excision isomorphisms. The conclusion
follows since a colimit of isomorphisms is an isomorphism.
\end{proof}
\begin{proof}[Step 4] {\em The result holds if $U$ is an open subset of a coordinate 
neighborhood.}\\
We may take $M=\bR^n$. If $U$ is a convex subset of $\bR^n$, then $U$ is homeomorphic 
to $\bR^n$ and Step 1 applies. Since the intersection of two convex sets is convex,
it follows by induction from Step 2 that the conclusion holds for any finite union of
convex open subsets of $\bR^n$. Any open subset $U$ of $\bR^n$ is the union of countably
many convex open subsets. By ordering them and letting $U_i$ be the union of the first $i$,
we see that the conclusion for $U$ follows from Step 3.
\end{proof}
\begin{proof}[Step 5] {\em The result holds for any open subset $U$ of $M$}.\\
We may as well take $M=U$. By Step 3, we may apply Zorn's lemma to conclude that there is
a maximal open subset $V$ of $M$ for which the conclusion holds. If $V$ is not all of $M$,
say $x\not\in V$, we may choose a coordinate chart $U$ such that $x\in U$. 
By Steps 2 and 4, the result holds for $U\cup V$, contradicting the maximality of $V$.
\end{proof}
This completes the proof of the Poincar\'e duality theorem. \end{proof}

\section{The orientation cover}

There is an orientation cover\index{orientation cover} of a manifold that helps illuminate the 
notion of orientability. For the moment, we relax the requirement that the total
space of a cover be connected. Here we take homology with integer coefficients.

\begin{prop}
Let $M$ be a connected $n$-manifold. Then there is a $2$-fold cover $p:\tilde{M}\rtarr M$
such that $\tilde{M}$ is connected if and only if $M$ is not orientable.\index{orientable} 
\end{prop}
\begin{proof}
Define $\tilde{M}$ to be the set of pairs $(x,\al)$, where $x\in M$ and where
$\al\in H_n(M,M-x)\iso \bZ$ is a generator. Define $p(x,\al)=x$. If $U\subset M$ is
open and $\be\in H_n(M,M-U)$ is a fundamental class of $M$ at $U$, define
$$\langle U,\be\rangle = \sset{(x,\al)|x\in U \tand \be  \  \text{maps to}\  \al}.$$
The sets $\langle U,\be\rangle$ form a base for a topology on $\tilde{M}$. In fact, if 
$(x,\al)\in \langle U,\be\rangle \cap \langle V,\ga\rangle$, we can choose a coordinate
neighborhood $W\subset U\cap V$ such that $x\in W$. There is a unique class $\al'\in H_n(M,M-W)$
that maps to $\al$, and both $\be$ and $\ga$ map to $\al'$. Therefore
$$\langle W,\al'\rangle \subset \langle U,\be\rangle \cap \langle V,\ga\rangle.$$
Clearly $p$ maps $\langle U,\be\rangle$ homeomorphically onto $U$ and 
$$p^{-1}(U) = \langle U,\be\rangle \cup \langle U,-\be\rangle.$$
Therefore $\tilde{M}$ is an $n$-manifold and $p$ is a $2$-fold cover. Moreover, $\tilde{M}$
is oriented. Indeed, if $U$ is a coordinate chart and $(x,\al)\in \langle U,\be\rangle$,
then the following maps all induce isomorphisms on passage to homology:
$$\diagram
(\tilde{M},\tilde{M}-\langle U,\be\rangle) \dto  & (M,M-U) \dto \\
(\tilde{M},\tilde{M}-(x,\al)) & (M,M-x) \\
(\langle U,\be\rangle, \langle U,\be\rangle -(x,\al)) \uto \rto^(0.6)p_(0.6){\iso} & (U,U-x). \uto \\
\enddiagram$$
Via the diagram, $\be\in H_n(M,M-U)$ specifies an element 
$\tilde{\be}\in H_n(\tilde{M},\tilde{M}-\langle U,\be\rangle)$, and $\tilde{\be}$ is 
independent of the choice of $(x,\al)$. These classes are easily seen to specify an
orientation of $\tilde{M}$. Essentially by definition, an orientation of $M$ is a 
cross section $s: M \rtarr \tilde{M}$: if $s(U) = \langle U,\be\rangle$, then these
$\be$ specify an orientation. Given one section $s$, changing the signs of the $\be$
gives a second section $-s$ such that $\tilde{M}= \im(s)\amalg \im(-s)$, showing that
$\tilde{M}$ is not connected if $M$ is oriented. 
\end{proof}

The theory of covering spaces gives the following consequence.

\begin{cor} If $M$ is simply connected, or if $\pi_1(M)$ contains no subgroup of
index $2$, then $M$ is orientable. If $M$ is orientable, then $M$ admits exactly
two orientations.
\end{cor}
\begin{proof}
If $M$ is not orientable, then $p_*(\pi_1(\tilde{M}))$ is a subgroup of $\pi_1(M)$
of index $2$. This implies the first statement, and the second statement is clear.
\end{proof}

We can use homology with coefficients in a commutative ring $R$ to construct an
analogous $R$-orientation cover.\index{Rorientation cover@$R$-orientation cover} It 
depends on the units of $R$. For
example, if $R=\bZ_2$, then the $R$-orientation cover is the identity map of $M$
since there is a unique unit in $R$. This reproves the obvious fact that any 
manifold is $\bZ_2$-oriented. The evident ring homomorphism $\bZ\rtarr R$ induces
a natural homomorphism $H_*(X;\bZ)\rtarr H_*(X;R)$, and we see immediately that
an orientation of $M$ induces an $R$-orientation of $M$ for any $R$.

\vspace{.1in}

\begin{center}
PROBLEMS
\end{center}
\begin{enumerate}
\item  Prove: there is no homotopy equivalence  
$f: \bC P^{2n}\rtarr \bC P^{2n}$ that reverses orientation 
(induces multiplication by $-1$ on  $H_{4n}(\bC P^{2n})$).
\end{enumerate}

In the problems below,  $M$ is assumed to be a compact connected 
$n$-manifold (without boundary), where  $n\geq 2$. 
\begin{enumerate}
\item[2.]  Prove that if  $M$ is a Lie group, then $M$ is orientable.
\item[3.]  Prove that if $M$ is orientable, then  $H_{n-1}(M; \bZ)$  is a free Abelian group.
\item[4.]  Prove that if  $M$ is not orientable, then the torsion subgroup of  $H_{n-1}(M; \bZ)$ is 
cyclic of order  $2$  and  $H_n(M; \bZ_q)$  is zero if  $q$  is odd and is cyclic 
of order  $2$ if  $q$  is even. (Hint: use universal coefficients and the transfer
homomorphism of the orientation cover.)
\item[5.]  Let  $M$ be oriented with fundamental class  $z$.  Let  $f: S^n\rtarr M$ be a 
map such that  $f_*(i_n) = qz$,  where  $i_n \in H_n(S^n; \bZ)$ is the fundamental 
class and $q \neq 0$.  
\begin{enumerate}
\item[(a)] Show that  $f_*: H_*(S^n; \bZ_p) \rtarr H_*(M; \bZ_p)$ is an isomorphism 
if $p$ is a prime that does not divide $q$.
\item[(b)]  Show that multiplication by  $q$  annihilates  $H_i(M; \bZ)$  if  
$1 \leq  i \leq n-1$.
\end{enumerate}
\item[6.]
\begin{enumerate}
\item[(a)]  Let  $M$ be a compact $n$-manifold.  Suppose that $M$ is 
homotopy equivalent to  $\SI Y$ for some connected based space $Y$.  Deduce that $M$  
has the same integral homology groups as  $S^n$.  (Hint: use the vanishing of cup products
on $\tilde H^*(\SI Y)$ and Poincar\'e duality, treating the cases $M$ orientable 
and $M$ non-orientable separately.)
\item[(b)] Deduce that $M$ is homotopy equivalent to $S^n$. Does it follow that $Y$ is homotopy
equivalent to $S^{n-1}$? 
\end{enumerate}
\item[7.]* Essay: The singular cohomology  $H^*(M;\bR)$ is isomorphic to the de Rham 
cohomology of $M$.  Why is this plausible?  Sketch proof?
\end{enumerate}

\clearpage

\thispagestyle{empty}

\chapter{The index of manifolds; manifolds with boundary}

The Poincar\'e duality theorem imposes strong constraints on the Euler 
characteristic of a manifold. It also leads to new invariants, most
notably the index. Moreover, there is a relative version of Poincar\'e 
duality in the context of manifolds with boundary, and this leads to
necessary algebraic conditions on the cohomology of a manifold that must be 
satisfied if it is to be a boundary. In particular, the index of a compact
oriented $4n$-manifold $M$ is zero if $M$ is a boundary.  We shall later 
outline the theory of cobordism, which leads to necessary {\em and sufficient} 
algebraic conditions for a manifold to be a boundary.

\section{The Euler characteristic of compact manifolds}

The Euler characteristic\index{Euler characteristic!of a space}  $\ch (X)$ of a space with 
finitely generated homology is defined by
$$\ch (X) = \textstyle{\sum}_i (-1)^i \ \text{rank}\ H_i(X;\bZ).$$
The universal coefficient theorem implies that
$$\ch (X) = \textstyle{\sum}_i (-1)^i \dim H_i(X;F)$$
for {\em any} field of coefficients $F$. Examination of the relevant short
exact sequences shows that 
$$\ch (X) = \textstyle{\sum}_i (-1)^i \ \text{rank}\  C_i(X;\bZ)$$
for {\em any} decomposition of $X$ as a finite CW complex. The verifications
of these statements are immediate from earlier exercises.

Now consider a compact oriented $n$-manifold. Recall that we take it for granted
that $M$ can be decomposed as a finite CW complex, so that each $H_i(M;\bZ)$ is 
finitely generated. By the universal coefficient theorem and Poincar\'e duality,
we have 
$$H_i(M;F)\iso H^i(M;F)\iso H_{n-i}(M;F)$$
for any field $F$. We may take $F=\bZ_2$, and so dispense with the requirement 
that $M$ be oriented. If $n$ is odd, the summands of $\ch(M)$ cancel in pairs,
and we obtain the following conclusion.

\begin{prop} If $M$ is a compact manifold of odd dimension, then $\ch(M)=0$.
\end{prop}

If $n=2m$ and $M$ is oriented, then
$$\ch(M) = \textstyle{\sum}_{i=0}^{m-1} (-1)^i 2 \dim H_i(M) + (-1)^m \dim H_m(M)$$
for any field $F$ of coefficients. Let us take $F=\bQ$. Of course, we can replace
homology by cohomology in the definition and formulas for $\ch(M)$. The middle
dimensional cohomology group $H^m(M)$ plays a particularly important role. Recall
that we have the cup product pairing\index{cup product pairing}
$$\ph: H^m(M)\ten H^m(M) \rtarr \bQ$$
specified by $\ph(\al,\be) = \langle\al\cup\be,z\rangle$.  This pairing is nonsingular.
Since $\al\cup \be =(-1)^m\be\cup\al$, it is skew symmetric if $m$ is odd and is 
symmetric if $m$ is even. When $m$ is odd, we obtain the following conclusion.

\begin{prop}
If $M$ is a compact oriented $n$-manifold, where $n\equiv 2\ \text{mod}\ 4$, then
$\ch(M)$ is even.
\end{prop}
\begin{proof} It suffices to prove that $\dim H^{m}(M)$ is even, where $n=2m$, and this is 
immediate from the following algebraic observation.
\end{proof}

\begin{lem}
Let $F$ be a field of characteristic $\neq 2$, $V$ be a finite dimensional vector 
space over $F$, and  $\ph: V\times V \rtarr F$ be a nonsingular skew symmetric bilinear form. 
Then $V$ has a basis $\sset{x_1,\ldots\!,x_r, y_1,\ldots\!,y_r}$ such that $\ph(x_i,y_i)=1$ for 
$1\leq i\leq r$ and $\ph(z,w)=0$ for all other pairs of basis elements $(z,w)$. Therefore 
the dimension of $V$
is even.
\end{lem}
\begin{proof}
We proceed by induction on $\dim V$, and we may assume that $V\neq 0$. Since 
$\ph(x,y)=-\ph(y,x)$, $\ph(x,x)=0$ for all 
$x\in V$. Choose $x_1\neq 0$. Certainly there exists $y_1$ such that $\ph(x_1,y_1)=1$, 
and $x_1$ and $y_1$ are then linearly independent. Define
$$W=\sset{x|\ph(x,x_1) = 0 \tand \ph(x,y_1)=0}\subset V.$$
That is, $W$ is the kernel of the homomorphism $\ps: V\rtarr F\times F$ specified
by $\ps(x)= (\ph(x,x_1),\ph(x,y_1))$. Since 
$\ps(x_1)=(0,1)$ and $\ps(y_1)=(-1,0)$, $\ps$ is an epimorphism. Thus $\dim W = \dim V -2$. 
Since $\ph$ restricts to a nonsingular skew symmetric bilinear form on $W$,
the conclusion follows from the induction hypothesis.
\end{proof}

\section{The index of compact oriented manifolds}

To study manifolds of dimension $4k$, we consider an analogue for symmetric bilinear
forms of the previous algebraic lemma. Since we will need to take square roots, we
will work over $\bR$.

\begin{lem}
Let $V$ be a finite dimensional real vector space and $\ph: V\times V \rtarr \bR$ be a 
nonsingular symmetric bilinear form. Define $q(x)=\ph(x,x)$. Then $V$ has a basis 
$\sset{x_1,\ldots\!,x_r, y_1,\ldots\!,y_s}$ such that $\ph(z,w)=0$ for all
pairs $(z,w)$ of distinct basis elements, $q(x_i)=1$ for $1\leq i\leq r$ 
and $q(y_j)=-1$ for $1\leq j\leq s$. The number $r-s$ is an invariant of $\ph$, called the 
signature\index{signature} of $\ph$.
\end{lem}
\begin{proof}
We proceed by induction on $\dim V$, and we may assume that $V\neq 0$. Clearly $q(rx)=r^2q(x)$. 
Since we can take square roots in $\bR$, we can choose $x_1\in V$ such that $q(x_1)=\pm 1$. Define 
$\ps: V\rtarr \bR$ by $\ps(x)=\ph(x,x_1)$ and let $W=\ker \ps$. Since $\ps(x_1)=\pm 1$, $\ps$
is an epimorphism and $\dim W=\dim V-1$. Since $\ph$ restricts to a nonsingular symmetric bilinear 
form on $W$, the existence of a basis as specified follows directly from the induction hypothesis.
Invariance means that the integer $r-s$ is independent of the choice of basis on which 
$q$ takes values $\pm 1$, and we leave the verification to the reader.
\end{proof}

\begin{defn} Let $M$ be a compact oriented $n$-manifold. If $n=4k$, define the index\index{index} 
of $M$, denoted $I(M)$, to be the signature of the cup product form
$H^{2k}(M;\bR)\ten H^{2k}(M;\bR)\rtarr \bR$. If $n\,\not\!\equiv\,0 \ \text{mod}\ 4$, define 
$I(M)=0$. 
\end{defn}

The Euler characteristic and index are related by the following congruence.

\begin{prop} For any compact oriented $n$-manifold, $\ch(M)\equiv I(M)\ \text{mod}\ 2$.
\end{prop}
\begin{proof} If $n$ is odd, then $\ch(M)=0$ and $I(M)=0$. If $n\equiv 2\ \text{mod}\ 4$,
then $\ch(M)$ is even and $I(M)=0$. If $n=4k$, then $I(M) = r-s$,
where $r+s = \dim H^{2k}(M;\bR) \equiv \ch(M)\ \text{mod}\ 2$. 
\end{proof}

Observe that the index of $M$ changes sign if the orientation of $M$ is reversed. We
write $-M$ for $M$ with the reversed orientation, and then $I(-M)=-I(M)$.
We also have the following algebraic identities. Write $H^*(M)=H^*(M;\bR)$.

\begin{lem} If $M$ and $M'$ are compact oriented $n$-manifolds, then 
$$I(M\amalg M')=I(M)+I(M'),$$
where $M\amalg M'$ is given the evident orientation induced from those of $M$ and $M'$.
\end{lem}
\begin{proof}
There is nothing to prove unless $n=4k$, in which case 
$$H^{2k}(M\amalg M')=H^{2k}(M)\times H^{2k}(M').$$
Clearly the cup product of an element of $H^*(M)$ with an element of $H^*(M')$ is zero, and
the cup product form on $H^{2k}(M\amalg M')$ is given by 
$$\ph((x,x'),(y,y')) = \ph(x,y)+\ph(x',y')$$
for $x,y\in H^{2k}(M)$ and $x',y'\in H^{2k}(M')$. The conclusion follows since the signature of
a sum of forms is the sum of the signatures.
\end{proof}

\begin{lem} Let $M$ be a compact oriented $m$-manifold and $N$ be a compact oriented $n$-manifold.
Then
$$I(M\times N)=I(M)\cdot I(N),$$
where $M\times N$ is given the orientation induced from those of $M$ and $N$.
\end{lem}
\begin{proof}
We must first make sense of the induced orientation on $M\times N$. For CW pairs $(X,A)$ and $(Y,B)$,
we have an identification of CW complexes
$$ (X\times Y)/(X\times B\cup A\times Y)\iso (X/A)\sma (Y/B)$$
and therefore an isomorphism
$$ C_*(X\times Y,\, X\times B\cup A\times Y)\iso C_*(X,A)\ten C_*(Y,B).$$
This implies a relative K\"{u}nneth theorem\index{Kunneth 
theorem@K\"unneth theorem!relative} for arbitrary pairs $(X,A)$ and $(Y,B)$. For 
subspaces $K\subset M$ and $L\subset N$,
$$(M\times N, M\times N-K\times L) = (M\times N, M\times (N-L) \cup (M-K)\times N).$$
In particular, for points $x\in M$ and $y\in Y$,
$$(M\times N, M\times N-(x,y)) = (M\times N, M\times (N-y) \cup (M-x)\times N).$$
Therefore fundamental classes $z_K$ of $M$ at $K$ and $z_L$ of $N$ at $L$ determine a 
fundamental class $z_{K\times L}$ of $M\times N$ at $K\times L$. In particular, the image
under $H_m(M)\ten H_n(N)\rtarr H_{m+n}(M\times N)$ of the tensor product of fundamental
classes of $M$ and $N$ is a fundamental class of $M\times N$. 

Turning to the claimed product formula, we see that there is nothing to prove
unless $m+n=4k$, in which case
$$H^{2k}(M\times N)=\sum_{i+j=2k} H^i(M)\ten H^j(N).$$
The cup product form is given by
$$\ph(x\ten y, x'\ten y')
= (-1)^{(\deg y)(\deg x')+ mn}\langle x\cup x',z_M\rangle \langle y\cup y',z_N\rangle$$
for $x, x'\in H^*(M)$ and $y,y'\in H^*(N)$. If $m$ and $n$ are odd, then the signature
of this form is zero. If $m$ and $n$ are even, then this form is the sum of the tensor 
product of the cup product forms on the middle dimensional cohomology groups of $M$ and $N$ 
and a form
whose signature is zero. Here, if $m$ and $n$ are congruent to 2 mod 4, the signature
is zero since the lemma of the previous section implies that the signature of the tensor 
product of two skew symmetric forms is zero. When $m$ and $n$ are congruent to 0 mod
4, the conclusion holds since the signature of the tensor product of two symmetric forms 
is the product of their signatures. We leave the detailed verifications of these algebraic
statements as exercises for the reader.
\end{proof}

\section{Manifolds with boundary}

Let $\bH^n=\sset{(x_1,\ldots\!,x_n)|x_n\geq 0}$ be the upper half-plane in $\bR^n$. Recall that an
$n$-manifold with boundary\index{manifold with boundary} is a Hausdorff space $M$ having a 
countable basis of open sets
such that every point of $M$ has a neighborhood homeomorphic to an open subset of $\bH^n$. 
A point $x$ is an interior point if it has a neighborhood homeomorphic to an open subset
of $\bH^n-\pa \bH^n\iso \bR^n$; otherwise it is a boundary point. It is a fact called 
``invariance of domain''\index{invariance of domain} that if $U$ and $V$ are homeomorphic 
subspaces of $\bR^n$ and 
$U$ is open, then $V$ is open. Therefore, a homeomorphism of an open subspace of 
$\bH^n$ onto an open subspace of $\bH^n$ carries boundary points to boundary points. 

We denote the boundary\index{boundary of a manifold} of an $n$-manifold $M$ by $\pa M$. Thus 
$M$ is a manifold without boundary if $\pa M$ is empty; $M$  is said to 
be closed\index{closed manifold} if, in addition, 
it is compact.  The space $\pa M$ is an $(n-1)$-manifold without boundary. 

It is a fundamental question in topology to determine which closed manifolds are boundaries.
The question makes sense with varying kinds of extra structure. For example, we can ask whether 
or not a smooth (= differentiable) closed manifold is the boundary of a smooth manifold (with 
the induced smooth structure). Numerical invariants in algebraic topology give criteria. One 
such criterion is given by the following consequence of the Poincar\'e duality theorem.
Remember that $\ch(M)=0$ if $M$ is a closed manifold of odd dimension. 

\begin{prop} If $M=\pa W$, where $W$ is a compact $(2m+1)$-manifold, then $\ch(M)=2\ch (W)$.
\end{prop}
\begin{proof}
The product $W\times I$ is a $(2m+2)$-manifold with
$$\pa(W\times I) = (W\times \sset{0}) \cup (M\times I) \cup (W\times \sset{1}).$$
Let $U=\pa(W\times I)-(W\times \sset{1})$ and $V=\pa(W\times I)-(W\times \sset{0})$.
Then $U$ and $V$ are open subsets of $\pa(W\times I)$. Clearly $U$ and $V$
are both homotopy equivalent to $W$ and $U\cap V$ is homotopy equivalent to $M$.
We have the Mayer-Vietoris sequence
$$\diagram
 H_{i+1}(U\cup V) \rto \ddouble & H_i(U\cap V)\rto \dto^{\iso} & H_i(U)\oplus H_i(V) 
\rto \dto^{\iso} & H_i(U\cup V) \ddouble\\
 H_{i+1}(\pa(W\times I)) \rto & H_i(M)\rto  & H_i(W)\oplus H_i(W) 
\rto & H_i(\pa(W\times I)).\\
\enddiagram$$
Therefore $2\ch(W)=\ch(M)+\ch(\pa(W\times I))$. However, $\ch(\pa(W\times I))=0$ since
$\pa(W\times I)$ is a closed manifold of odd dimension.
\end{proof}

\begin{cor} If $M=\pa W$ for a compact manifold $W$, then $\ch(M)$ is even.
\end{cor}

For example, since $\ch(\bR P^{2m})=1$ and $\ch(\bC P^n)=n+1$, this criterion
shows that $\bR P^{2m}$ and $\bC P^{2m}$ cannot be boundaries. Notice that
we have proved that these are not boundaries of topological manifolds, let
alone of smooth ones.

\section{Poincar\'e duality for manifolds with boundary}

The index gives a more striking criterion: if a closed oriented $4k$-manifold $M$ is the 
boundary of a (topological) manifold, then $I(M)=0$. To prove this,  we must first obtain
a relative form of the Poincar\'e duality theorem applicable to manifolds with boundary.

We let $M$ be an $n$-manifold with boundary, $n>0$, throughout this section, and we let
$R$ be a given commutative ring. We say that $M$ 
is $R$-orientable\index{Rorientable@$R$-orientable}
(or orientable\index{orientable} if $R=\bZ$) 
if its interior $\cir{M}=M-\pa M$ is $R$-orientable; similarly, an 
$R$-orientation\index{Rorientation@$R$-orientation} of $M$ is an 
$R$-orientation of its interior. To study these notions, we shall need the following 
result, which is intuitively clear but is somewhat technical to prove. In the case of
smooth manifolds, it can be seen in terms of inward-pointing unit vectors of the normal
line bundle of the embedding $\pa M\rtarr M$. 

\begin{thm}[Topological collaring]\index{topological collar}  There is an open neighborhood 
$V$ of $\pa M$ in $M$ such that the identification $\pa M = \pa M\times \sset{0}$ extends 
to a homeomorphism $V\iso \pa M\times [0,1)$. 
\end{thm}

It follows that the inclusion $\cir{M}\rtarr M$ is a homotopy equivalence and the 
inclusion $\pa M\rtarr M$ is a cofibration. We take homology with coefficients 
in $R$ in the next two results.

\begin{prop}
An $R$-orientation of $M$ determines an $R$-orientation of $\pa M$. 
\end{prop}
\begin{proof}
Consider a coordinate chart $U$ of a point $x\in \pa M$. If $\dim M= n$, then $U$ is 
homeomorphic to an open half-disk in $\bH^n$. Let $V=\pa U = U\cap \pa M$ and let
$y\in \cir{U}=U-V$. We have the following chain of isomorphisms:
\begin{eqnarray*}
H_n(\cir{M},\cir{M}-\cir{U}) & \iso & H_n(\cir{M},\cir{M}-y) \\
& \iso & H_n(M,M-y)  \\
& \iso & H_n(M,M-\cir{U})\\
& \overto{\pa} & H_{n-1}(M-\cir{U},M-U) \\
& \iso & H_{n-1}(M-\cir{U},(M-\cir{U})-x)\\
& \iso & H_{n-1}(\pa M,\pa M-x)\\
& \iso & H_{n-1}(\pa M,\pa M-V).
\end{eqnarray*}
The first and last isomorphisms are restrictions of the sort that enter into the 
definition of an $R$-orientation, and the third isomorphism is similar. We see by
use of a small boundary collar that the inclusion $(\cir{M},\cir{M}-y) \rtarr (M,M-y)$ 
is a homotopy equivalence, and that gives the second isomorphism. The connecting 
homomorphism is that of the triple $(M,M-\cir{U},M-U)$ and is an isomorphism since
$H_*(M,M-U)\iso H_*(M,M)=0$. The isomorphism that follows comes from the observation that
the inclusion $(M-\cir{U})-x \rtarr M-U$ is a homotopy equivalence, and the next to last
isomorphism is given by excision of $\cir{M}-\cir{U}$. The conclusion is an easy consequence 
of these isomorphisms.
\end{proof}

\begin{prop} If $M$ is compact and $R$-oriented and $z_{\pa M}\in H_{n-1}(\pa M)$ is 
the fundamental class determined by the induced $R$-orientation on $\pa M$, then there 
is a  unique element $z\in H_n(M,\pa M)$ such that $\pa z = z_{\pa M}$; $z$ is called
the $R$-fundamental class\index{Rfundamental class@$R$-fundamental class} determined 
by the $R$-orientation of $M$.
\end{prop}
\begin{proof}
Since $\cir{M}$ is a non-compact manifold without boundary and $\cir{M}\rtarr M$ is a
homotopy equivalence, $H_n(M)\iso H_n(\cir{M})=0$ by the vanishing theorem. Therefore
$\pa: H_n(M,\pa M)\rtarr H_{n-1}(\pa M)$
is a monomorphism. Let $V$ be a 
boundary collar and let $N=M-V$. Then $N$ is a closed subspace and a deformation
retract of the $R$-oriented open manifold $\cir{M}$, and we have
$$H_n(\cir{M},\cir{M}-N)\iso H_n(M,M-\cir{M}) = H_n(M,\pa M).$$
Since $M$ is compact, $N$ is a compact subspace of $\cir{M}$. Therefore the $R$-orientation 
of $\cir{M}$ determines a fundamental class in $H_n(\cir{M},\cir{M}-N)$. Let $z$ be its
image in $H_n(M,\pa M)$. Then $z$ restricts to a generator of
$H_n(M,M-y)\iso H_n(\cir{M},\cir{M}-y)$ for every $y\in \cir{M}$. Via naturality diagrams and the
chain of isomorphisms in the previous proof, we see that $\pa z$ restricts to a 
generator of $H_{n-1}(\pa M,\pa M-x)$ for all $x\in \pa M$ and is the fundamental
class determined by the $R$-orientation of $\pa M$.
\end{proof}

\begin{thm}[Relative Poincar\'e duality]\index{Poincare duality theorem@Poincar\'e duality
theorem!relative} Let $M$ be a compact $R$-oriented $n$-\linebreak
manifold
with $R$-fundamental class $z\in H_n(M,\pa M;R)$. Then, with coefficients taken in
any $R$-module $\pi$, capping with $z$ specifies duality isomorphisms
$$ D: H^p(M,\pa M)\rtarr H_{n-p}(M) \ \ \tand \ \ D:H^p(M)\rtarr H_{n-p}(M,\pa M).$$
\end{thm}
\begin{proof}
The following diagram commutes by inspection of definitions:
$$\diagram
H^{p-1}(\pa M) \rto \dto_D & H^p(M,\pa M) \rto \dto^D 
& H^p(M) \rto \dto^D & H^p(\pa M) \dto^D\\
H_{n-p}(\pa M) \rto & H_{n-p}(M) \rto 
& H_{n-p}(M,\pa M) \rto & H_{n-p-1}(\pa M).\\
\enddiagram$$
Here $D$ for $\pa M$ is obtained by capping with $\pa z$ and is an isomorphism.
By the five lemma, it suffices to prove that $D: H^p(M)\rtarr H_{n-p}(M,\pa M)$
is an isomorphism. To this end, let $N = M\cup_{\pa M} M$ be the ``double''\index{double
of a manifold} of 
$M$ and let $M_1$ and $M_2$ be the two copies of $M$ in $N$. Clearly $N$ is a
compact manifold without boundary, and it is easy to see that $N$ inherits an
$R$-orientation from the orientation on $M_1$ and the negative of the orientation
on $M_2$. Of course, $\pa M = M_1\cap M_2$. If $U$ is the union of $M_1$ and a 
boundary collar in $M_2$ and $V$ is the union of $M_2$ and a boundary collar in
$M_1$, then we have a Mayer-Vietoris sequence for the triad $(N;U,V)$. Using the 
evident equivalences of $U$ with $M_1$, $V$ with $M_2$, and $U\cap V$ with $\pa M$,
this gives the exact sequence in the top row of the following commutative diagram.
The bottom row is the exact sequence of the pair $(N,\pa M)$, and the isomorphism 
results from the homeomorphism $N/\pa M\iso (M_1/\pa M) \vee (M_2/\pa M)$; we 
abbreviate $N_1=(M_1,\pa M)$ and $N_2=(M_2,\pa M)$:
$$\diagram
H^p(N) \rto \dto_D & H^p(M_1) \oplus H^p(M_2) \rto^{\ps} \dto^{D\oplus D} 
& H^p(\pa M) \dto^D \rto^{\DE} & H^{p+1}(N) \dto^D \\
H_{n-p}(N) \rto \ddouble & H_{n-p}(N_1)\oplus H_{n-p}(N_2) \rto \dto^{\iso}
& H_{n-p-1}(\pa M) \ddouble \rto & H_{n-p-1}(N) \ddouble \\
H_{n-p}(N) \rto & H_{n-p}(N,\pa M) \rto 
& H_{n-p-1}(\pa M)  \rto & H_{n-p-1}(N). \\
\enddiagram$$
The top left square commutes by naturality. In the top middle square, we have
$\ps(x,y)=i_1^*(x)-i_2^*(y)$, where $i_1: \pa M\rtarr M_1$ and $i_2: \pa M\rtarr M_2$
are the inclusions. Since $D$ for $M_2$ is the negative of $D$ for $M_1$ under the
identifications with $M$, the commutativity of this square follows from the relation
$D\com i^* = \pa\com D: H^p(M)\rtarr H_{n-p-1}(\pa M)$, $i: \pa M\rtarr M$, which holds
by inspection of definitions. For the top right square, $\DE$ is the the top composite
in the diagram
$$\diagram
H^p(\pa M)\rto^(0.3){\de} \dto_D & H^{p+1}(M_1,\pa M)\iso H^{p+1}(N,M_2)\dto^D \rto & H^{p+1}(N)\dto^D \\
H_{n-p-1}(\pa M) \rto_{{i_1}_*} & H_{n-p-1}(M_1) \rto & H_{n-p-1}(N).
\enddiagram$$
The right square commutes by naturality, and $D\com \de = {i_1}_*\com D$ by inspection
of definitions. By the five lemma, since the duality maps $D$ for $N$ and $\pa M$ are 
isomorphisms, both maps $D$ between direct summands must be isomorphisms. The conclusion follows.
\end{proof}

\section{The index of manifolds that are boundaries}

We shall prove the following theorem.

\begin{thm} If $M$ is the boundary of a compact oriented $(4k+1)$-manifold, then
$I(M)=0$.
\end{thm}

We first give an algebraic criterion for the vanishing of the signature of a form
and then show that the cup product form on the middle dimensional cohomology of $M$ 
satisfies the criterion. 

\begin{lem} Let $W$ be a $n$-dimensional subspace of a $2n$-dimensional real vector
space $V$. Let $\ph: V\times V\rtarr \bR$ be a nonsingular symmetric bilinear form
such that $\ph: W\times W\rtarr \bR$ is identically zero. Then the signature of $\ph$
is zero.
\end{lem}
\begin{proof}
Let $r$ and $s$ be as in the definition of the signature. Then $r+s=2n$ and we must 
show that $r=s$. We prove that $r\geq n$. Applied to the form $-\ph$, this will also
give that $s\geq n$, implying the conclusion. We proceed by induction on $n$. Let
$\sset{x_1,\ldots\!,x_n,z_1,\ldots\!,z_n}$ be a basis for $V$, where $\sset{x_1,\ldots\!,x_n}$
is a basis for $W$. Define $\tha: V\rtarr \bR^n$ and $\ps: V\rtarr \bR^n$ by 
$$\tha(x) = (\ph(x,x_1),\ldots\!,\ph(x,x_n)) \  \tand \ \ps(x) = (\ph(x,z_1),\ldots\!,\ph(x,z_n)).$$
Since $\ph$ is nonsingular, $\ker{\tha}\cap\ker{\ps}=0$. Since $\ker{\tha}$ and $\ker{\ps}$ each
have dimension at least $n$, neither can have dimension more than $n$ and $\tha$ and $\ps$ must
both be epimorphisms. Choose $y_1$ such that $\tha(y_1)=(1,0,\ldots\!,0)$. Let 
$q(x)=\ph(x,x)$ and note that $q(x)=0$ if $x\in W$. Since $q(x_1)=0$ and $\ph(x_1,y_1)=1$,
$q(ax_1+y_1) = 2a+q(y_1)$ for $a\in\bR$. Taking $a=(1-q(y_1))/2$, we find $q(ax_1+y_1)=1$. If 
$n=1$, this gives $r\geq 1$ and completes the proof. If $n>1$, define $\om: V\rtarr \bR^2$ by
$\om(x)=(\ph(x,x_1),\ph(x,y_1))$. Since $\om(x_1)=(0,1)$ and $\om(y_1)=(1,q(y_1))$, $\om$ is
an epimorphism. Let $V'=\ker\om$ and let $W'\subset V'$ be the span of $\sset{x_2,\ldots\!,x_n}$.
The restriction of $\ph$ to $V'$ satisfies the hypothesis of the lemma, and the induction
hypothesis together with the construction just given imply that $r\geq n$. 
\end{proof}

Take homology and cohomology with coefficients in $\bR$.

\begin{lem} Let $M=\pa W$, where $W$ is a compact oriented $(4k+1)$-manifold,
and let $i: M\rtarr W$ be the inclusion. Let 
$\ph: H^{2k}(M)\ten H^{2k}(M)\rtarr \bR$ be the cup product form.  Then the image 
of $i^*: H^{2k}(W)\rtarr H^{2k}(M)$ is a subspace of half the dimension of $H^{2k}(M)$ 
on which $\ph$ is identically zero.
\end{lem}
\begin{proof}
Let $z\in H_{4k+1}(W,M)$ be the fundamental class. For $\al,\be\in H^{2k}(W)$, 
$$\ph(i^*(\al),i^*(\be))=\langle i^*(\al\cup\be),\pa z\rangle 
=\langle \al\cup\be,i_*\pa z\rangle =0$$
since $i_*\pa=0$ by the long exact sequence of the pair $(W,M)$. Thus $\ph$ is
identically zero on $\im i^*$. The commutative diagram with exact rows
$$\diagram
H^{2k}(W) \rto^{i^*} \dto_D  & H^{2k}(M) \rto^{\de} \dto^D & H^{2k+1}(W,M) \dto^D\\
H_{2k+1}(W,M) \rto_{\pa} & H_{2k}(M) \rto_{i_*} & H_{2k}(W)
\enddiagram$$
implies that $H^{2k}(M)\iso \im i^*\oplus \im\de\iso \im i^*\oplus \im i_*$. Since
$i^*$ and $i_*$ are dual homomorphisms, $\im i^*$ and $\im i_*$ are dual vector spaces
and thus have the same dimension.
\end{proof}

\vspace{.1in}

\begin{center}
PROBLEMS
\end{center}

Let  $M$  be a compact connected $n$-manifold with boundary $\pa M$, where $n\geq 2$.
\begin{enumerate}
\item  Prove:  $\pa M$ is not a retract of $M$.
\item  Prove: if  $M$  is contractible, then  $\pa M$  has the homology of a sphere.
\item  Assume that  $M$  is orientable.  Let  $n = 2m+1$  and let  $K$ be the kernel of the 
homomorphism  $H_m(\pa M) \rtarr H_m(M)$ induced by the inclusion, where homology is taken 
with coefficients in a field.  Prove:  $\dim<EFBFBD><EFBFBD>\,H_m(\pa M)<EFBFBD>=<EFBFBD>2\dim<EFBFBD><EFBFBD>\,K$.
\end{enumerate}

Let  $n = 3$  in the rest of the problems.

\begin{enumerate}
\item[4.] Prove: if  $M$  is orientable,  $\pa M$  is empty, and  $H_1(M; \bZ) = 0$, then  $M$  has 
the same homology groups as a  $3$-sphere.
\item[5.]  Prove: if  $M$  is nonorientable and  $\pa M$  is empty, then  $H_1(M;\bZ)$ is infinite.
\end{enumerate}

(Hint for the last three problems: use the standard classification of closed  $2$-manifolds 
and think about first homology groups.)

\begin{enumerate}
\item[6.]  Prove: if  $M$  is orientable and  $H_1(M;\bZ) = 0$,  then  $\pa M$ is a disjoint union 
of $2$-spheres.
\item[7.] Prove: if $M$ is orientable,  $\pa M \neq\ph$,  and  $\pa M$ contains no  $2$-spheres, 
then  $H_1(M;\bZ)$  is infinite. 
\item[8.] Prove: if  $M$ is nonorientable and  $\pa M$ contains no  $2$-spheres and no projective 
planes, then  $H_1(M;\bZ)$  is infinite.
\end{enumerate}

\chapter{Homology, cohomology, and $K(\pi,n)$s}

We have given an axiomatic definition of ordinary homology and cohomology, and we have shown
how to realize the axioms by means of either cellular or singular chain and cochain complexes.
We here give a homotopical way of constructing ordinary theories that makes no use of chains, 
whether cellular or singular. We also show how to construct cup and cap products homotopically.
This representation of homology and cohomology in terms of Eilenberg-Mac\,Lane spaces is the 
starting point of the modern approach to homology and cohomology theory, and we shall indicate
how theories that do not satisfy the dimension axiom can be represented.  We shall also describe 
Postnikov systems, which give a way to approximate general (simple) spaces by weakly equivalent
spaces built up out of Eilenberg-Mac\,Lane spaces. This is conceptually dual to the way that CW 
complexes allow the approximation of spaces by weakly equivalent spaces built up out of spheres. 
Finally, we present the important notion of cohomology operations and 
relate them to the cohomology of Eilenberg-Mac\,Lane spaces. 

\section{$K(\pi,n)$s and homology}

Recall that a reduced homology theory on based CW complexes is a 
sequence of functors 
$\tilde{E}_q$ from the homotopy category of based CW complexes to the category of Abelian 
groups. Each $\tilde{E}_q$ must satisfy the exactness and additivity axioms, and there must 
be a natural suspension isomorphism. Up to isomorphism, ordinary reduced homology with
coefficients in $\pi$ is characterized as the unique such theory that satisfies the 
dimension axiom: $\tilde{E}_0(S^0) = \pi$ and $\tilde{E}_q(S^0) = 0$ if $q\neq 0$. We
proceed to construct such a theory homotopically.

For based spaces $X$ and $Y$, we let $[X,Y]$ denote the set of based homotopy classes of based 
maps $X\rtarr Y$. Recall that we require Eilenberg-Mac\,Lane spaces $K(\pi,n)$ to have the
homotopy\index{Eilenberg-Mac\,Lane space} types of CW complexes and that, up to homotopy equivalence, there is a unique such
space for each $n$ and $\pi$. By a result of Milnor, if $X$ has the homotopy type of a CW
complex, then so does $\OM X$. By the Whitehead theorem, we therefore have a homotopy equivalence 
$$\tilde{\si}: K(\pi,n)\rtarr \OM K(\pi,n+1).$$
This map is the adjoint of a map
$$\si: \SI K(\pi,n) \rtarr K(\pi,n+1).$$
We may take the smash product of the map $\si$ with a based CW complex $X$ and use the 
suspension homomorphism on homotopy groups to obtain maps
\begin{eqnarray*}
\pi_{q+n}(X\sma K(\pi,n)) & \overto{\SI} & \pi_{q+n+1}(\SI(X\sma K(\pi,n)))  \\
              &     =  & \pi_{q+n+1}(X\sma \SI K(\pi,n)) \overto{(\id\sma\si)_*}  
\pi_{q+n+1}(X\sma K(\pi,n+1)).
\end{eqnarray*}

\begin{thm}
For CW complexes $X$, Abelian groups $\pi$ and integers $n\geq 0$, there are natural 
isomorphisms\index{homology theory!ordinary}
$$ \tilde{H}_q(X;\pi)\iso \colim_{n}\pi_{q+n}(X\sma K(\pi,n)).$$
\end{thm}

It suffices to verify the axioms, and the dimension axiom is clear. If $X=S^0$, 
then $X\sma K(\pi,n)=K(\pi,n)$. Here the homotopy groups in the colimit system are 
zero if $q\neq 0$, and, if $q=0$, the colimit runs over a sequence of isomorphisms 
between copies of $\pi$. 

The verifications of the rest of the axioms are exercises in the use of the homotopy excision 
and Freudenthal suspension theorems, and it is worthwhile to carry out these exercises in 
greater generality.

\begin{defn} A prespectrum\index{prespectrum} is a sequence of based spaces $T_n$, $n\geq 0$, and 
based maps $\si: \SI T_n \rtarr T_{n+1}$. 
\end{defn}

The example at hand is the Eilenberg-Mac\,Lane prespectrum $\sset{K(\pi,n)}$. Another 
example is the ``suspension prespectrum''\index{suspension prespectrum} $\sset{\SI^n X}$ of 
a based space $X$; the required maps $\SI(\SI^nX) \rtarr \SI^{n+1}X$ are the evident identifications. When $X=S^0$, this 
is called the sphere prespectrum.\index{sphere prespectrum}

\begin{thm} Let $\sset{T_n}$ be a prespectrum such that $T_n$ is $(n-1)$-connected and
of the homotopy type of a CW complex for each $n$. Define\index{homology theory!reduced}
$$\tilde{E}_q(X) = \colim_{n}\pi_{q+n}(X\sma T_n),$$
where the colimit is taken over the maps 
$$
\pi_{q+n}(X\sma T_n) \overto{\SI} \pi_{q+n+1}(\SI(X\sma T_n)) \iso
              \pi_{q+n+1}(X\sma \SI T_n) \overto{\id\sma\si}  
\pi_{q+n+1}(X\sma T_{n+1}).
$$
Then the functors $\tilde{E}_q$ define a reduced homology theory on based CW complexes.
\end{thm}
\begin{proof}
Certainly the $\tilde{E}$  are well defined functors from the homotopy category of based 
CW complexes to the category of Abelian groups. We must verify the exactness, additivity,
and suspension axioms. Without loss of generality, we may take the $T_n$ to be CW complexes 
with one vertex and no other cells of dimension less than $n$. Then $X\sma T_n$ is a quotient 
complex of $X\times T_n$, and it too has one vertex and no other cells of dimension less than $n$. 
In particular, it is $(n-1)$-connected.

If $A$ is a subcomplex of $X$, then the homotopy excision theorem implies that the quotient map 
$$(X\sma T_n,A\sma T_n) \rtarr ((X\sma T_n)/(A\sma T_n),*)\iso ((X/A)\sma T_n,*)$$
is a $(2n-1)$-equivalence. We may restrict to terms with $n>q-1$ in calculating $\tilde{E}_q(X)$,
and, for such $q$, the long exact sequence of homotopy groups of the pair $(X\sma T_n,A\sma T_n)$ 
gives that the sequence
$$\pi_{q+n}(A\sma T_n) \rtarr \pi_{q+n}(X\sma T_n) \rtarr \pi_{q+n}((X/A)\sma T_n)$$
is exact. Since passage to colimits preserves exact sequences, this proves the exactness
axiom.

We need some preliminaries to prove the additivity axiom.

\begin{defn} Define the weak product\index{weak product} $\prod^{w}_{\, _i} Y_i$ of a set of based 
spaces $Y_i$ to be 
the subspace of $\prod_i Y_i$ consisting of those points all but finitely many of whose coordinates
are basepoints. 
\end{defn}

\begin{lem} For a set of based spaces $\sset{Y_i}$, the canonical map
$$\textstyle{\sum}_i \pi_q(Y_i) \rtarr \pi_q(\textstyle{\prod}^w_{\, i} Y_i)$$
is an isomorphism.
\end{lem}
\begin{proof} The homotopy groups of $\prod^w_{\, i} Y_i$ are the colimits of the homotopy 
groups of the finite subproducts of the $Y_i$, and the conclusion follows.
\end{proof}

\begin{lem} If $\sset{Y_i}$ is a set of based CW complexes, then $\prod^w_{\, i}Y_i$ 
is a CW complex whose cells are the cells of the finite subproducts of the $Y_i$. 
If each $Y_i$ has a single vertex and no $q$-cells for $q<n$, then
the $(2n-1)$-skeleton of $\prod^w_{\, i} Y_i$ coincides with the $(2n-1)$-skeleton of $\bigvee_i Y_i$. 
Therefore the inclusion $\bigvee_i Y_i\rtarr \prod^w_{\, i}Y_i$ is a $(2n-1)$-equivalence. 
\end{lem}

Returning to the proof of the additivity axiom, suppose given based  CW complexes $X_i$ 
and consider the natural map
$$(\textstyle{\bigvee}_i X_i)\sma T_n \iso \textstyle{\bigvee}_i (X_i\sma T_n) 
\rtarr \textstyle{\prod}^w_{\, i} (X_i\sma T_n).$$
It induces isomorphisms on $\pi_{q+n}$ for $q<n-1$, and the additivity axiom follows.

Finally, we must prove the suspension axiom. We have the suspension map
$$\pi_{q+n}(X\sma T_n)\overto{\SI}\pi_{q+n+1}(\SI(X\sma T_n))\iso \pi_{q+n+1}((\SI X)\sma T_n).$$
By the Freudenthal suspension theorem, it is an isomorphism for $q<n-1$. Keeping track of suspension
coordinates and their permutation, we easily check that these maps commute with the maps defining
the colimit systems for $X$ and for $\SI X$. Therefore they induce a natural suspension isomorphism
$$\tilde{E}_q(X) \iso \tilde{E}_{q+1}(\SI X).$$
This completes the proof of the theorem.
\end{proof}

\begin{exmp} Applying the theorem to the sphere prespectrum, we find that the stable
homotopy groups\index{homotopy groups!stable} $\pi_q^s(X)$ give the values of a reduced 
homology theory; it is called ``stable homotopy theory.''\index{stable homotopy theory}
\end{exmp}

\section{$K(\pi,n)$s and cohomology}

The homotopical description of ordinary cohomology theories is both simpler and more
important to the applications than the homotopical description of ordinary homology 
theories. 

\begin{thm}
For CW complexes $X$, Abelian groups $\pi$, and integers $n\geq 0$, there are natural 
isomorphisms\index{cohomology theory!ordinary}
$$ \tilde{H}^n(X;\pi)\iso [X,K(\pi,n)].$$
\end{thm}

The dimension axiom is built into the definition of $K(\pi,n)$, as we see by taking
$X=S^0$. As in homology, it is worthwhile to carry out the verification of the remaining 
axioms  in greater generality. We first state some properties of the functor
$[-,Z]$ on based CW complexes that is ``represented''\index{represented functor} by a 
based space $Z$.

\begin{lem}
For any based space $Z$, the functor $[X,Z]$ from based CW complexes $X$ to pointed 
sets satisfies the following properties.
\begin{itemize}
\item HOMOTOPY \ \ If $f\htp g: X\rtarr Y$, then $f^*=g^*: [Y,Z]\rtarr [X,Z]$.
\item EXACTNESS \ \ If $A$ is a subcomplex of $X$, then the sequence
$$[X/A,Z]\rtarr [X,Z]\rtarr [A,Z]$$
is exact.
\item ADDITIVITY \ \ If $X$ is the wedge of a set of based CW complexes $X_i$, then
the inclusions $X_i\rtarr X$ induce an isomorphism
$$ [X,Z] \rtarr \textstyle{\prod} [X_i,Z].$$
\end{itemize}
\end{lem}

If $Z$ has a multiplication $\ph: Z\times Z\rtarr Z$ such that the basepoint $*$ of $Z$ is 
a two-sided unit up to homotopy, so that $Z$ is an ``$H$-space,''\index{Hspace@$H$-space} 
then $\ph$ induces an ``addition''
$$[X,Z]\times [X,Z]\rtarr [X,Z].$$
The trivial map $X\rtarr Z$ acts as zero. If $Z$ is homotopy associative, in the sense 
that there is a homotopy between the maps given on elements by $(xy)z$ and by $x(yz)$, 
then the addition is associative. If, further, $Z$ is homotopy commutative, in the sense
that there is a homotopy between the maps given on elements by $xy$ and by $yx$, then
this addition is commutative. We say that $Z$ is ``grouplike'' if there is a map 
$\chi: Z\rtarr Z$ such that $\ph(\id\times\,\chi)\DE: Z\rtarr Z$ is homotopic to the trivial 
map, and then $\chi_*:[X,Z]\rtarr [X,Z]$ sends an element $x\in [X,Z]$ to $x^{-1}$.

\begin{lem}
If $Z$ is a grouplike homotopy associative and commutative $H$-space, then the functor 
$[X,Z]$ takes values in Abelian groups.
\end{lem}

Actually, the existence of inverses can be deduced if $Z$ is only ``grouplike'' in the
weaker sense that $\pi_0(X)$ is a group, but we shall not need the extra generality.
Now consider the multiplication on a loop space $\OM Y$ given by composition of loops. 
Our proof that $\pi_1(Y)$ is a group and $\pi_2(Y)$ is an Abelian group amounts to a
proof of the following result. 

\begin{lem} For any based space $Y$, $\OM Y$ is a grouplike homotopy associative $H$-space 
and $\OM^2 Y$ is a grouplike homotopy associative and commutative $H$-space.
\end{lem}

Recall too that we have 
$$[\SI X,Y]\iso [X,\OM Y]$$
for any based spaces $X$ and $Y$. 

\begin{defn} An $\OM$-prespectrum\index{prespectrum!Omega@$\OM$} is a 
sequence of based spaces $T_n$ and weak
homotopy equivalences $\tilde{\si}: T_n \rtarr \OM T_{n+1}$.
\end{defn}

It is usual, but unnecessary, to require the $T_n$ to have the homotopy types of
CW complexes, in which case the $\tilde{\si}$ are homotopy equivalences.
Specialization of the observations above leads to the following fundamental fact.

\begin{thm}
Let $\sset{T_n}$ be an $\OM$-prespectrum. Define
\[
\tilde{E}^q(X) = \left 
                 \{ \begin{array}{ll} 
        [X,T_q] & \mbox{if $q\geq 0$} \\

        [X,\Omega^{-q}T_{0}] & \mbox{if $q<0$.} \end{array} 
\right.
\]
Then the functors $\tilde{E}^q$ define a reduced cohomology theory on based CW complexes.
\end{thm}
\begin{proof}
We have already verified the exactness and additivity axioms, and the weak
equivalences $\tilde{\si}$ induce the suspension isomorphisms:
$$\tilde{E}^q(X)=[X,T_q]\rtarr [X,\OM T_{q+1}]\iso [\SI X, T_{q+1}]=\tilde{E}^{q+1}(\SI X).\qed$$
\renewcommand{\qed}{}\end{proof}

It is a consequence of a general result called the Brown representability theorem\index{Brown
representability theorem}
that every reduced cohomology theory is represented in this fashion by an $\OM$-prespectrum.

\section{Cup and cap products}

Changing notations, let $A$ and $B$ be Abelian groups and $X$ and $Y$ be based spaces. We have 
an external product 
$$ \tilde{H}^p(X;A)\ten \tilde{H}^q(Y;B) \rtarr \tilde{H}^{p+q}(X\sma Y;A\ten B).$$
Indeed, if $X$ and $Y$ are based CW complexes, then we have an isomorphism of cellular chain 
complexes
$$ \tilde{C}_*(X)\ten \tilde{C}_*(Y)\iso \tilde{C}_*(X\sma Y).$$
On passage to cochains with coefficients in $A$, $B$, and $A\ten B$, this induces a
homomorphism of cochain complexes
$$ \tilde{C}^*(X;A)\ten \tilde{C}^*(Y;B)\rtarr \tilde{C}^*(X\sma Y;A\ten B).$$
In turn, this induces the cited product on passage to cohomology. With $X=Y$, we can apply the 
diagonal $\DE: X\rtarr X\sma X$ and any homomorphism $A\ten B\rtarr C$ to obtain a cup product
$$ \tilde{H}^p(X;A)\ten \tilde{H}^q(X;B)\rtarr \tilde{H}^{p+q}(X;C).$$
When $X=X'_+$ for an unbased space $X'$, this gives the cup product on the unreduced 
cohomology of $X'$. 

We can obtain these external products and therefore their induced cup products\index{cup product} 
homotopically. The smash product of maps gives a pairing
$$[X,K(A,p)]\ten [Y,K(B,q)] \rtarr [X\sma Y,K(A,p)\sma K(B,q)].$$
Therefore, to obtain an external product, we need only obtain a suitable map
$$\phi_{p,q}: K(A,p)\sma K(B,q)\rtarr K(A\ten B,p+q).$$
Such a map may be interpreted as an element of $\tilde{H}^{p+q}(K(A,p)\sma K(B,q);A\ten B)$.
Since the space $K(A,p)\sma K(B,q)$ is $(p+q-1)$-connected, the universal 
coefficient, K\"unneth, and Hurewicz theorems give isomorphisms
\begin{eqnarray*}
\tilde{H}^{p+q}(K(A,p)\sma K(B,q);A\ten B) & \iso & \Hom(\tilde{H}_{p+q}(K(A,p)\sma K(B,q)),A\ten B)\\
            & \iso & \Hom(\tilde{H}_{p}(K(A,p))\ten \tilde{H}_q(K(B,q)),A\ten B)\\
            & \iso & \Hom(\pi_{p}(K(A,p))\ten \pi_q(K(B,q)),A\ten B) \\
            & = & \Hom(A\ten B,A\ten B).
\end{eqnarray*}
Therefore the identity homomorphism on the group $A\ten B$ gives rise to the required map $\ph_{p,q}$.
Arguing similarly, it is easy to check that the system of maps $\sset{\ph_{p,q}}$ is associative,
commutative, and unital in the sense that the following diagrams are homotopy commutative.
Indeed, translating back along isomorphisms of the form just displayed, each of the diagrams 
translates to an elementary algebraic identity.
$$\diagram
K(A,p)\sma K(B,q)\sma K(C,r)\dto_{\id\sma\ph} \rto^{\ph\sma\id} 
& K(A\ten B,p+q)\sma K(C,r) \dto^{\ph}\\
K(A,p)\sma K(B\ten C,q+r) \rto_{\ph} & K(A\ten B\ten C, p+q+r),\\
\enddiagram$$
$$\diagram
K(A,p)\sma K(B,q)\dto_{t} \rto^{\ph} 
& K(A\ten B,p+q) \dto^{K(t,p+q)}\\
K(B,q)\sma K(A,p) \rto_{\ph} & K(B\ten A, p+q),\\
\enddiagram$$
and
$$\diagram
S^0\sma K(A,p) \dto_{i\sma\id} \rdouble  & K(A,p) \ddouble \\
K(\bZ,0)\sma K(A,p) \rto_{\ph} & K(\bZ\ten A,p),\\
\enddiagram$$
where $i: S^0\rtarr \bZ=K(\bZ,0)$ takes $0$ to $0$ and $1$ to $1$. 
The associativity, graded commutativity, and unital properties 
of the cup product follow.

The cup products on cohomology defined in terms of cellular cochains and in terms of the homotopical
representation of cohomology agree. To see this, observe that the identity homomorphism of
$A$ specifies a fundamental class $$\io_p\in \tilde{H}^p(K(A,p),A)$$ via the isomorphisms
$$\Hom(A,A) \iso \Hom(\pi_p(K(A,p)),A) \iso \Hom(\tilde{H}_p(K(A,p)),A)\iso \tilde{H}^p(K(A,p);A).$$
A moment's thought shows that the two cup products will agree on arbitrary pairs of cohomology
classes if they agree when applied to $\io_p\ten \io_q$ for all $p$ and $q$. We may take our 
Eilenberg-Mac\,Lane spaces to be CW complexes and give their smash product the induced CW
structure. Considering representative cycles for generators of our groups as images under
the Hurewicz homomorphism of representative maps $S^p\rtarr K(A,p)$,  we find that the
required agreement follows from the canonical identifications $S^p\sma S^q\iso S^{p+q}$. 

We can also construct cap products homotopically. To do so, it is convenient to bring function
spaces into play, using the obvious isomorphisms
$$ [X,Y] \iso \pi_0F(X,Y)$$
and evaluation maps
$$\epz: F(X,Y)\sma X \rtarr Y.$$
We wish to construct the cap product\index{cap product}
$$\tilde{H}^p(X;A)\ten \tilde{H}_{n}(X;B)\rtarr \tilde{H}_{n-p}(X;A\ten B),$$
and it is equivalent to construct
$$ \pi_0(F(X,K(A,p)))\ten \colim_q\pi_{n+q}(X\sma K(B,q))
\rtarr \colim_r\pi_{n-p+r}(X\sma K(A\ten B,r)).$$
Changing the variable of the second colimit by setting $r=p+q$ and recalling the algebraic
fact that tensor products commute with colimits, we can rewrite this as 
\begin{small}
$$\colim_q (\pi_0(F(X,K(A,p)))\ten \pi_{n+q}(X\sma K(B,q)))
\rtarr \colim_q\pi_{n+q}(X\sma K(A\ten B,p+q)).$$
\end{small}
Thus it suffices to define maps
$$\pi_0(F(X,K(A,p)))\ten \pi_{n+q}(X\sma K(B,q))
\rtarr \pi_{n+q}(X\sma K(A\ten B,p+q)).$$
These are given by the following composites:
$$\diagram
\pi_0(F(X,K(A,p)))\ten \pi_{n+q}(X\sma K(B,q)) \dto^{\sma} \\
\pi_{n+q}(F(X,K(A,p))\sma X\sma K(B,q)) \dto^{(\id\sma\DE\sma\id)_*} \\
\pi_{n+q}(F(X,K(A,p))\sma X\sma X\sma K(B,q)) \dto^{(\epz\sma\id)_*} \\
\pi_{n+q}(K(A,p)\sma X\sma K(B,q)) \dto^{(\id\sma\ph_{p,q})_*(t\sma \id)_*} \\
\pi_{n+q}(X\sma K(A\ten B, p+q)).\\
\enddiagram$$

Similarly, we can construct the evaluation pairing\index{evaluation pairing}
$$\tilde{H}^n(X;A)\ten \tilde{H}_n(X;B)\rtarr A\ten B$$
homotopically. It is obtained by passage to colimits over $q$ from the composites
$$\diagram
\pi_0(F(X,K(A,n)))\ten \pi_{n+q}(X\sma K(B,q)) \dto^{\sma} \\
\pi_{n+q}(F(X,K(A,n))\sma X\sma K(B,q)) \dto^{(\epz\sma\id)_*} \\
\pi_{n+q}(K(A,n)\sma K(B,q)) \dto^{(\ph_{n,q})_*} \\
\pi_{n+q}(K(A\ten B, n+q)) = A\ten B.\\
\enddiagram$$

The following formula relating the cup and cap products to the evaluation pairing
was central to our discussion of Poincar\'e duality:
$$\langle \al \cup \be, x\rangle = \langle \be,\al \cap x\rangle\in R,$$
where $\al\in \tilde{H}^p(X;R)$, $\be\in \tilde{H}^q(X;R)$, and $x\in \tilde{H}_{p+q}(X;R)$
for a commutative ring $R$. It is illuminating to rederive this from our homotopical descriptions 
of these products. In fact, a straightforward diagram 
chase shows that this formula is a direct consequence of the following elementary facts, where
$X$, $Y$, and $Z$ are based spaces. First, the following diagram commutes:

$$\diagram
F(X,Y)\sma F(X,Z)\sma X \dto_{(\sma)\sma\id} \rrto^{\id\sma\id\sma\DE} 
& & F(X,Y)\sma F(X,Z)\sma X\sma X \dto^{(\sma)\sma\id\sma\id}\\
F(X\sma X,Y\sma Z)\sma X \dto_{F(\DE,\id)\sma\id} \rrto^{\id\sma\DE} 
& & F(X\sma X,Y\sma Z)\sma X\sma X \dto^{\epz}\\
F(X,Y\sma Z)\sma X \rrto_{\epz} & & Y\sma Z.\\
\enddiagram$$
Second, the right vertical composite in the diagram coincides with the common composite
in the commutative diagram
$$\diagram
F(X,Y)\sma F(X,Z)\sma X\sma X \dto_{\id\sma t\sma \id} \rrto^{t\sma\id} 
& & F(X,Z)\sma F(X,Y)\sma X\sma X \dto^{\id\sma\epz\sma\id} \\
F(X,Y)\sma X \sma F(X,Z)\sma X \dto_{\epz\sma\epz} & & F(X,Z)\sma Y\sma X \dto^{\id\sma t}\\
Y\sma Z & Z\sma Y \lto^t & F(X,Z)\sma X\sma Y \lto^(0.6){\epz\sma\id}.\\
\enddiagram$$

The observant reader will see a punch line here: everything in this section applies equally
well to the homology and cohomology theories represented by $\OM$-prespectra. A little more
precisely, thinking of the case when $A=B=C$ is a commutative ring in the discussion above,
we see by use of the product on $A$ that we have a well behaved system of product maps 
$$\ph_{p,q}: K(A,p)\sma K(A,q)\rtarr K(A,p+q).$$
We have analogous cup and cap products and an evaluation pairing for the theories represented 
by any $\OM$-prespectrum $\sset{T_n}$ with such a system 
of product maps $$\ph_{p,q}:T_p\sma T_q\rtarr T_{p+q}.$$

\section{Postnikov systems}

We have implicitly studied the represented functors $k(X)=[X,Y]$ by 
decomposing $X$ into cells. This led in particular to the calculation of 
ordinary represented cohomology $[X,K(\pi,n)]$ by means of cellular chains. 
There is an Eckmann-Hilton dual way of studying $[X,Y]$ by decomposing $Y$ 
into ``cocells.'' We briefly describe this decomposition of spaces into their 
``Postnikov systems'' here. 

This decomposition answers a natural question: how close are the homotopy 
groups of a CW complex $X$ to being a complete set of invariants for its 
homotopy type?  Since $\prod_{n} K(\pi_n(X),n)$ has the same homotopy groups 
as $X$ but is generally not weakly homotopy equivalent to it, some added information 
is needed. If $X$ is simple, it turns out that the homotopy groups together 
with an inductively defined sequence of cohomology classes give a complete 
set of invariants.

Recall that a connected space  $X$  is said to be simple\index{simple space} if  $\pi_{1}(X)$  is 
Abelian and acts trivially on $\pi_{n}(X)$ for $n\geq 2$. A Postnikov system\index{Postnikov
system} for a simple based space  $X$  consists of based spaces $X_n$ together 
with based maps 
\[ \alpha _{n}: X \rtarr  X_{n} \ \ \tand \ \  p_{n+1}: X_{n+1} \rtarr X_{n}, \]
$n\geq 1$,  such that $p_{n+1}\com\alpha _{n+1} = \alpha _{n}$, $X_{1}$ is an 
Eilenberg-Mac\,Lane space $K({\pi}_{1}(X),1)$, $p_{n+1}$  is the fibration induced 
from the path space fibration over an Eilenberg-Mac\,Lane space $K({\pi}_{n+1}(X),n+2)$ 
by a map  
$$k^{n+2}: X_{n} \rtarr K({\pi}_{n+1}(X),n+2),$$ 
and $\alpha_{n}$ induces an isomorphism  
${\pi}_{q}(X) \rightarrow {\pi}_{q}(X_{n})$  for  $q\leq n$.
It follows that ${\pi}_{q}(X_{n})=0$  for  $q>n$.  The system can be displayed diagrammatically
as follows:
$$\diagram
& \vdots \dto & \\
& X_{n+1} \dto^{p_{n+1}} \rto^(0.29){k^{n+3}} & K(\pi_{n+2}(X),n+3) \\
X \urto^{\al_{n+1}} \ddrto^{\al_1} \rto^{\al_n} & X_n \rto^(0.27){k^{n+2}} \dto & K(\pi_{n+1}(X),n+2) \\
& \vdots \dto & \\
& X_1 \rto^(0.33){k^3} & K(\pi_{2}(X),3). & \\
\enddiagram$$
Our requirement that Eilenberg-Mac\,Lane spaces 
have the homotopy types of CW complexes implies (by a result of Milnor) 
that each  $X_{n}$  has the homotopy type of a CW complex.  The maps  $\alpha _{n}$  
induce a weak equivalence  $X \rightarrow \lim X_{n}$,  but the inverse limit 
generally will not have the homotopy type of a CW complex. The  ``$k$-invariants'' $\sset{k^{n+2}}$
\index{kinvariants@$k$-invariants} that specify the system are to be regarded as cohomology classes
\[ k^{n+2}\in H^{n+2}(X_{n};{\pi}_{n+1}(X)). \]
These classes together with the homotopy groups $\pi_{n}(X)$  specify the 
weak homotopy type of $X$. We outline the proof of the following theorem.
         
\begin{thm}  
A simple space  $X$  of the homotopy type of a CW complex has a 
Postnikov system.
\end{thm}
\begin{proof}
Assume inductively that  $\alpha _{n}: X \rightarrow  X_{n}$  has been constructed.  
A consequence of the homotopy excision theorem shows that the cofiber $C(\alpha _{n})$  
is  $(n+1)$-connected and satisfies 
\[ {\pi} _{n+2}(C(\alpha_{n}))={\pi} _{n+1}(X). \]
More precisely, the canonical map  $\et: F(\alpha _{n}) \rightarrow  \Omega C(\alpha _{n})$  
induces 
an isomorphism on  ${\pi} _{q}$  for  $q\leq n+1$.  We construct  
\[ j: C(\alpha _{n}) \rightarrow     K({\pi} _{n+1}(X),n+2) \]
by inductively attaching cells to $C(\alpha _{n})$ to kill its higher
homotopy groups.  We take the composite of  $j$  and the inclusion
$X_{n} \subset  C(\alpha _{n})$  to be the  $k$-invariant  
$$k^{n+2}: X_n \rtarr K(\pi_{n+1}(X),n+2).$$  
By our definition of a Postnikov system, we must define $X_{n+1}$ to be the 
homotopy fiber of $k^{n+2}$.  Thus its points are pairs  $(\omega ,x)$  consisting 
of a path  $\omega : I\rightarrow  K({\pi}_{n+1}(X),n+2)$  and a point  $x\in X_{n}$  
such that  $\omega (0)=*$  and  $\omega (1)=k^{n+2}(x)$. The map  
$p_{n+1}: X_{n+1} \rightarrow  X_{n}$  is given by 
$p_{n+1}(\omega ,x)=x$,  and the map  $\alpha _{n+1}: X \rightarrow  X_{n+1}$  
is given by  $\alpha _{n+1}(x)=(\omega (x),\alpha_{n}(x))$,  where  
$\omega (x)(t) = j(x,1-t)$,  $(x,1-t)$  being a point on the cone  
$CX \subset C(\alpha _{n})$.  Clearly  $p_{n+1}\com\alpha _{n+1} = \alpha _{n}$.  
It is evident that  $\alpha _{n+1}$ induces an isomorphism on  
${\pi}_{q}$  for  $q\leq n$,  and a diagram chase shows that this also holds 
for  $q=n+1$.
\end{proof}

\section{Cohomology operations}

Consider a ``represented functor''\index{represented functor} $k(X)=[X,Z]$ and another 
contravariant functor 
$k'$ from the homotopy category of based CW complexes to the category of sets. The 
following simple observation actually applies to represented functors on arbitrary 
categories. We shall use it to describe cohomology operations, but it also applies to 
describe many other invariants in algebraic topology, such as the characteristic classes 
of vector bundles. 

\begin{lem}[Yoneda]\index{Yoneda lemma} There is a canonical bijection between natural 
transformations 
$\PH: k\rtarr k'$ and elements $\ph\in k'(Z)$.
\end{lem}
\begin{proof}
Given $\PH$, we define $\ph$ to be $\PH(\id)$, where $\id\in k(Z)=[Z,Z]$ is the
identity map. Given $\ph$, we define $\PH: k(X)\rtarr k'(X)$ by the formula
$\PH(f)=f^*(\ph)$. Here $f$ is a map $X\rtarr Z$, and it induces 
$f^*=k'(f):k'(Z)\rtarr k'(X)$. It is simple to check that these are inverse
bijections.
\end{proof}

We are interested in the case when $k'$ is also represented, say $k'(X)=[X,Z']$.

\begin{cor} There is a canonical bijection between natural transformations 
$\PH: [-,Z]\rtarr [-,Z']$ and elements $\ph\in [Z,Z']$.
\end{cor}

\begin{defn} Suppose given cohomology theories $\tilde{E}^*$ and $\tilde{F}^*$. A cohomology 
operation\index{cohomology operation} of type $q$ and degree $n$ is a natural transformation 
$\tilde{E}^q\rtarr \tilde{F}^{q+n}$. A stable cohomology 
operation\index{cohomology operation!stable} of degree $n$ is a sequence 
$\sset{\PH^q}$ of cohomology operations of type $q$ and degree $n$ such that the following 
diagram commutes for each $q$ and each based space $X$:
$$\diagram
\tilde{E}^q(X)\rto^{\PH^q} \dto_{\SI} & \tilde{E}^{q+n}(X) \dto^{\SI} \\
\tilde{E}^{q+1}(\SI X) \rto_(0.45){\PH^{q+1}} & \tilde{E}^{q+1+n}(\SI X). \\
\enddiagram$$
We generally abbreviate notation by setting $\PH^q=\PH$.
\end{defn}

In general, cohomology operations are only natural transformations of set-valued functors.
However, stable operations are necessarily homomorphisms of cohomology groups, as the
reader is encouraged to check.

\begin{thm}
Cohomology operations $\tilde{H}^q(-;\pi)\rtarr \tilde{H}^{q+n}(-;\rh)$ are in canonical
bijective correspondence with elements of $\tilde{H}^{q+n}(K(\pi,q);\rh)$.
\end{thm}
\begin{proof}
Translate to the represented level, apply the previous corollary, and translate back.
\end{proof}

This seems very abstract, but it has very concrete consequences. To determine all cohomology
operations, we need only compute the cohomology of all Eilenberg-Mac\,Lane spaces. We have
described an explicit construction of these spaces as topological Abelian groups
in Chapter 16 \S5, and this construction leads to an inductive method of computation. We briefly 
indicate a key example of how this works, without proofs. 

\begin{thm} For $n\geq 0$, there are stable cohomology operations
$$Sq^n: H^q(X;\bZ_2)\rtarr H^{q+n}(X;\bZ_2),$$
called the Steenrod operations.\index{Steenrod operations} They satisfy the following properties.
\begin{enumerate}
\item[(i)] $Sq^0$ is the identity operation.
\item[(ii)] $Sq^n(x)=x^2$ if $n=\text{\em deg}\,x$ and $Sq^n(x)=0$ if $n> \text{\em deg}\,x$.
\item[(iii)] The Cartan formula\index{Cartan formula} holds:
$$Sq^n(xy)= \sum_{i+j=n}Sq^i(x)Sq^j(y).$$
\end{enumerate}
\end{thm}

In fact, the Steenrod operations are uniquely characterized by the stated properties.
There are also formulas, called the Adem relations,\index{Adem relations} describing $Sq^iSq^j$, 
as a linear combination of operations $Sq^{i+j-k}Sq^k$, $2k\leq i$, when $0<i<2j$; explicitly,
$$Sq^iSq^j = \sum_{0\leq k\leq [i/2]}
 \left(\begin{array}{c}j-k-1\\i-2k\end{array}\right) Sq^{i+j-k}Sq^k.$$
It turns out that the Steenrod operations generate all mod $2$ cohomology operations. In fact, 
the identity map of $K(\bZ_2,q)$ specifies a fundamental class $\io_q\in H^q(K(\bZ_2,q);\bZ_2)$, 
and the following theorem holds.

\begin{thm} $H^*(K(\bZ_2,q);\bZ_2)$ is a polynomial algebra whose generators are certain 
iterates of Steenrod operations applied to the fundamental class $\io_q$. Explicitly,
writing $Sq^I = Sq^{i_1}\cdots Sq^{i_j}$ for a sequence of positive integers
$I=\sset{i_1,\ldots\!,i_j}$, the generators are the $Sq^I\io_q$ for those sequences $I$ 
such that $i_{r}\geq 2 i_{r+1}$ for $1\leq r < j$ and $i_1<i_2+\cdots +i_j+q$. 
\end{thm}

\vspace{.1in}

\begin{center}
PROBLEMS
\end{center}
\begin{enumerate}
\item For Abelian groups $\pi$ and $\rh$, show that  $[K(\pi,n),K(\rh,n)]\iso \Hom(\pi,\rh)$.  
(Hint: use the natural isomorphism $[X,K(\rh,n)]\iso \tilde{H}^n(X;\rh)$ and universal
coefficients.)
\item 
\begin{enumerate}
\item[(a)] Let $f: \pi\rtarr \rh$ be a homomorphism of Abelian groups. Construct  cohomology
operations $f^*: H^q(X;\pi)\rtarr H^q(X;\rh)$ for all $q$.
\item[(b)] Let $0\rtarr \pi\overto{f}\rh\overto{g}\si\rtarr 0$ be an exact sequence of
Abelian groups. Construct cohomology operations $\be: H^q(X;\si)\rtarr H^{q+1}(X;\pi)$ for
all $q$ such that the following is a long exact sequence:
$$ \cdots \rtarr H^q(X;\pi)\overto{f^*} H^q(X;\rh)\overto{g^*} H^q(X;\si)
\overto{\be} H^{q+1}(X;\pi)\rtarr \cdots.$$
The $\be$ are called Bockstein operations.\index{Bockstein operation}
\end{enumerate}
\item Using the calculation of $H^*(K(\bZ,2);\bZ_2)$ stated in the text, prove that 
$Sq^1: H^q(X;\bZ_2)\rtarr H^{q+1}(X;\bZ_2)$ coincides with
the Bockstein operation associated to the short exact sequence 
$0\rtarr \bZ_2 \rtarr \bZ_4 \rtarr \bZ_2 \rtarr 0$.
\item Prove that each $\PH^q$ of a stable cohomology operation $\sset{\PH^q}$ is a natural
homomorphism.
\item Write $H^*(\bR P^{\infty};\bZ_2)=\bZ_2[\al]$, $\deg \al=1$. Compute $Sq^i(\al^j)$
for all $i$ and $j$.
\end{enumerate}

\chapter{Characteristic classes of vector bundles}

Some of the most remarkable applications of algebraic topology result from the
translation of problems in geometric topology into problems in homotopy theory.
The essential intermediary in many of these translations is the theory of vector
bundles. We here explain the classification theorem for vector bundles and its 
relationship to the theory of characteristic classes. 
The reader is assumed to be familiar with the tangent and normal bundles of smooth 
manifolds and to be reasonably well acquainted with the definitions and elementary 
properties of vector bundles in general. 

\section{The classification of vector bundles}

Let $E$ be a (real) vector bundle\index{vector bundle} over a base space $B$.  Thus we are given a
projection $p: E\rtarr B$ such that, for each $b\in B$, the fiber $p^{-1}(b)$ is a copy of 
$\bR^n$ for some $n$. In the case of non-connected base spaces,
the fibers over points in different components may have different dimension. We say that 
$p$ is an $n$-plane bundle\index{nplane bundle@$n$-plane bundle} if all fibers have dimension $n$. 
For each $U$ in some open cover of $B$, there is a homeomorphism (a ``coordinate chart'')
$\ph_U: U\times \bR^n\rtarr p^{-1}(U)$ over $U$ that restricts to a linear isomorphism on each 
fiber. We shall require our open covers to be numerable, as can always be arranged when
$B$ is paracompact. For a second vector bundle $q: D\rtarr A$, a map $(g,f): D\rtarr E$ of 
vector bundles is a pair of maps $f: A\rtarr B$ and $g: D\rtarr E$ such that 
$p\com g = f\com q$ and $g: q^{-1}(a) \rtarr p^{-1}(f(a))$ is linear for all $a\in A$. 
This gives the category of vector bundles. A map 
$(g,f)$ of vector bundles is an isomorphism 
if and only if $f$ is a homeomorphism and $g$ restricts to an isomorphism on each fiber. 
We are mainly interested in the subcategories of $n$-plane bundles and maps that are 
linear isomorphisms on fibers. We say that two vector bundles over $B$ are equivalent 
if they are isomorphic over $B$, so that
there is an isomorphism $(g,\id)$ between them. We let $\sE_n(B)$\index{Ean@$\sE_n(-)$} denote 
the set of 
equivalence classes of $n$-plane bundles over $B$. (That this really is a well defined set
will emerge shortly.) If $f: A\rtarr B$ is a continuous map, then the pullback of $f$ and 
a vector bundle $p: E\rtarr B$ is a vector bundle $f^*E$ over $A$. Moreover, a bundle $D$ 
over $A$ is equivalent to $f^*E$ if
and only if there is a map $(g,f): D\rtarr E$ that is an isomorphism on fibers.

Thus we have a contravariant set-valued functor $\sE_n(-)$ on spaces. 
Vector bundles should be thought of as rather rigid geometric objects, and the equivalence 
relation between them preserves that rigidity. Nevertheless, equivalence classes of $n$-plane
bundles can be classified homotopically. This is a crucial starting point for the 
translation of geometric problems to homotopical ones. In turn, the starting point of the 
classification theorem is the observation that the functor $\sE_n(-)$, like homology
and cohomology, is homotopy invariant in the sense that it factors through the homotopy
category $h\sU$. In less fancy language, this amounts to the following result.

\begin{prop}
The pullbacks of an $n$-plane bundle $p:E\rtarr B$ along homotopic maps $f_0,f_1: A\rtarr B$
are equivalent.
\end{prop}
\begin{proof}[Sketch proof]
Let $h: A\times I\rtarr B$ be a homotopy $f_0\htp f_1$. Then the restrictions of $h^*E$  over
$A\times\sset{0}$ and $A\times\sset{1}$ can be identified with $f_0^*E$ and $f_1^*E$. Thus we
change our point of view and consider a general $n$-plane bundle $p: E\rtarr B\times I$. It 
suffices to show that the restrictions $E_0$ and $E_1$ of $E$ over $B\times\sset{0}$ and 
$B\times\sset{1}$ are equivalent. Define $r: B\times I\rtarr B\times I$ by $r(b,t)=(b,1)$. 
We claim that there is a map $g: E\rtarr E$ such that $(g,r)$ is a map of vector bundles. It
follows that $E$ is equivalent to $r^*E$, and it is easy to see that $r^*E$ is isomorphic to
the bundle $E_1\times I$. The restriction of $g$ to $E_0$ will be an equivalence to $E_1$. 
To construct $g$, one first proves, using the compactness of $I$, that there is a numerable 
open cover $\sO$ of $B$ such that the restriction of $E$ to $U\times I$ is trivial for all 
$U\in\sO$. One then uses trivializations $\ph_U: U\times I\times \bR^n\rtarr p^{-1}(U\times I)$ 
together with functions $\la_U: B\rtarr I$ such that $\la_U^{-1}(0,1]=U$ to construct $g$
by gradually pushing the bundle to the right along neighborhoods where it is trivial.
\end{proof}

It can be verified on general abstract nonsense grounds, using Brown's representability 
theorem, that the functor $\sE_n(-)$ is representable in the form $[-,BO(n)]$ for some space
$BO(n)$. It is far more useful to have an explicit concrete construction of the relevant 
``classifying space''\index{classifying space} $BO(n)$.\index{BOn@$BO(n)$} More precisely, we 
think of ``$BO(n)$'' as specifying a homotopy 
type of spaces, and we want an explicit representative of the homotopy type. Here $[X,Y]$ denotes 
unbased homotopy classes of maps. We construct a particular $n$-plane bundle $\ga_n: E_n \rtarr BO(n)$,
called the ``universal $n$-plane bundle.''\index{universal n-plane bundle@universal $n$-plane bundle}
By pulling back $\ga_n$ along (homotopy classes of) 
maps $f: B\rtarr BO(n)$, we obtain a natural transformation of functors $[-,BO(n)]\rtarr \sE_n(-)$. 
We show that this natural transformation is a natural isomorphism of functors by showing how to 
construct a map $(g,f)$, unique up to homotopy, from any given $n$-plane bundle $E$ over any space 
$B$ to the universal $n$-plane bundle $E_n$; it is in this sense that $E_n$ is ``universal.'' 

Let $V_n(\bR^q)$ be the Stiefel\index{Stiefel variety} variety of orthonormal $n$-frames in 
$\bR^q$. Its points
are $n$-tuples of orthonormal vectors in $\bR^q$, and it is topologized as a subspace of
$(\bR^q)^n$ or, equivalently, as a subspace of $(S^{q-1})^n$. It is a compact manifold.
Let $G_n(\bR^q)$ be the Grassmann\index{Grassmann variety} variety of $n$-planes in $\bR^q$. 
Its points are the
$n$-dimensional subspaces of $\bR^q$. Sending an $n$-tuple of orthonormal vectors to the 
$n$-plane they span gives a surjective function $V_n(\bR^q)\rtarr G_n(\bR^q)$, and we 
topologize $G_n(\bR^q)$ as a quotient space of $V_n(\bR^q)$. It too is a compact manifold.
For example, $V_1(\bR^q)=S^{q-1}$ and $G_1(\bR^q)= \bR P^{q-1}$. The standard inclusion
of $\bR^q$ in $\bR^{q+1}$ induces inclusions $V_n(\bR^q)\subset V_n(\bR^{q+1})$ and
$G_n(\bR^q)\subset G_n(\bR^{q+1})$. We define $V_n(\bR^{\infty})$ and $G_n(\bR^{\infty})$
to be the unions of the $V_n(\bR^q)$ and $G_n(\bR^q)$, with the topology of the union.
We define the classifying space $BO(n)$ to be $G_n(\bR^{\infty})$.

Let $E_n^q$ be the subbundle of the trivial bundle $G_n(\bR^q)\times \bR^q$ whose points 
are the pairs $(x,v)$ such that $v$ is a vector in the plane $x$; denote the projection 
of $E_n^q$ by $\ga_n^q$, so that $\ga_n^q(x,v)=x$. When $n=1$, $\ga_1^q$
is called the ``canonical line bundle''\index{canonical line bundle} over $\bR P^{q-1}$. 
We may let $q$ go to infinity. 
We let $E_n=E_n^{\infty}$ and let $\ga_n=\ga_n^{\infty}: E_n\rtarr BO(n)$. This is our
universal bundle, and it is not hard to verify that it is indeed an $n$-plane bundle. We
must explain why it is universal. (Technically, it is usual to assume that base spaces are 
paracompact, but the restriction to numerable systems of coordinate charts in our
definition of vector bundles allows the use of general base spaces.)

\begin{thm}\index{classification theorem!for real $n$-plane bundles} The natural 
transformation $\PH: [-,BO(n)]\rtarr \sE_n(-)$ obtained by sending the
homotopy class of a map $f: B\rtarr BO(n)$ to the equivalence class of the $n$-plane
bundle $f^*E_n$ is a natural isomorphism of functors.
\end{thm}
\begin{proof}[Sketch proof]
To illustrate ideas, let $M$ be a smooth compact $n$-manifold smoothly embedded in $\bR^q$ and 
let $\ta(M)$ be its tangent bundle. The tangent plane $\ta_x$ at a point $x\in M\subset \bR^q$ 
is then embedded as an affine plane through $x$ in $\bR^q$. Translating to a plane through the
origin by subtracting $x$ from each vector, we 
obtain a point $f(x)\in G_n(\bR^q)$ and an isomorphism $g_x: \ta_x\rtarr (\ga_n^q)^{-1}(f(x))$.
The $g_x$ glue together to give a map $(g,f)$ of bundles from $E(\ta(M))$ to $E_n^q$; it is called 
the Gauss map\index{Gauss map} of the tangent bundle of $M$. Similarly, using the orthogonal 
complements of tangent planes, we obtain the Gauss map $E(\nu) \rtarr E_{q-n}^q$ of the normal 
bundle $\nu$ of the embedding of $M$ in $\bR^q$. 

For a general $n$-plane bundle $p: E\rtarr B$, we must construct a map $(g,f): E\rtarr E_n$
of vector bundles that is an isomorphism on fibers; it will follow that $E$ is equivalent
to $f^*E_n$, thus showing that $\PH$ is surjective. It suffices to construct a map 
$\hat{g}: E\rtarr \bR^{\infty}$ that is a linear
monomorphism on fibers, since we can then define $f(e)$ to be the image under $\hat{g}$ of 
the fiber through $e$ and can define $g(e)=(f(e),\hat{g}(e))$. One first shows that one can construct
a countable numerable cover of coordinate charts from a general numerable cover of coordinate
charts. Using trivializations $\ph_U: U\times \bR^n\rtarr p^{-1}(U)$ and functions
$\la_U: B \rtarr I$ such that $U=\la_U^{-1}(0,1]$, we define $\hat{g}_U: E\rtarr \bR^n$ by
$$ \hat{g}_U(e)= \la_U (p(e))\cdot p_2(\ph_U^{-1}(e))$$
for $e\in p^{-1}(U)$, where $p_2: U\times \bR^n\rtarr \bR^n$ is the projection,  and 
$\hat{g}_U(e)=0$ for $e\not\!{\in}\, p^{-1}(U)$. Taking $\bR^{\infty}$ to be the sum of 
countably many copies of $\bR^n$, we then define $\hat{g}=\sum g_U$. 

To show that $\PH$ is injective, we must show further that the resulting classifying map $f$ is 
unique up to homotopy, and for this it suffices to show that any two maps $(g_0,f_0)$ and $(g_1,f_1)$
of vector bundles from $E$ to $E_n$ that are isomorphisms on fibers are bundle homotopic. These 
bundle maps are determined by their second coordinates $\hat{g}_0$ and $\hat{g}_1$, which are maps 
$E\rtarr \bR^{\infty}$. 
Provided that $\hat{g}_0(e)$ is not a negative multiple of $\hat{g}_1(e)$ for any $e$, we obtain a 
homotopy $\hat{h}: \hat{g}_0\htp\hat{g}_1$ by setting 
$$\hat{h}(e,t) = (1-t)\hat{g}_0(e)+t\hat{g}_1(e).$$
The proviso ensures that $\hat{h}$ is a monomorphism on fibers, and $\hat{h}$ determines the 
required bundle homotopy $E\times I\rtarr E_n$. For the general case, let 
$i_0$ and $i_1$ be the linear isomorphisms from $\bR^{\infty}$ to itself that
send the $q$th standard basis element $e_q$ to $e_{2q}$ and $e_{2q-1}$, respectively. The 
composites $i_0\com \hat{g}_0$ and $i_1\com \hat{g}_1$ determine bundle maps $k_0$ and $k_1$ from
$E$ to $E_n$, and the construction just given applies to give bundle homotopies from $g_0$ to
$k_0$, from $k_0$ to $k_1$, and from $k_1$ to $g_1$. 
\end{proof}

\section{Characteristic classes for vector bundles}

\begin{defn} Let $k^*$ be a cohomology theory, such as $H^*(-;\pi)$ for an Abelian group $\pi$.
A characteristic class\index{characteristic class} $c$ of degree $q$ for $n$-plane bundles is a 
natural assignment
of a cohomology class $c(\xi)\in k^q(B)$ to bundles $\xi$ with base space $B$. Thus, if
$(g,f)$ is a map from a bundle $\ze$ over $A$ to a bundle $\xi$ over $B$, so that $\ze$ is
equivalent to $f^*\xi$, then  $f^*c(\xi)=c(\ze)$. Clearly $c(\xi)=c(\xi')$ if $\xi$ is
equivalent to $\xi'$.
\end{defn}

Since the functor $\sE_n$ is represented by $BO(n)$, the Yoneda lemma\index{Yoneda lemma} specializes 
to give the following result.

\begin{lem} Evaluation on $\ga_n$ specifies a canonical bijection between characteristic
classes of $n$-plane bundles and elements of $k^*(BO(n))$.
\end{lem}

The formal similarity to the definition of cohomology operations is obvious, and we shall
illustrate how to exploit this similarity in the following sections. Clearly calculation of 
$k^*(BO(n))$ determines all characteristic classes. Moreover, the behavior of characteristic classes 
with respect
to operations on bundles can be determined by calculating the maps on cohomology induced by 
maps between classifying spaces. We are particularly interested in Whitney sums of bundles. We have
the evident Cartesian product, or external sum, of an $m$-plane bundle over $A$ and an $n$-plane 
bundle over $B$; it is an $(m+n)$-plane bundle over $A\times B$. The internal sum, or 
Whitney sum,\index{Whitney sum} of two bundles over the same base space $B$ is obtained by pulling 
back their external sum along the diagonal map of $B$. 

For example, let $\epz$ denote the trivial line bundle over any space. We have the operation that 
sends an $n$-plane bundle $\xi$ over $B$ to the $(n+1)$-plane bundle $\xi\oplus \epz$ over $B$. 
There is a classifying map 
$$i_n: BO(n)\rtarr BO(n+1)$$
that is characterized up to homotopy by $i_n^*(\ga_{n+1})=\ga_n\oplus \epz$. 
If we have a characteristic class $c$ on $(n+1)$-plane bundles, then 
$$i_n^*c(\ga_{n+1})=c(\ga_n\oplus \epz),$$
and this leads by naturality to a description of 
$c(\xi\oplus\epz)$ for general $n$-plane bundles $\xi$. To give an
explicit description of $i_n$, we may think of $BO(n+1)$ as $G_{n+1}(\bR^{\infty}\oplus \bR)$;
precisely, we use an isomorphism between $\bR^{\infty}\oplus \bR$ and $\bR^{\infty}$ to define
a homeomorphism $G_{n+1}(\bR^{\infty}\oplus \bR)\iso G_{n+1}(\bR^{\infty})$, and we check that 
the homotopy class of this homeomorphism is independent of the choice of isomorphism. We then
define $i_n$ on $G_n(\bR^{\infty})$ by sending an $n$-plane $x$ in $\bR^{\infty}$ to the 
$(n+1)$-plane $x\oplus \bR$.

Similarly, we have a classifying map
$$p_{m,n}: BO(m)\times BO(n)\rtarr BO(m+n)$$
that is characterized up to homotopy by
$p_{m,n}^*(\ga_{m+n})=\ga_m\times \ga_n$. If we have a characteristic class $c$ on $(m+n)$-plane 
bundles, then 
$$p_{m,n}^*c(\ga_{m+n})=c(\ga_m\times \ga_n),$$ 
and this leads
by naturality to a description of $c(\ze\times\xi)$ for general $m$-plane bundles $\ze$ and 
$n$-plane bundles $\xi$. To give an explicit description of $p_{m,n}$, we may think of $BO(m+n)$ as 
$G_{m+n}(\bR^{\infty}\oplus \bR^{\infty})$; precisely, we use an isomorphism between
$\bR^{\infty}\oplus \bR^{\infty}$ and $\bR^{\infty}$ to define
a homeomorphism $G_{m+n}(\bR^{\infty}\oplus \bR^{\infty})\iso G_{m+n}(\bR^{\infty})$, and we check 
that the homotopy class of this homeomorphism is independent of the choice of isomorphism.
We then define $p_{m,n}$ on $G_m(\bR^{\infty})\times G_n(\bR^{\infty})$ by sending $(x,y)$
to $x\oplus y$, where $x$ is an $m$-plane in $\bR^{\infty}$ and $y$ is an 
$n$-plane in $\bR^{\infty}$.

We bring this down to earth by describing all characteristic classes in mod $2$ cohomology.
In fact, we have the following equivalent pair of theorems. The first uses the
language of characteristic classes, while the second describes $H^*(BO(n);\bZ_2)$ together
with the induced maps $i^*_n$ and $p_{m,n}^*$. 

\begin{thm} For $n$-plane bundles $\xi$ over base spaces $B$, $n\geq 0$, there are 
characteristic classes $w_i(\xi)\in H^i(B;\bZ_2)$, $i\geq 0$, called the Stiefel-Whitney
classes.\index{Stiefel-Whitney classes} They satisfy and are uniquely characterized by the 
following axioms.
\begin{enumerate}
\item $w_0(\xi)=1$ and $w_i(\xi) = 0$ if $i>\text{\em dim}\,\xi$.
\item $w_1(\ga_1)\neq 0$, where $\ga_1$ is the universal line bundle over $\bR P^{\infty}$.
\item $w_i(\xi\oplus \epz)= w_i(\xi)$.
\item $w_i(\ze\oplus \xi)= \sum_{j=0}^i w_j(\ze)\cup w_{i-j}(\xi)$.
\end{enumerate}
Every mod $2$ characteristic class for $n$-plane bundles can be written uniquely as a polynomial
in the Stiefel-Whitney classes $\sset{w_1,\ldots\!,w_n}$.
\end{thm}

\begin{thm} For $n\geq 1$, there are elements $w_i\in H^i(BO(n);\bZ_2)$, $i\geq 0$, called the
Stiefel-Whitney classes. They satisfy and are uniquely characterized by the following
axioms.
\begin{enumerate}
\item $w_0=1$ and $w_i = 0$ if $i>n$.
\item $w_1\neq 0$ when $n=1$.
\item $i_n^*(w_i)=w_i$.
\item $p_{m,n}^*(w_i) = \sum_{j=0}^i w_j\ten w_{i-j}$.
\end{enumerate}
The mod $2$ cohomology $H^*(BO(n);\bZ_2)$ is the polynomial algebra $\bZ_2[w_1,\ldots\!,w_n]$.
\end{thm}

For the uniqueness, suppose given another collection of classes $w'_i$ for all $n\geq 1$ that 
satisfy the stated properties. Since $BO(1)=\bR P^{\infty}$, $w_1=w'_1$ is the unique non-zero 
element of $H^1(\bR P^{\infty};\bZ_2)$. Therefore $w_i=w'_i$ for all $i$ when $n=1$, and we 
assume that this is true for all $m<n$. Visibly $i_{n-1}^*$ is an
isomorphism in degrees less than $n$, and this implies that $w_i=w'_i$ in $H^i(BO(n);\bZ_2)$
for $i<n$. It is less visible but easily checked that the $p_{m,n}^*$ are all monomorphisms
in all degrees. Since $p_{1,n-1}^*(w_n)=p_{1,n-1}^*(w'_n)$, this implies that $w_n=w'_n$.

\section{Stiefel-Whitney classes of manifolds}

It is convenient to consider $H^{**}(X)=\prod_iH^i(X)$ and to write its elements as formal 
sums $\sum x_i$, $\deg\,x_i = i$. In practice, we usually impose conditions that guarantee
that the sum is finite. We define the total
Stiefel-Whitney class\index{total Stiefel-Whitney class} $w(\xi)$ of a vector bundle $\xi$ to 
be $\sum w_i(\xi)$; here the sum is clearly finite. Note in particular that $w(\epz^q)=1$, where 
$\epz^q$ is the trivial $q$-plane bundle. With this notation, we have the formula
$$w(\ze\oplus \xi) = w(\ze)\cup w(\xi).$$ 

It is usual to write $w_i(M)=w_i(\ta(M))$ and $w(M)=w(\ta(M))$ for a smooth compact manifold $M$.
Suppose that $M$ immerses in $\bR^q$ with normal bundle $\nu$. Then $\ta(M)\oplus \nu \iso \epz^q$
and we have the ``Whitney duality formula''\index{Whitney duality formula}
$$ w(M)\cup w(\nu) = 1,$$
which shows how to calculate tangential Stiefel-Whitney classes in terms of normal
Stiefel-Whitney classes, and conversely. This formula can be used to prove non-immersion 
results when we know $w(M)$.  If $M$ has dimension $n$, then $\nu$ has dimension $q-n$ 
and must satisfy $w_i(\nu)=0$ if $i>q-n$. Calculation of $w_i(\nu)$ from the Whitney duality 
formula can lead to a contradiction if $q$ is too small. 

One calculation is immediate. Since the normal bundle of the standard embedding 
$S^q\rtarr \bR^{q+1}$ is trivial, $w(S^q)=1$. A manifold is said to be 
parallelizable\index{parallelizable manifold} if its tangent bundle is trivial.
For some manifolds $M$, we can show that $M$ is not parallelizable by showing that
one of its Stiefel-Whitney classes is non-zero, but this strategy fails for $M=S^q$.

We describe some standard computations in the cohomology of projective spaces that give 
less trivial examples. Write $\ze_q$ for the canonical line bundle\index{canonical line
bundle} over $\bR P^{q}$ in this 
section. (We called it $\ga_1^{q+1}$ before.)  The total space of $\ze_q$ consists of 
pairs $(x,v)$, where $x$ is a line in $\bR^{q+1}$
and $v$ is a point on that line. This is a subbundle of the trivial  $(q+1)$-plane 
bundle $\epz^{q+1}$, and we write $\ze_q^{\perp}$ for the complementary bundle whose
points are pairs $(x,w)$ such that $w$ is orthogonal to the line $x$. Thus
$$\ze_q\oplus \ze_q^{\perp}\iso\epz^{q+1}.$$

Write $H^*(\bR P^q;\bZ_2) =\bZ_2[\al]/(\al^{q+1})$,
$\deg\al =1$. Thus $\al=w_1(\ze_q)$. Since $\ze_q$ is a line bundle, $w_i(\ze_q)=0$ for 
$i>1$. The formula $w(\ze_q)\cup w(\ze_q^{\perp})=1$ implies that
$$w(\ze_q^{\perp}) = 1+\al +\cdots + \al^{q}.$$

We can describe $\ta(\bR P^q)$ in terms of $\ze_q$. Consider a point $x\in S^q$ and write 
$(x,v)$ for a typical vector in the tangent plane of $S^q$ at $x$. Then $x$ is orthogonal
to $v$ in $\bR^{q+1}$ and $(x,v)$ and $(-x,-v)$ have the same image in $\ta(\bR P^q)$. If 
$L_x$ is the line through $x$, then this image point determines and is determined by the 
linear map $f: L_x\rtarr L_x^{\perp}$ that sends $x$ to $v$. Starting from this, it is 
easy to check that $\ta(\bR P^q)$ is isomorphic to the bundle $\Hom(\ze_q,\ze_q^{\perp})$.
As for any line bundle, we have $\Hom(\ze_q,\ze_q)\iso\epz$ since the identity homomorphisms
of the fibers specify a cross-section. Again, as for any bundle over a smooth manifold, a
choice of Euclidean metric determines an isomorphism $\Hom(\ze_q,\epz)\iso \ze_q$. These
facts give the following calculation of $\ta(\bR P^q)\oplus\epz$:
\begin{eqnarray*}
\ta(\bR P^q)\oplus\epz & \iso & \Hom(\ze_q,\ze_q^{\perp})\oplus\Hom(\ze_q,\ze_q) \\
& \iso & \Hom(\ze_q,\ze_q^{\perp}\oplus \ze_q) \iso \Hom(\ze_q,\epz^{q+1}) \\
& \iso & (q+1)\Hom(\ze_q,\epz)\iso (q+1)\ze_q.
\end{eqnarray*}
Therefore 
$$w(\bR P^q) = w((q+1)\ze_q) = w(\ze_q)^{q+1} 
= (1+\al)^{q+1}=\sum_{0\leq i\leq q}
 \left(\begin{array}{c}q+1\\i\end{array}\right) \al^i.$$
Explicit computations are obtained by computing mod $2$ binomial coefficients.

For example, $w(\bR P^q)=1$ if and only if $q=2^k-1$ for some $k$ (as the reader should 
check) and therefore $\bR P^q$ can be parallelizable only if $q$ is of this form. If $\bR^{q+1}$ 
admits a bilinear product without zero divisors, then it is not hard to prove that
$\ta(\bR P^{q})\iso \Hom(\ze_q,\ze_q^{\perp})$ admits $q$ linearly independent cross-sections 
and is therefore trivial. We conclude that $\bR^{q+1}$ can admit such a product only if $q+1=2^k$
for some $k$. The real numbers, complex numbers, quaternions, and Cayley numbers show that there
is such a product for $q+1=1$, $2$, $4$, and $8$. As we shall explain in the next chapter, these 
are in fact the only $q$ for which $\bR^{q+1}$ admits such a product.

While the calculation of $w(\bR P^q)$ just given is quite special, there is a remarkable general
recipe, called the ``Wu formula,'' for the computation of $w(M)$ in terms of Poincar\'e duality
and the Steenrod operations in $H^*(M;\bZ_2)$. In analogy with $w(M)$, we define the total
Steenrod square of an element $x$ by $Sq(x)=\sum_i Sq^i(x)$.\index{total Steenrod operation} 

\begin{thm}[Wu formula]\index{Wu formula} Let $M$ be a smooth closed $n$-manifold with 
fundamental class
$z\in H_n(M;\bZ_2)$. Then the total Stiefel-Whitney class $w(M)$ is equal to $Sq(v)$, where
$v=\sum v_i\in H^{**}(M;\bZ_2)$ is the unique cohomology class such that
$$\langle v\cup x,z\rangle = \langle Sq(x),z \rangle $$
for all $x\in H^*(M;\bZ_2)$. Thus, for $k\geq 0$, $v_k\cup x = Sq^k(x)$ for all 
$x\in H^{n-k}(M;\bZ_2)$, and 
$$w_k(M)=\sum_{i+j=k}Sq^i(v_j).$$
\end{thm}

Here the existence and uniqueness of $v$ is an easy exercise from the Poincar\'e duality
theorem. The basic reason that such a formula holds is that the Stiefel-Whitney classes 
can be defined in terms of the Steenrod operations, as we shall see shortly. The Wu formula 
implies that the Stiefel-Whitney classes are homotopy invariant: if $f:M\rtarr M'$ is a 
homotopy equivalence between smooth closed $n$-manifolds, then 
$f^*: H^*(M';\bZ_2)\rtarr H^*(M;\bZ_2)$ satisfies $f^*(w(M'))=w(M)$. In fact, the conclusion 
holds for any map $f$, not necessarily a homotopy equivalence, that induces an isomorphism in 
mod $2$ cohomology. Since the tangent bundle of $M$ depends on its smooth structure, this 
is rather surprising.

\section{Characteristic numbers of manifolds}

Characteristic classes determine important numerical invariants of manifolds, called their
characteristic numbers.

\begin{defn} Let $M$ be a smooth closed $R$-oriented $n$-manifold with fundamental class 
$z\in H_n(M;R)$. For a characteristic class $c$ of degree $n$, define the tangential 
characteristic number\index{characteristic class}\index{characteristic number!tangential}
\index{characteristic number!normal} $c[M]\in R$ by $c[M] = \langle c(\ta(M)),z \rangle$. 
Similarly, define the normal characteristic number $c[\nu(M)]$ by
$c[\nu(M)] = \langle c(\nu(M)),z \rangle$, where $\nu(M)$ is the normal bundle associated
to an embedding of $M$ in $\bR^q$ for $q$ sufficiently large. (These numbers are well defined
because any two embeddings of $M$ in $\bR^q$ for large $q$ are isotopic and have equivalent
normal bundles.)
\end{defn}

In particular, if $r_i$ are integers such that $\sum ir_i=n$, then the monomial
$w_1^{r_1}\cdots w_n^{r_n}$ is a characteristic class of degree $n$, and all mod $2$
characteristic classes of degree $n$ are linear combinations of these. Different 
manifolds can have the same Stiefel-Whitney numbers.\index{Stiefel-Whitney numbers} In fact, 
we have the following observation.

\begin{lem} If $M$ is the boundary of a smooth compact $(n+1)$-manifold $W$, then
all tangential Stiefel-Whitney numbers of $M$ are zero.
\end{lem}
\begin{proof}
Using a smooth tubular neighborhood, we see that there is an inward-pointing normal 
vector field along $M$ that spans a trivial bundle $\epz$ such that 
$$\ta(W)|_M\iso \ta(M)\oplus \epz.$$
Therefore, if $i:M\rtarr W$ is the inclusion, then $i^*(w_j(W))=w_j(M)$. Let $f$ be
a polynomial in the $w_j$ of degree $n$. Recall that the fundamental class of $M$
is $\pa z$, where $z\in H_{n+1}(W,M)$ is the fundamental class of the pair $(W,M)$. We have
$$\langle f(M),\pa z \rangle
= \langle i^*f(W),\pa z \rangle
= \langle f(W), i_*\pa z \rangle = 0$$
since $i_*\pa = 0$ by the long exact homology sequence of the pair.
\end{proof}

\begin{lem} 
All tangential Stiefel-Whitney numbers\index{Stiefel-Whitney numbers!tangential} of a 
smooth closed manifold $M$ are zero if
and only if all normal Stiefel-Whitney numbers\index{Stiefel-Whitney numbers!normal} of 
$M$ are zero.
\end{lem}
\begin{proof} The Whitney duality formula implies that every $w_i(M)$ is a polynomial
in the $w_i(\nu(M))$ and every $w_i(\nu(M))$ is a polynomial in the $w_i(M)$.
\end{proof}

We shall explain the following amazing result of Thom in the last chapter.

\begin{thm}[Thom] If $M$ is a smooth closed $n$-manifold all of whose normal 
Stiefel-Whitney numbers are zero, then $M$ is the boundary of a smooth
$(n+1)$-manifold.
\end{thm}

Thus we need only compute the Stiefel-Whitney numbers of $M$ to determine whether
or not it is a boundary. By Wu's formula, the computation only requires knowledge
of the mod $2$ cohomology of $M$, with its Steenrod operations. In practice, it might 
be fiendishly difficult to actually construct a manifold with 
boundary $M$ geometrically.

\section{Thom spaces and the Thom isomorphism theorem}

There are several ways to construct the Stiefel-Whitney classes. The most illuminating
one depends on a simple, but fundamentally important, construction on vector bundles, 
namely their ``Thom spaces.'' This construction will also be at the heart of the proof of
Thom's theorem in the last chapter.

\begin{defn} Let $\xi: E\rtarr B$ be an $n$-plane bundle. Apply one-point compactification to each
fiber of $\xi$ to obtain a new bundle $Sph(E)$\index{Sph(E)@$Sph(E)$} over $B$ whose fibers are 
spheres $S^n$ with 
given basepoints, namely the points at $\infty$. These basepoints specify a cross-section
$B\rtarr  Sph(E)$. Define the Thom space\index{Thom space} $T\xi$ to be the quotient space
$T(\xi)= Sph(E)/B$. That is, $T(\xi)$ is obtained from $E$ by applying fiberwise one-point
compactification and then identifying all of the points at $\infty$ to a single basepoint 
(denoted $\infty$). Observe that this construction is functorial with respect to maps of 
vector bundles. 
\end{defn}

\begin{rem} If we give the bundle $\xi$ a Euclidean metric and let $D(E)$ and $S(E)$
denote its unit disk bundle and unit sphere bundle, then there is an evident
homeomorphism between $T\xi$ and the quotient space $D(E)/S(E)$. In turn, $D(E)/S(E)$
is homotopy equivalent to the cofiber of the inclusion $S(E)\rtarr D(E)$ and therefore
to the cofiber of the projection $S(E)\rtarr B$.
\end{rem}

If the bundle $\xi$ is trivial, so that $E=B\times \bR^n$, then $ Sph(E)=B\times S^n$. 
Quotienting out $B$ amounts to the same thing as giving $B$ a disjoint basepoint and then 
forming the smash product $B_+\sma S^n$. That is, in this case the Thom complex is $\SI^nB_+$. 
Therefore, for any cohomology theory $k^*$, 
$$k^q(B)=\tilde{k}^q(B_+) \iso \tilde{k}^{n+q}(T\xi).$$
There is a conceptual way of realizing this isomorphism. For any $n$-plane bundle $\xi: E\rtarr B$,
we have a projection $\xi:  Sph(E)\rtarr B$ and a quotient map $\pi:  Sph(E)\rtarr T\xi$.
We can compose their product with the diagonal map of $Sph(E)$ to obtain a composite map
$$  Sph(E)\rtarr  Sph(E)\times  Sph(E) \rtarr B\times T\xi.$$
This sends all points at $\infty$ to points of $B\times \sset{\infty}$. Therefore it factors through
a map
$$ \DE: T\xi\rtarr B_+\sma T\xi,$$
which is called the ``Thom diagonal.''\index{Thom diagonal} For a commutative ring $R$, we can 
use $\DE$ to define a cup product
$$ H^p(B;R)\ten \tilde{H}^q(T\xi;R) \rtarr \tilde{H}^{p+q}(T\xi;R).$$
When the bundle $\xi$ is trivial, we let $\mu\in \tilde{H}^n(B_+\sma S^n;R)$ be the suspension of
the identity element $1\in H^0(B;R)$, and we find that $x\rtarr x\cup \mu$ specifies the
suspension isomorphism $H^q(B;R)\iso \tilde{H}^{n+q}(B_+\sma S^n;R) = \tilde{H}^{n+q}(T\xi;R)$.

Now consider a general bundle $\xi$. On neighborhoods $U$ of $B$ over which $\xi$ is trivial,
we have $H^q(U;R)\iso \tilde{H}^{n+q}(T(\xi|_U);R)$. The isomorphism depends on the trivialization
$\ph_U: U\times \bR^n\rtarr \xi^{-1}(U)$. It is natural to ask if these isomorphisms patch
together to give a global isomorphism $H^q(B_+)\rtarr \tilde{H}^{n+q}(T\xi)$. This should look
very similar to the problem of patching local fundamental classes to obtain a global one; that
is, it looks like a question of orientation. This leads to the following definition and theorem.
For a point $b\in B$, let $S^n_b$ be the one-point compactification of the fiber $\xi^{-1}(b)$;
since $S^n_b$ is the Thom space of $\xi|_b$, we have a canonical map $i_b: S^n_b\rtarr T\xi$.

\begin{defn} Let $\xi: E\rtarr B$ be an $n$-plane bundle.
An $R$-orientation,\index{Rorientation@$R$-orientation} or Thom class,\index{Thom class} of 
$\xi$ is an element $\mu\in \tilde{H}^n(T\xi;R)$ such that, for every point $b\in B$, 
$i_b^*(\mu)$ is a generator of the free $R$-module $\tilde{H}^n(S^n_b)$.
\end{defn}
 
We leave it as an instructive exercise to verify that an $R$-orientation of a closed $n$-manifold
$M$ determines and is determined by an $R$-orientation of its tangent bundle $\ta(M)$. 

\begin{thm}[Thom isomorphism theorem]\index{Thom isomorphism} Let $\mu\in \tilde{H}^n(T\xi;R)$ 
be a Thom class for an $n$-plane bundle $\xi: E\rtarr B$. Define
$$\PH: H^q(B;R)\rtarr \tilde{H}^{n+q}(T\xi;R)$$
by $\PH (x)=x\cup \mu$. Then $\PH$ is an isomorphism.
\end{thm}
\begin{proof}[Sketch Proof] When $R$ is a field, this can be proved by an inductive Mayer-Vietoris
sequence argument. To exploit inverse images of open subsets of $B$, it is convenient to observe 
that, by easy homotopy and excision arguments,
$$\tilde{H}^*(T\xi)\iso H^*(Sph(E),B)\iso H^*(Sph(E),Sph(E)_0)\iso H^*(E,E_0),$$ 
where $E_0$ and $Sph(E)_0$ are the subspaces of $E$ and $Sph(E)$ obtained by deleting $\sset{0}$ 
from each fiber. Use of a field ensures that the cohomology of the relevant direct limits is the 
inverse limit of the cohomologies. An
alternative argument that works for general $R$ can be obtained by first showing that one can
assume that $B$ is a CW complex, by replacing $\xi$ by its pullback along a CW approximation of $B$,
and then proceeding by induction over the restrictions of $\xi$ to the skeleta of $B$; one point
is that the restriction of $\xi$ to any cell is trivial and another is that the cohomology
of $B$ is the inverse limit of the cohomologies of its skeleta. However, much the best proof
from the point of view of anyone seriously interested in algebraic topology is to apply the 
Serre spectral sequence of the bundle $Sph(E)$. The Serre spectral sequence\index{Serre spectral
sequence} is a device for
computing the cohomology of the total space $E$ of a fibration from the cohomologies of its base
$B$ and fiber $F$. It measures the cohomological deviation of $H^*(E)$ from $H^*(B)\ten H^*(F)$.
In the present situation, the existence of a Thom class ensures that there is no deviation for 
the sphere bundle $Sph(E)\rtarr B$, so that 
$$H^*(Sph(E);R)\iso H^*(B;R)\ten H^*(S^n;R).$$ 
The section given by the points at $\infty$ induces an isomorphism of $H^*(B;R)\ten H^0(S^n;R)$ 
with $H^*(B;R)$, and the quotient map $Sph(E)\rtarr T\xi$ induces an isomorphism of 
$\tilde{H}^*(T\xi;R)$ with $H^*(B;R)\ten H^n(S^n;R)$.
\end{proof}

Just as in orientation theory for manifolds, the question of orientability depends on the
structure of the units of the ring $R$, and this leads to the following conclusion.

\begin{prop} Every vector bundle admits a unique $\bZ_2$-orientation. 
\end{prop}

This can be proved along with the Thom isomorphism theorem by a Mayer-Vietoris argument.

\section{The construction of the Stiefel-Whitney classes}

We indicate two constructions of the Stiefel-Whitney classes. Each has distinct advantages over 
the other. First, taking the characteristic class point of view, we define the Stiefel-Whitney 
classes\index{Stiefel-Whitney classes} in terms of the Steenrod operations by setting
$$w_i(\xi) = \PH^{-1}Sq^i\PH(1) = \PH^{-1}Sq^i\mu.$$
Naturality is obvious. Axiom 1 is immediate from the relations $Sq^0=\id$ and 
$Sq^i(x)=0$ if $i> \deg\,x$. For axiom 2, we use the following observation.

\begin{lem} There is a homotopy equivalence $j: \bR P^{\infty}\rtarr T\ga_1$.
\end{lem}
\begin{proof}
$T\ga_1$ is homeomorphic to $D(\ga_1)/S(\ga_1)$. Here $S(\ga_1)$ is the infinite sphere 
$S^{\infty}$, which is the universal cover of $\bR P^{\infty}$ and is therefore 
contractible. The zero section $\bR P^{\infty}\rtarr D(\ga_1)$ and the quotient map 
$D(\ga_1)\rtarr T\ga_1$ are homotopy equivalences, and their composite is the required 
homotopy equivalence $j$.  
\end{proof}

Since $Sq^1(x)=x^2$ if $\deg\,x=1$, the lemma implies that $Sq^1$ is non-zero on the Thom class 
of $\ga_1$, verifying axiom 2. For axiom 3, we easily check that
$T(\xi\oplus\epz)\iso \SI T(\xi)$ for any vector bundle $\xi$ and that the Thom class of 
$\xi\oplus\epz$ is the suspension of the Thom class of $\xi$. Thus axiom 3 follows from the 
stability of the Steenrod operations. For axiom 4, we easily check that, for any vector bundles 
$\ze$ and $\xi$, $T(\ze\times \xi)\iso T\ze\sma T\xi$ and the Thom class of $\ze\times \xi$ is the 
tensor product of the Thom classes of $\ze$ and $\xi$. Interpreting the Cartan formula for the 
Steenrod operations externally in the cohomology of products and therefore of smash products,
we see that it implies axiom 4. That is, the properties that axiomatize the Steenrod operations 
directly imply the properties that axiomatize the Stiefel-Whitney classes. 

We next take the classifying space point of view. As we shall explain in \S8, 
passage from topological groups to their classifying spaces is a product-preserving 
functor, at least up to homotopy. We may embed $(\bZ_2)^n = O(1)^n$ in $O(n)$ as the
subgroup of diagonal matrices. The classifying space $BO(1)$ is $\bR P^{\infty}$,
and we obtain a map
$$\om: (\bR P^{\infty})^n \htp B(O(1)^n) \rtarr BO(n) $$
upon passage to classifying spaces. The symmetric group $\SI_n$ is contained in $O(n)$ 
as the subgroup of permutation matrices, and the diagonal subgroup $O(1)^n$ is closed 
under conjugation by symmetric matrices. Application of the classifying space functor 
to conjugation by permutation matrices induces the corresponding permutation of the factors 
of $BO(1)^n$, and it induces the identity map on $BO(n)$. Indeed, up to homotopy, inner 
conjugation by an element of $G$ induces the identity map on $BG$ for any topological
group $G$. 

By the K\"unneth theorem, we see that
$$H^*((\bR P^{\infty})^n;\bZ_2) 
= \ten_{i=1}^n H^*(\bR P^{\infty};\bZ_2) =\bZ_2[\al_1,\ldots\!,\al_n],$$
where the generators $\al_i$ are of degree one. The symmetric group $\SI_n$ acts on this 
cohomology ring by permuting the variables $\al_i$. The subring 
$H^*((\bR P^{\infty})^n;\bZ_2)^{\SI_n}$ of elements invariant under 
the action is the polynomial algebra on the elementary symmetric functions 
$\si_i$, $1\leq i\leq n$, in the variables $\al_i$. Here 
$$\si_i = \textstyle{\sum} \al_{j_1}\cdots\al_{j_i},\ \ 1\leq j_1 < \cdots < j_n,$$ 
has degree $i$. The induced map $\om^*: H^*(BO(n);\bZ_2)\rtarr H^*((\bR P^{\infty})^n;\bZ_2)$ 
takes values in $H^*((\bR P^{\infty})^n;\bZ_2)^{\SI_n}$. We shall give a general reason why
this is so in \S8.  The resulting map
$$\om^*: H^*(BO(n);\bZ_2)\rtarr H^*((\bR P^{\infty})^n;\bZ_2)^{\SI_n}$$
is a ring homomorphism between polynomial algebras on generators of the same degrees. It
turns out to be a monomorphism and therefore an isomorphism. We redefine the Stiefel-Whitney 
classes by letting $w_i$ be the unique element such that $\om^*(w_i)=\si_i$ for $1\leq i\leq n$ 
and defining $w_0=1$ and $w_i=0$ for $i>n$. Then axioms 1 and 2 for the Stiefel-Whitney 
classes are obvious, and we derive axioms 3 and 4 from algebraic properties of elementary 
symmetric functions. 

One advantage of this approach is that, since we know the Steenrod 
operations on $H^*(\bR P^{\infty};\bZ_2)$ and can read them off on $H^*((\bR P^{\infty})^n;\bZ_2)$ 
by the Cartan formula, it leads to a purely algebraic calculation of the Steenrod 
operations in $H^*(BO(n);\bZ_2)$. Explicitly, the following ``Wu formula''\index{Wu formula} holds:
$$ Sq^i(w_j) = \sum_{t=0}^i\left(\begin{array}{c}j+t-i-1\\t\end{array}\right) w_{i-t}w_{j+t}.$$

\section{Chern, Pontryagin, and Euler classes}

The theory of the previous sections extends appropriately to complex vector bundles
and to oriented real vector bundles. The proof of the classification theorem for complex 
$n$-plane bundles works in exactly the same way as for real $n$-plane bundles, using 
complex Grassmann varieties. For oriented real $n$-plane bundles, we use the Grassmann 
varieties\index{Grassmann variety!of oriented $n$-planes} of oriented $n$-planes, 
the points of which are planes $x$ together with a chosen 
orientation. In fact, the fundamental groups of the real Grassmann varieties are $\bZ_2$, 
and their universal covers are their orientation covers. These covers are the oriented 
Grassmann varieties $\tilde{G}_n(\bR^q)$. We write $BU(n) = G_n(\bC^{\infty})$\index{BUn@$BU(n)$} 
and $BSO(n) = \tilde{G}_n(\bR^{\infty})$,\index{BSOn@$BSO(n)$} and we construct universal 
complex $n$-plane bundles\index{universal n-plane bundle@universal $n$-plane bundle!complex}
\index{universal n-plane bundle@universal $n$-plane bundle!oriented}
$\ga_n: EU_n\rtarr BU(n)$ and oriented $n$-plane bundles $\tilde{\ga}_n: \tilde{E}_n\rtarr BSO(n)$
as in the first section. Let $\sE U_n(B)$\index{EUkn(-)@$\sE U_n(B)$} denote the set of equivalence 
classes of complex $n$-plane
bundles over $B$ and let $\tilde{\sE}_n(B)$\index{EanBa@$\tilde{\sE}_n(B)$} denote the set of 
equivalence classes of oriented
real $n$-plane bundles over $B$; it is required that bundle maps $(g,f)$ be orientation preserving,
in the sense that the induced map of Thom spaces carries the orientation of the target bundle to the
orientation of the source bundle. The universal bundle $\tilde{\ga_n}$ has a canonical orientation
which determines an orientation on $f^*\tilde{E}_n$ for any map $f: B\rtarr BSO(n)$. 

\begin{thm}\index{classification theorem!for complex $n$-plane bundles} The natural 
transformation $\PH: [-,BU(n)]\rtarr \sE U_n(-)$ obtained by sending the
homotopy class of a map $f: B\rtarr BU(n)$ to the equivalence class of the $n$-plane
bundle $f^*EU_n$ is a natural isomorphism of functors.
\end{thm}

\begin{thm}\index{classification theorem!for oriented $n$-plane bundles} The natural 
transformation $\PH: [-,BSO(n)]\rtarr \tilde{\sE}_n(-)$ obtained by 
sending the homotopy class of a map $f: B\rtarr BSO(n)$ to the equivalence class of the
oriented $n$-plane bundle $f^*\tilde{E}_n$ is a natural isomorphism of functors.
\end{thm}

The definition of characteristic classes for complex $n$-plane bundles and for oriented
real $n$-plane bundles in a cohomology theory $k^*$ is the same as for real $n$-plane bundles, 
and the Yoneda lemma applies.

\begin{lem} Evaluation on $\ga_n$ specifies a canonical bijection between characteristic
classes of complex $n$-plane bundles and elements of $k^*(BU(n))$.
\end{lem}

\begin{lem} Evaluation on $\tilde{\ga}_n$ specifies a canonical bijection between characteristic
classes of oriented $n$-plane bundles and elements of $k^*(BSO(n))$.
\end{lem}

Clearly we have a $2$-fold cover $\pi_n: BSO(n)\rtarr BO(n)$. The mod $2$ characteristic 
classes for oriented $n$-plane bundles are as one might expect from this. Continue to write
$w_i$ for $\pi^*(w_i)\in H^i(BSO(n);\bZ_2)$; here $w_1=0$ since $BSO(n)$ is simply connected.

\begin{thm} $H^*(BSO(n);\bZ_2) \iso \bZ_2[w_2,\ldots\!,w_n]$. 
\end{thm}

If we regard a complex $n$-plane bundle as a real $2n$-plane bundle, then the complex structure 
induces a canonical orientation. By the Yoneda lemma, the resulting natural transformation 
$r: \sE U_n(-)\rtarr \tilde{\sE}_n(-)$ is represented by a map $r: BU(n)\rtarr BSO(2n)$. Explicitly, ignoring its complex structure, we may identify 
$\bC^{\infty}$ with $\bR^{\infty}\oplus\bR^{\infty}\iso \bR^{\infty}$ and so regard a complex 
$n$-plane in $\bC^{\infty}$ as
an oriented $2n$-plane in $\bR^{\infty}$. Similarly, we may complexify real bundles fiberwise 
and so obtain a natural transformation $c: \sE_n(-)\rtarr \sE U_n(-)$. It is represented by a 
map $c: BO(n)\rtarr BU(n)$. Explicitly, identifying $\bC^{\infty}$ with 
$\bR^{\infty}\ten_{\bR}{\bC}$, we may complexify an $n$-plane in $\bR^{\infty}$ to obtain an 
$n$-plane in $\bC^{\infty}$.

The Thom space\index{Thom space!of a complex bundle} of a complex or oriented real vector bundle 
is the Thom space of its underlying real vector bundle. We obtain characteristic classes in 
cohomology with any coefficients by
applying cohomology operations to Thom classes, but it is rarely the case that the resulting
characteristic classes generate all characteristic classes: the cases $H^*(BO(n);\bZ_2)$ and
$H^*(BSO(n);\bZ_2)$ are exceptional. Characteristic classes constructed in this fashion satisfy
homotopy invariance properties that fail for general characteristic classes. 

In the complex case, with integral coefficients, we have a parallel to our second
approach to Stiefel-Whitney classes that leads to a description of $H^*(BU(n);\bZ)$ in 
terms of Chern classes. We may embed $(S^1)^n = U(1)^n$ in $U(n)$ as the
subgroup of diagonal matrices. The classifying space $BU(1)$ is $\bC P^{\infty}$,
and we obtain a map
$$\om: (\bC P^{\infty})^n \htp B(U(1)^n) \rtarr BU(n) $$
upon passage to classifying spaces. The symmetric group $\SI_n$ is contained in $U(n)$ 
as the subgroup of permutation matrices, and the diagonal subgroup $U(1)^n$ is closed 
under conjugation by symmetric matrices. Application of the classifying space functor 
to conjugation by permutation matrices induces the corresponding permutation of the factors 
of $BU(1)^n$, and it induces the identity map on $BU(n)$. 

By the K\"unneth theorem, we see that
$$H^*((\bC P^{\infty})^n;\bZ) 
= \ten_{i=1}^n H^*(\bC P^{\infty};\bZ) =\bZ[\be_1,\ldots\!,\be_n],$$
where the generators $\be_i$ are of degree two. The symmetric group $\SI_n$ acts on this 
cohomology ring by permuting the variables $\be_i$. The subring 
$H^*((\bC P^{\infty})^n;\bZ)^{\SI_n}$ of elements invariant under 
the action is the polynomial algebra on the elementary symmetric functions 
$\si_i$, $1\leq i\leq n$, in the variables $\be_i$. Here 
$$\si_i = \textstyle{\sum} \be_{j_1}\cdots\be_{j_i},\ \ 1\leq j_1 < \cdots < j_n,$$ 
has degree $2i$. The induced map $\om^*: H^*(BU(n);\bZ)\rtarr H^*((\bC P^{\infty})^n;\bZ)$ 
takes values in $H^*((\bC P^{\infty})^n;\bZ)^{\SI_n}$. The resulting map
$$\om^*: H^*(BU(n);\bZ)\rtarr H^*((\bC P^{\infty})^n;\bZ)^{\SI_n}$$
is a ring homomorphism between polynomial algebras on generators of the same degrees. It
turns out to be a monomorphism and thus an isomorphism when tensored with any field, and it is
therefore an isomorphism. We define the Chern classes by letting $c_i$, $1\leq i\leq n$, be the 
unique element such that $\om^*(c_i)=\si_i$. 

\begin{thm} For $n\geq 1$, there are elements $c_i\in H^{2i}(BU(n);\bZ)$, $i\geq 0$, called the
Chern classes.\index{Chern classes} They satisfy and are uniquely characterized by the following
axioms.
\begin{enumerate}
\item $c_0=1$ and $c_i = 0$ if $i>n$.
\item $c_1$ is the canonical generator of $H^2(BU(1);\bZ)$ when $n=1$.
\item $i_n^*(c_i)=c_i$.
\item $p_{m,n}^*(c_i) = \sum_{j=0}^i c_j\ten c_{i-j}$.
\end{enumerate}
The integral cohomology $H^*(BU(n);\bZ)$ is the polynomial algebra $\bZ[c_1,\ldots\!,c_n]$.
\end{thm}

Here we take axiom 1 as a definition and we interpret axiom 2 as meaning that $c_1$
corresponds to the identity map of $\bC P^{\infty}$ under the canonical identification of 
$[\bC P^{\infty},\bC P^{\infty}]$ with $H^2(\bC P^{\infty};\bZ)$. Axioms 3 and 4 can
be read off from algebraic properties of elementary symmetric functions. The theorem admits
an immediate interpretation in terms of characteristic classes. Observe that, since $H^*(BU(n);\bZ)$
is a free Abelian group, the theorem remains true precisely as stated with $\bZ$ replaced by any
other commutative ring of coefficients $R$. We continue to write $c_i$ for the image of $c_i$
in $H^*(BU(n);R)$ under the homomorphism induced by the unit $\bZ\rtarr R$ of the ring $R$.

The reader deserves to be warned about a basic inconsistency in the literature.

\begin{rem} With the discussion above, $c_1(\ga_1^{n+1})$ is the canonical generator of
$H^2(\bC P^n;\bZ)$, where $\ga_1^{n+1}$ is the canonical line bundle
\index{canonical line bundle} of lines in $\bC^{n+1}$
and points on the line. This is the standard convention in algebraic topology. In algebraic 
geometry, it is more usual to define Chern classes so that the first Chern class of the dual 
of $\ga_1^{n+1}$ is the canonical generator of $H^2(\bC P^n;\bZ)$. With this convention, the 
$n$th Chern class would be $(-1)^nc_n$. It is often unclear in the literature which convention 
is being followed.
\end{rem}

Turning to oriented real vector bundles, we define the Pontryagin and Euler classes as 
follows, taking cohomology with coefficients in any commutative ring $R$.

\begin{defn} Define the Pontryagin classes\index{Pontryagin classes} $p_i\in H^{4i}(BO(n);R)$ by 
$$p_i = (-1)^ic^*(c_{2i}),$$ 
$c^*: H^{4i}(BU(n);R)\rtarr H^{4i}(BO(n);R)$; also write $p_i$ for
$\pi_n^*(p_i)\in H^{4i}(BSO(n);R)$. 
\end{defn}

\begin{defn} Define the Euler class\index{Euler class} $e(\xi)\in H^n(B;R)$ of an $R$-oriented 
$n$-plane bundle 
$\xi$ over the base space $B$ by $e(\xi)=\PH^{-1}\mu^2$, where $\mu\in H^n(T\xi;R)$ is the Thom 
class. Giving the universal oriented $n$-plane bundle over $BSO(n)$ the $R$-orientation 
induced by its integral orientation, this defines the Euler class $e\in H^n(BSO(n);R)$.
\end{defn}

If $n$ is odd, then $2\mu^2=0$ and thus $2e=0$. If $R=\bZ_2$, then $Sq^n(\mu)=\mu^2$ and
thus $e=w_n$. The name ``Euler class'' is justified by the following classical result, 
which well illustrates the kind of information that characteristic numbers can 
encode.\footnote{See Corollary 11.12 of Milnor and Stasheff {\em Characteristic Classes}
for a proof.}

\begin{thm} If $M$ is a smooth closed oriented manifold, then the characteristic number
$e[M]=\langle e(\ta(M)),z\rangle\in \bZ$ is the Euler characteristic\index{Euler characteristic} 
of $M$.
\end{thm}

The evident inclusion $T^n \iso SO(2)^n\rtarr SO(2n)$ is a maximal torus, and it induces
a map $BT^n\rtarr BSO(2n)$. A calculation shows that $e$ restricts to the $n$th 
elementary symmetric polynomial $\be_1\cdots\be_n$. The cited inclusion factors through 
the homomorphism $U(n)\rtarr SO(2n)$, hence $BT^n\rtarr BSO(2n)$ factors through
$r: BU(n)\rtarr BSO(2n)$. This implies another basic fact about the Euler class. 

\begin{prop} $r^*: H^*(BSO(2n);\bZ)\rtarr H^*(BU(n);\bZ)$ sends $e$ to $c_n$. 
\end{prop}

The presence of $2$-torsion makes the description of the integral cohomology rings of $BO(n)$
and $BSO(n)$ quite complicated, and these rings are almost never used in applications.
Rather, one uses the mod $2$ cohomology rings and the following description of the cohomology
rings that result by elimination of $2$-torsion. 

\begin{thm} Take coefficients in a ring $R$ in which $2$ is a unit. Then
$$H^*(BO(2n)) \iso H^*(BO(2n+1))\iso H^*(BSO(2n+1)) \iso R[p_1,\ldots\!,p_n]$$
and
$$H^*(BSO(2n))\iso R[p_1,\ldots\!,p_{n-1},e], \, \, \text{with}\, \, e^2=p_n.$$
\end{thm}

\section{A glimpse at the general theory}

We should place the theory of vector bundles in a more general context. We have
written $BO(n)$, $BU(n)$, and $BSO(n)$ for certain ``classifying spaces'' in this
chapter, but we defined a classifying space $BG$ for any topological group
$G$ in Chapter 16 \S5. In fact, the spaces here are homotopy equivalent to the spaces of 
the same name that we defined there, and we here explain why. 

Consider bundles $\xi: Y\rtarr B$ with fiber $G$. For spaces $U$ in a numerable open cover 
$\sO$ of $B$, there are homeomorphisms $\ph: U\times G\rtarr p^{-1}(U)$ such 
that $p\com \ph= \pi_1$. We say that $Y$ is a principal $G$-bundle
\index{principal G-bundle@principal $G$-bundle} if $Y$ has a free
right action by $G$, $B$ is the orbit space $Y/G$, $\xi$ is the quotient map, and the 
$\ph$ are maps of right $G$-spaces. We say that $\xi: Y\rtarr B$ is a universal 
principal $G$-bundle\index{universal principal G-bundle@universal principal $G$-bundle}
if $Y$ is a contractible space. In particular, for any topological group $G$
whose identity element is a nondegenerate basepoint, such as any Lie group $G$, the map 
$p: EG\rtarr BG$ constructed in Chapter 16 \S5 is a universal principal $G$-bundle. The 
classification
theorem below implies that the base spaces of any two universal principal $G$-bundles are 
homotopy equivalent, and it is usual to write $BG$ for any space\index{classifying space}
in this homotopy type. Observe that the long exact sequence of homotopy groups of a universal 
principal $G$-bundle gives isomorphisms $\pi_q(BG)\iso \pi_{q-1}(G)$ for $q\geq 1$. 

We have implicitly constructed other examples of universal principal $G$-bundles when 
$G$ is $O(n)$, $U(n)$, or $SO(n)$. To see this, consider $V_n(\bR^q)$.
Write $\bR^q=\bR^n\times \bR^{q-n}$ and note that this fixes embeddings of $O(n)$ and 
$O(q-n)$ in the orthogonal group $O(q)$. Of course, $O(q)$ acts on vectors in $\bR^q$ 
and thus on $n$-frames. Consider the fixed $n$-frame $x_0=\sset{e_1,\ldots\!,e_n}$. Any other 
$n$-frame can be obtained from this one by the action of an element of $O(q)$, and 
the isotropy group of $x_0$ is $O(q-n)$. Thus the action of $O(q)$ is transitive, and 
evaluation on $x_0$ induces a homeomorphism $O(q)/O(q-n) \rtarr V_n(\bR^q)$ of $O(q)$-spaces.
The action of $O(n)\subset O(q)$ is free, and passage to orbits gives a homeomorphism
$O(q)/O(n)\times O(q-n) \rtarr G_n(\bR^q)$. It is intuitively clear and not hard to prove
that the colimit over $q$ of the inclusions $O(q-n)\rtarr O(q)$ is a homotopy equivalence
and that this implies the contractibility of $V_n(\bR^{\infty})$. We deduce that
$V_n(\bR^{\infty})$ is a universal principal $O(n)$-bundle. We have analogous universal
principal $U(n)$-bundles and $SO(n)$-bundles.

There is a classification theorem\index{classification theorem!for principal $G$-bundles} for 
principal $G$-bundles. Let $\sP G(B)$\index{PG(B)@$\sP G(B)$} denote the set
of equivalence classes of principal $G$-bundles over $B$, where two principal $G$-bundles
over $B$ are equivalent if there is a $G$-homeomorphism over $B$ between them.
Via pullback of bundles, this is a contravariant set-valued functor on the homotopy category 
of spaces.

\begin{thm} Let $\ga: Y\rtarr Y/G$ be any universal principal $G$-bundle. The natural 
transformation $\PH: [-,Y/G]\rtarr \sP G(-)$ obtained by sending the
homotopy class of a map $f: B\rtarr Y/G$ to the equivalence class of the principal
$G$-bundle $f^*Y$ is a natural isomorphism of functors.
\end{thm}

Now let $F$ be any space on which $G$ acts effectively from the left. Here an action
is effective\index{effective group action} if $gf=f$ for every $f\in F$ implies $g=e$. For a principal 
$G$-bundle $Y$, let $G$ act on $Y\times F$ by $g(y,f)=(yg^{-1},gf)$ and let $Y\times_G F$ be the
orbit space $(Y\times F)/G$. With the correct formal definition of a fiber bundle 
with group $G$ and fiber $F$, every such fiber bundle $p: E \rtarr B$ is equivalent to 
one of the form $Y\times_G F\rtarr Y/G\iso B$ for some principal $G$-bundle $Y$ over $B$;
moreover $Y$ is uniquely determined up to equivalence. 

In fact, the ``associated principal
$G$-bundle''\index{associated principal $G$-bundle} $Y$ can be constructed 
as the function space of all maps $\ps:F\rtarr E$ such that $\ps$ is an admissible 
homeomorphism onto some fiber $F_b=p^{-1}(b)$. Here admissibility means that the composite
of $\ps$ with the homeomorphism $F_b\rtarr F$ determined by a coordinate chart
$\ph: U\times F\overto{\iso} p^{-1}(U)$, $b\in U$, coincides with action by some element 
of $G$. The left action of $G$ on $F$ induces a right action of $G$ on $Y$; this action
is free because the given action on $F$ is effective. The projection $Y\rtarr B$
sends $\ps$ to $b$ when $\ps: F\overto{\iso} F_b$, and it factors through a homeomorphism
$Y/G\rtarr B$. $Y$ inherits local triviality from $p$, and the evaluation map $Y\times F\rtarr E$ 
induces an equivalence of bundles $Y\times_G F\rtarr E$. 

We conclude that, for any $F$, $\sP G(B)$ 
is naturally isomorphic to the set of equivalence classes of bundles with group $G$ and fiber 
$F$ over $B$. Fiber bundles with group $O(n)$ and fiber $\bR^n$ are real $n$-plane bundles,
fiber bundles with group $U(n)$ and fiber $\bC^n$ are complex $n$-plane bundles, and fiber
bundles with group $SO(n)$ and fiber $\bR^n$ are oriented real $n$-plane bundles. Thus the
classification theorems of the previous sections could all be rederived as special cases
of the general classification theorem for principal $G$-bundles stated in this section.

In our discussion of Stiefel-Whitney and Chern classes, we used that passage to
classifying spaces is a product-preserving functor, at least up to homotopy. 
For the functoriality, if $f: G\rtarr H$ is a homomorphism of topological groups, then 
consideration of the way bundles are constructed by gluing together coordinate charts shows 
that a principal $G$-bundle $\xi: Y\rtarr B$ naturally gives rise to a principal $H$-bundle 
$f_*Y\rtarr B$. This construction is represented on the classifying space level by a map 
$Bf: BG\rtarr BH$. 

In fact, if $EG\rtarr BG$ and $EH\rtarr BH$ are universal principal 
bundles, then any map $\tilde{f}: EG\rtarr EH$ such that $\tilde{f}(xg)=\tilde{f}(x)f(g)$ 
for all $x\in EG$ and $g\in G$ induces a map in the homotopy class $Bf$ on passage to orbits. 
For example, if $f: G\rtarr G$ is given by conjugation by $\ga\in G$, $f(g) = \ga^{-1}g\ga$, 
then $\tilde{f}(x) = x\ga$ satisfies this equivariance property and therefore $Bf$ is homotopic
to the identity. This explains why inner conjugations induce the identity map on passage to 
classifying spaces, as we used in our discussion of Stiefel-Whitney and Chern classes.

If $EG\rtarr BG$ and $EG'\rtarr BG'$ are universal principal $G$ and $G'$ bundles, then
$EG\times EG'$ is a contractible space with a free action by $G\times G'$. The orbit
space is $BG\times BG'$, and this shows that $BG\times BG'$ is a choice for the 
classifying space $B(G\times G')$ and is therefore homotopy equivalent to any other choice.

The explicit construction of $BG$ given in Chapter 16 \S5 is functorial in $G$ on the point-set 
level and not just up to homotopy, and it is product preserving in the strong sense that 
the projections induce a homeomorphism $B(H\times G)\iso BH\times BG$. 

\vspace{.1in}

\begin{center}
PROBLEMS
\end{center}
\begin{enumerate}
\item Verify that $w(\bR P^q)=1$ if and only if $q=2^k-1$ for some $k$.
\item Prove that $\bR P^{2^k}$ cannot immerse in $\bR^{2^{k+1}-2}$. (By the Whitney embedding
theorem, any smooth closed $n$-manifold immerses in $\bR^{2n-1}$, so this is a best possible
non-immersion result.)
\item Prove that all tangential Stiefel-Whitney numbers of $\bR P^{q}$ are zero if and only 
if $q$ is odd.
\item* Try to construct a smooth compact manifold whose boundary is $\bR P^{3}$.
\item Prove that a smooth closed $n$-manifold $M$ is $R$-orientable if and only its tangent
bundle is $R$-orientable. 
\end{enumerate}

\chapter{An introduction to $K$-theory}

The first generalized cohomology theory to be discovered was $K$-theory, and it plays a 
vital role in the connection of algebraic topology to analysis and algebraic geometry. 
The fact that it is a generalized cohomology theory is a consequence of the Bott
periodicity theorem, which is one of the most important and influential theorems in all of
topology. We give some basic information about $K$-theory and, following Adams and
Atiyah, we explain how the Adams operations in $K$-theory allow a quick solution to the 
``Hopf invariant one problem.''  One implication is the purely algebraic theorem that the 
only possible dimensions
of a real (not necessarily associative) division algebra are 1, 2, 4, and 8. We shall only
discuss complex $K$-theory, although there is a precisely analogous construction of real
$K$-theory $KO$. From the point of view of algebraic topology, real $K$-theory is a 
substantially more powerful invariant, but complex $K$-theory is usually more relevant to 
applications in other fields.

\section{The definition of $K$-theory}

Except where otherwise noted, we work with complex vector bundles throughout this chapter. 
Dimension will mean complex dimension and line bundles will mean complex line bundles. 
We consider the set $Vect(X)$\index{Vect(X)@$Vect(X)$} of equivalence classes of vector bundles 
over a space $X$. We assume unless otherwise specified that $X$ is compact. We remind the reader
that vector bundles can have different dimension over different components of $X$. The set 
$Vect(X)$ forms an Abelian monoid (= semi-group) under Whitney sum, and it forms a semi-ring 
with multiplication given by the (internal) tensor product of vector bundles over $X$. 

There is a standard construction, called the Grothendieck construction,\index{Grothendieck 
construction}
of an Abelian group $G(M)$ associated to an Abelian monoid $M$: one takes the quotient of 
the free Abelian group generated by the elements of $M$ by the subgroup generated by the
set of elements of the form $m+n-m\oplus n$, where $\oplus$ is the sum in $M$. The evident
morphism of Abelian monoids $i: M\rtarr G(M)$ is universal: for any homomorphism of monoids
$f: M\rtarr G$, where $G$ is an Abelian group, there is a unique homomorphism of groups
$\tilde{f}: G(M)\rtarr G$ such that $\tilde{f}\com i=f$. If $M$ is a semi-ring, then its
multiplication induces a multiplication on $G(M)$ such that $G(M)$ is a ring, called the
Grothendieck ring\index{Grothendieck ring} of $M$. If the semi-ring $M$ is commutative, then 
the ring $G(M)$ is commutative.

\begin{defn} The $K$-theory\index{K-theory@$K$-theory} of $X$, denoted $K(X)$,\index{K(X)@$K(X)$} 
is the Grothendieck ring of the 
semi-ring $Vect(X)$. An element of $K(X)$ is called a virtual bundle\index{virtual bundle} over $X$.
We write $[\xi]$ for the element of $K(X)$ determined by a vector bundle $\xi$. 
\end{defn}

Since $\epz$ is the identity element for the product in $K(X)$, it is standard to write 
$q=[\epz^q]$, where $\epz^q$ is the $q$-dimensional trivial bundle. For vector bundles
over a based space $X$, we have the function $d: Vect(X)\rtarr \bZ$ that sends a vector
bundle to the dimension of its restriction to the component of the basepoint $*$. Since
$d$ is a homomorphism of semi-rings, it induces a dimension function\index{dimension function}
$d: K(X)\rtarr \bZ$, which is a homomorphism of rings. Since $d$ is an isomorphism
when $X$ is a point, $d$ can be identified with the induced map $K(X)\rtarr K(*)$. 

\begin{defn} The reduced $K$-theory\index{K-theory@$K$-theory!reduced} 
$\tilde{K}(X)$\index{K(X)a@$\tilde K(X)$} of a based 
space $X$ is the kernel
of $d: K(X)\rtarr \bZ$. It is an ideal of $K(X)$ and thus a ring without identity.
Clearly $K(X)\iso \tilde{K}(X)\times \bZ$.
\end{defn}

We have a homotopical interpretation of these definitions, and it is for this that we
need $X$ to be compact. By the classification
theorem, we know that $\sE U_n(X)$ is naturally isomorphic to $[X_+,BU(n)]$; we have
adjoined a disjoint basepoint because we are thinking cohomologically and want the
brackets to denote based homotopy classes of maps. We have maps $i_n: BU(n)\rtarr BU(n+1)$.
With our construction of classifying spaces via Grassmannians, these maps are inclusions,
and we define $BU$ to be the colimit of the $BU(n)$, with the topology of the union. 

We say that bundles $\ze$ and $\xi$ are stably equivalent\index{stably equivalent bundles} if, 
for a sufficiently large $q$, 
the bundles $\ze\oplus \epz^{q-m}$ and $\xi\oplus \epz^{q-n}$ are equivalent, where 
$m=d(\ze)$ and $n=d(\xi)$. Let $\sE U(X)$\index{EU(X)@$\sE U(X)$} be the set of stable 
equivalence classes of vector 
bundles over $X$. If $X$ is connected, or if we restrict attention to vector bundles that are 
$n$-plane bundles for some $n$, then $\sE U$ is isomorphic to $\colim \sE U_n(X)$, where the 
colimit is taken over the maps $\sE U_n(X)\rtarr \sE U_{n+1}(X)$ obtained by sending a bundle 
$\xi$ to $\xi\oplus \epz$. Since a map from a compact space $X$ into $BU$ has image in one of the 
$BU(n)$, and similarly for homotopies, we see that in this case $[X_+,BU]\iso \colim [X_+,BU(n)]$ 
and therefore
$$\sE U(X)\iso [X_+,BU].$$

A deeper use of compactness gives the following basic fact.

\begin{prop} If $\xi:E\rtarr X$ is a vector bundle over $X$, then there is a
bundle $\et$ over $X$ such that $\xi\oplus \et$ is equivalent to $\epz^q$ for some $q$.
\end{prop}
\begin{proof}[Sketch proof]
The space $\GA E$ of sections of $E$ is a vector space under fiberwise addition
and scalar multiplication. Using a partition of unity argument, one can show that there
is a finite dimensional vector subspace $V$ of $\GA(E)$ such that the map $g: X\times V\rtarr E$ 
specified by $g(x,s)=s(x)$ is an epimorphism of bundles over $X$. The resulting short exact 
sequence of vector bundles, like any other short exact sequence of vector bundles, splits as a 
direct sum, and the conclusion follows.
\end{proof}

\begin{cor} Every virtual bundle over $X$ can be written in the form $[\xi] - q$ for some 
bundle $\xi$ and non-negative integer $q$.
\end{cor}
\begin{proof}
Given a virtual bundle $[\om] -[\ze]$, where $\om$ and $\ze$ are bundles, choose $\et$ such that 
$\ze\oplus \et \iso \epz^q$ and let $\xi = \om\oplus \et$. Then $[\om] -[\ze] = [\xi] - q$ in $K(X)$. 
\end{proof}

\begin{cor} 
There is a natural isomorphism $\sE U(X)\rtarr \tilde{K}(X)$.
\end{cor}
\begin{proof}
Writing $\sset{\xi}$ for the stable equivalence class of a bundle $\xi$, the required 
isomorphism is given by the correspondence $\sset{\xi} \leftrightarrow [\xi] - d(\xi)$.
\end{proof}

\begin{cor} Give $\bZ$ the discrete topology. For compact spaces $X$, there is a 
natural isomorphism
$$K(X)\iso [X_+,BU\times \bZ].$$
For nondegenerately based compact spaces $X$, there is a natural isomorphism
$$ \tilde{K}(X)\iso [X,BU\times \bZ].$$
\end{cor}
\begin{proof}
When $X$ is connected, the first isomorphism sends $[\xi]-q$ to $(f,n-q)$, where $\xi$ is 
an $n$-plane bundle with classifying map $f: X \rtarr BU(n)\subset BU$. The isomorphism
for non-connected spaces follows since both functors send disjoint unions to Cartesian
products. The second isomorphism follows from the first since $d: K(X)\rtarr \bZ$ can be 
identified with the map $[X_+,BU\times \bZ]\rtarr [S^0,BU\times \bZ]$ induced by the 
cofibration $S^0\rtarr X_+$, and the latter has kernel $[X,BU\times \bZ]$ since $X_+/S^0=X$. 
\end{proof}

For general, non-compact, spaces $X$, it is best to define $K$-theory to mean represented 
$K$-theory. Here we implicitly apply CW approximation, or else use the definition in the 
following form.

\begin{defn}\index{K-theory@$K$-theory!represented} For a space $X$ of the homotopy type 
of a CW complex, define
$$ K(X) = [X_+,BU\times \bZ].$$
For a nondegenerately based space of the homotopy type of a CW complex, define
$$ \tilde{K}(X) = [X,BU\times \bZ].$$
\end{defn}

When $X$ is compact, we know that $K(X)$ is a ring. It is natural to expect this to remain
true for general $X$. That this is the case is a direct consequence of the following result,
which the reader should regard as an aside.

\begin{prop}
The space $BU\times \bZ$ is a ring space\index{ring space} up to homotopy. That is, there are 
additive and 
multiplicative $H$-space structures on $BU\times \bZ$ such that the associativity,
commutativity, and distributivity diagrams required of a ring commute up to homotopy.
\end{prop}
\begin{proof}[Indications of proof]
By passage to colimits over $m$ and $n$, the maps $p_{m,n}: BU(m)\times BU(n) \rtarr BU(m+n)$ 
induce an ``addition'' $\oplus: BU\times BU\rtarr BU$. In fact, we
can define $BU$ in terms of planes in any copy of $\bC^{\infty}$, and the explicit maps
$p_{m,n}$ of Chapter 23 \S2 pass to colimits to give 
$$G_{\infty}(\bC^{\infty})\times G_{\infty}(\bC^{\infty})
\rtarr G_{\infty}(\bC^{\infty}\oplus\bC^{\infty});$$
use of an isomorphism $\bC^{\infty}\oplus\bC^{\infty}\iso \bC^{\infty}$ gives the required
map $\oplus$, which is well defined, associative, and commutative up to homotopy; the
zero-dimensional plane provides a convenient basepoint $0$ with which to check that we have a zero
element up to homotopy. Using ordinary addition on $\bZ$, we obtain the additive $H$-space 
structure on $BU\times \bZ$. Tensor products of universal bundles give rise to classifying maps 
$q_{m,n}: BU(m)\times BU(n)\rtarr BU(mn)$. These do not pass to colimits so readily, since one 
must take into account the bilinearity of the tensor product, for example the relation 
$(\ga_m\oplus\epz)\ten \ga_n \iso (\ga_m\ten\ga_n)\oplus \ga_n$, and we merely affirm that, by
fairly elaborate arguments, one can pass to colimits to obtain a product on $BU\times \bZ$.
It actually factors through the smash product with respect to the basepoint $0$, since that
acts as zero for the tensor product, and it restricts to an $H$-space structure on 
$BO\times \sset{1}$ with basepoint $(0,1)$.
\end{proof}

The study of ring spaces such as this is a relatively new, and quite deep, part of algebraic
topology. However, the reader should feel reasonably comfortable with the additive 
$H$-space structure on $BU$.

\section{The Bott periodicity theorem}

There are various ways to state, and various ways to prove, this basic result. We describe 
several versions and implications. One starting point is the following calculation.
We have a canonical line bundle $\ga_1^2$ over $S^2\iso \bC P^1$; its points are pairs 
$(L,x)$, where $L$ is a line in $\bC^2$ and $x$ is a point on that line. We let 
$H=\Hom(\ga_1^2,\epz)$ denote its dual.

\begin{thm} $K(S^2)$ is generated as a ring by $[H]$ subject
to the single relation $([H]-1)^2=0$. Therefore, as Abelian groups,
$K(S^2)$ is free on the basis $\sset{1,[H]}$ and $\tilde{K}(S^2)$
is free on the basis $\sset{1-[H]}$.
\end{thm}
\begin{proof}[Indication of proof]
We think of $S^2$ as the one-point compactification of $\bC$ decomposed as the 
union of the unit disk $D$ and the complement $D'$ of the interior of $D$, so that 
$D\cap D'=S^1$. Any $n$-plane bundle over $S^2$ restricts to a trivial bundle
over $D$ and $D'$, and these trivial bundles restrict to the same bundle over $S^1$. Conversely,
an isomorphism $f$ from the trivial bundle over $S^1$ to itself gives a way to glue together the 
trivial bundles over $D$ and $D'$ to reconstruct a bundle over $S^2$. Say that two such 
``clutching functions''\index{clutching function} $f$ are equivalent if the bundles 
they give rise to are equivalent. 
A careful analysis of the form of the possible clutching functions $f$ leads to a canonical
example in each equivalence class and thus to the required calculation.
\end{proof}

For any pair of spaces $X$ and $Y$, we have a K\"unneth-type ring homomorphism\index{Kunneth
map@K\"unneth map}
$$\al: K(X)\ten K(Y)\rtarr K(X\times Y)$$
specified by $\al(x\ten y) = \pi_1^*(x)\pi_2^*(y)$.

\begin{thm}[Bott periodicity]\index{Bott periodicity} For compact spaces $X$,
$$\al: K(X)\ten K(S^2) \rtarr K(X\times S^2)$$
is an isomorphism.
\end{thm}
\begin{proof}[Indication of proof]
The restrictions to $X\times D$ and $X\times D'$ of a bundle over $X\times S^2$ are 
equivalent to pullbacks of bundles over $X$, and their further restrictions to $S^1$
are equivalent. Conversely, bundles $\ze$ and $\xi$ over $X$ together with an 
equivalence $f$ between the restrictions to $X\times S^1$ of the pullbacks of
$\ze$ and $\xi$ to $X\times D$ and $X\times D'$ determine a bundle over $X\times S^2$.
Again, a careful analysis, which is similar to that in the special case when $X=pt$, 
of the equivalence classes of the possible clutching data $(\ze,f,\xi)$
leads to the conclusion.
\end{proof}

The following useful observation applies to any representable functor, not just $K$-theory.

\begin{lem} For nondegenerately based spaces $X$ and $Y$, the
projections of $X\times Y$ on $X$ and on $Y$ and the quotient map $X\times Y\rtarr X\sma Y$
induce a natural isomorphism
$$\tilde K(X\sma Y)\oplus \tilde K(X) \oplus \tilde K(Y) \iso \tilde K(X\times Y),$$
and $\tilde K(X\sma Y)$ is the kernel of the map 
$\tilde K(X\times Y)\rtarr \tilde K(X)\oplus \tilde K(Y)$
induced by the inclusions of $X$ and $Y$ in $X\times Y$.
\end{lem}
\begin{proof}
The inclusion $X\wed Y\rtarr X\times Y$ is a cofibration with quotient $X\sma Y$, and $X$ and
$Y$ are retracts of $X\times Y$ via the inclusions and projections.
\end{proof}

It follows easily that the K\"unneth map\index{Kunneth
map@K\"unneth map} $\al: K(X)\ten K(Y)\rtarr K(X\times Y)$ induces a
reduced K\"unneth map $\be: \tilde K(X)\ten \tilde K(Y)\rtarr \tilde K(X\sma Y)$. We have
a splitting
$$ \tilde K(X)\ten \tilde K(Y) \oplus \tilde K(X) \oplus \tilde K(Y) \oplus \bZ \iso K(X)\ten K(Y)$$
that is compatible with the splitting of the lemma. Therefore the following reduced form of the 
Bott periodicity theorem is equivalent to the unreduced form that we have already stated.

\begin{thm}[Bott periodicity]\index{Bott periodicity} For nondegenerately based compact spaces $X$,
$$\be: \tilde K(X)\ten \tilde K(S^2) \rtarr \tilde K(X\sma S^2) = \tilde K(\SI^2 X)$$
is an isomorphism.
\end{thm}

Write $b=1-[H]\in\tilde K(S^2)$. Since $\tilde K(S^2)\iso \bZ$ with generator $b$, the theorem 
implies that multiplication by the ``Bott element'' $b$ specifies an isomorphism 
$$[X,BU\times \bZ]\iso \tilde K(X) \rtarr \tilde K(\SI^2 X)\iso [X,\OM^2(BU\times \bZ)]$$
for nondegenerately based compact spaces $X$. Here the addition in the source and target is derived
from the natural additive $H$-space structure on $BU\times \bZ$ on the left and the displayed
double loop space on the right. If we had this isomorphism for general non-compact
spaces $X$, we could apply it with $X=BU\times \bZ$ and see that it is induced by a homotopy
equivalence of $H$-spaces
$$\be: BU\times \bZ \rtarr \OM^2(BU\times\bZ).$$
In fact, one can deduce such a homotopy equivalence from the Bott periodicity theorem as just 
stated, but there are more direct proofs. On the right, the double loop space obviously depends 
only on the basepoint component $BU=BU\times\sset{0}$. Since $\pi_2(BU)=\bZ$, a little argument
with $H$-spaces shows that $\OM^2(BU\times \bZ)$ is equivalent as an $H$-space to 
$(\OM^2_0 BU)\times \bZ$, where $\OM^2_0 BU$ denotes the component of the basepoint in  
$\OM^2 BU$. Using the identity function on the factor $\bZ$, we see that what is
needed is an equivalence of $H$-spaces $\be: BU\rtarr \OM^2_0 BU$. In fact, it is easily deduced 
from the form of Bott periodicity that, up to homotopy, $\be$  must be the adjoint of the composite
$$\xymatrix{
\SI^2 BU = BU\sma S^2 \ar[r]^-{\id\sma b} & BU\sma BU \ar[r]^-{\ten} & BU.}$$

The infinite unitary group $U$ is defined to be the union of the unitary groups $U(n)$, where 
$U(n)$ is embedded in $U(n+1)$ as matrices with last row and column zero except for $1$ on the 
diagonal. Then $\OM BU$ is homotopy equivalent as an $H$-space to $U$. Since $\pi_1(U)=\bZ$ and 
the universal cover of $U$ is the infinite special unitary group $SU$, $\OM U$ is equivalent as 
an $H$-space to $(\OM SU)\times \bZ$. Therefore $\be$ may be viewed as a map
$BU\rtarr \OM SU$. Bott's original proof of the Bott periodicity theorem used the Grassmannian
model for $BU$ to write down an explicit map $\be$ in the required homotopy class and then used
Morse theory to prove that $\be$ is a homotopy equivalence. 

Bott's map $\be$ can also be proved to be a homotopy equivalence using only
basic algebraic topology. Since $BU$ and $\OM SU$ are simply connected spaces of the homotopy types 
of CW complexes, a relative version of the Hurewicz theorem called the Whitehead 
theorem\index{Whitehead theorem} shows that 
$\be$ will be a weak equivalence and therefore a homotopy equivalence if it induces an isomorphism 
on integral homology. Since $H^*(BU(n))=\bZ[c_1,\ldots\!,c_n]$, $H^*(BU)\iso \bZ[c_i|i\geq 1]$. The 
$H$-space structure on $BU$ is induced by the maps $p_{m,n}$, and we find that the map
$\ps: H^*(BU)\rtarr H^*(BU\times BU)\iso H^*(BU)\ten H^*(BU)$
induced by the product is given by $\ps(c_k)=\sum_{i+j=k}c_i\ten c_j$. A purely algebraic
dualization argument proves that, as a ring,
$$H_*(BU)\iso \bZ[\ga_i|i\geq 1],$$
where $\ga_i$ is the image of a generator of $H_{2i}(\bC P^{\infty})$ under the map
induced by the inclusion of $\bC P^{\infty}=BU(1)$ in $BU$. One can calculate $H_*(\OM SU)$
and see that it too is a polynomial algebra with an explicitly given generator in each 
even degree. A direct inspection of the map $\be$ shows that it carries generators to generators.

In any case, it should now be clear that we have a periodic $\OM$-prespectrum and therefore
a generalized cohomology theory represented by it.

\begin{defn} The $K$-theory $\OM$-prespectrum $KU$\index{KU@$KU$} has spaces 
$KU_{2i} = BU\times \bZ$ and
$KU_{2i+1}=U$ for all $i\geq 0$. The structure maps are given by the canonical homotopy
equivalence $U\htp \OM BU = \OM(BU\times \bZ)$ and the Bott equivalence $BU\times \bZ\htp \OM U$.
\end{defn}

We have a resulting reduced cohomology theory\index{K-theory@$K$-theory!periodic} on based 
spaces such that 
$\tilde K^{2i}(X) = \tilde K(X)$ and $\tilde K^{2i+1}(X) = \tilde K(\SI X)$
for all integers $i$. This theory has products that are induced by tensor products
of bundles over compact spaces and that are induced by suitable maps 
$\ph: KU_i\sma KU_j\rtarr KU_{i+j}$ in general, just as for the cup product in
ordinary cohomology. It is standard to view this simply as a $\bZ_2$-graded 
theory with groups $\tilde K^0(X)$ and $\tilde K^1(X)$. 

\section{The splitting principle and the Thom isomorphism}

Returning to our bundle theoretic construction of $K$-theory, with $X$ compact, we describe 
briefly some important generalizations of the Bott periodicity theorem. The reader should recall the 
Thom isomorphism theorem in ordinary cohomology from Chapter 23 \S5. We let $\xi: E\rtarr X$ be an 
$n$-plane bundle over $X$, fixed throughout this section. (We shall use the letters $E$ and $\xi$
more or less interchangeably.) Results for general vector bundles over non-connected spaces $X$
can be deduced by applying the results to follow to one component of $X$ at a time.

\begin{defn} Let $E_0$ be the zero section of $E$. Define
the projective bundle\index{projective bundle} $\pi: P(E)\rtarr X$ by letting the non-zero 
complex numbers 
act on $E-E_0$ by scalar multiplication on fibers and taking the orbit space under this action. 
Equivalently, the fiber $\pi^{-1}(x)\subset P(E)$ is the complex projective space of lines
through the origin in the fiber $\xi^{-1}(x)\subset E$.  
Define the canonical line bundle $L(E)$ over $P(E)$ to be the subbundle of the pullback 
$\pi^*E$ of $\xi$ along $\pi$ whose points are the pairs consisting of a line in a fiber of $E$ 
and a point on that line.  Let $Q(E)$ be the quotient bundle $\pi^*E/L(E)$ and let $H(E)$ denote 
the dual of $L(E)$.
\end{defn}

Observe that $P(\epz^2)=X\times \bC P^1$ is the trivial bundle over $X$ with fiber 
$\bC P^1\iso S^2$. The first version of Bott periodicity generalizes, with essentially 
the same proof by analysis of clutching data, to the following version. Regard $K(P(E))$
as a $K(X)$-algebra via $\pi^*: K(X)\rtarr K(P(E))$.

\begin{thm}[Bott periodicity]\index{Bott periodicity} Let $L$ be a line bundle over $X$ 
and let $H=H(L\oplus \epz)$. Then the $K(X)$-algebra $K(P(L\oplus\epz))$ is generated by the
single element $[H]$ subject to the single relation $([H]-1)([L][H]-1)=0$. 
\end{thm}

There is a further generalization to arbitrary bundles $E$. To place it in context, we shall
first explain a cohomological analogue that expresses a different approach to the Chern classes 
than the one that we sketched before. It will be based on a generalization to projective bundles 
of the calculation of $H^*(\bC P^n)$. The proofs of both results are intertwined with the proof
of the following ``splitting principle,'' which allows the deduction of explicit formulas about
general bundles from formulas about sums of line bundles.

\begin{thm}[Splitting principle]\index{splitting principle}
There is a compact space $F(E)$ and a map $p: F(E)\rtarr X$ such that $p^*E$ is a 
sum of line bundles over $F(E)$ and both $p^*: H^*(X;\bZ)\rtarr H^*(F(E);\bZ)$ and 
$p^*: K(X)\rtarr K(F(E))$ are monomorphisms. 
\end{thm}

This is an easy inductive consequence of the following result, which we shall refer 
to as the ``splitting lemma.''

\begin{lem}[Splitting lemma]\index{splitting lemma}
Both $\pi^*: H^*(X;\bZ)\rtarr H^*(P(E);\bZ)$ and
$\pi^*: K(X)\rtarr K(P(E))$ are monomorphisms. 
\end{lem}

\begin{proof}[Proof of the splitting principle]
The pullback $\pi^*E$ splits as the sum $L(E)\oplus Q(E)$. (The splitting is canonically 
determined by a choice of a Hermitian metric on $E$.) Applying this construction 
to the bundle $Q(E)$ over $P(E)$, we obtain a map
$\pi: P(Q(E))\rtarr P(E)$ with similar properties. We obtain the desired map
$p: F(E)\rtarr X$ by so reapplying the projective bundle construction $n$ times.
Explicitly, using a Hermitian metric on $E$, we find that the fiber $F(E)_x$ is the 
space of splittings of the fiber $E_x$ as a sum of $n$ lines,
and the points of the bundle $p^*E$ are $n$-tuples of vectors in given lines.
The splitting lemma implies the desired monomorphisms on cohomology and $K$-theory.
\end{proof}

\begin{thm} 
Let $x=c_1(L(E))\in H^2(P(E);\bZ)$. Then $H^*(P(E);\bZ)$ is the free $H^*(X;\bZ)$-module 
on the basis $\sset{1,x,\ldots\!,x^{n-1}}$, and the Chern classes\index{Chern classes} of 
$\xi$ are characterized 
by $c_0(\xi)=1$ and the formula 
$$\sum_{k=0}^n(-1)^kc_k(E)x^{n-k}=0.$$
\end{thm}
\begin{proof}[Sketch proof]
This is another case where the Serre spectral sequence shows that the bundle behaves
cohomologically as if it were trivial and the K\"unneth theorem applied. This gives
the structure of $H^*(P(E))$ as an $H^*(X)$-module. In particular, it implies the 
splitting lemma and thus the splitting principle in ordinary cohomology. It also implies 
that there must be some description of $x^n$ as a linear combination of the $x^k$ for $k<n$,
and the splitting principle may now be used to help determine that description. Write
$$x^n = \sum_{k=1}^n(-1)^{k+1} c'_k(E)x^{n-k}.$$
This defines characteristic classes $c'_k(E)$. One deduces that $c'_k(E)=c_k(E)$
by verifying that the $c'_k$ satisfy the axioms that characterize the Chern classes.
For a line bundle $E$, $L(E)=E$ and $c_1(E)=c'_1(E)$ by the definition of $x$. One 
first verifies by direct calculation that if $E=L_1\oplus \cdots \oplus L_n$ is a sum of 
line bundles, then $\prod_{1\leq k\leq n} (x-c_1(L_k))=0$. This implies that $c'_k(E)$ is 
the $k$th elementary symmetric polynomial in the $c_1(L_k)$. By the Whitney sum formula 
for the Chern classes, this implies that $c'_k(E)=c_k(E)$ in this case. The general case 
follows from the splitting principle. Indeed, we have a map $P(p^*E) \rtarr P(E)$ of 
projective bundles whose induced map on base spaces is $p: F(E)\rtarr X$. Writing 
$p^*E\iso L_1\oplus\cdots\oplus L_n$ and using the naturality of the classes $c'_k$, 
we have 
$$p^*(c'_k(E))=c'_k(L_1\oplus\cdots\oplus L_n) = \si_k(c_1(L_1),\ldots\!,c_k(L_n)).$$ 
It follows easily that the $c'_k$ satisfy the Whitney sum axiom for the Chern 
classes. Since the remaining axioms are clear, this implies that $c'_k=c_k$.
\end{proof}

The following analogue in $K$-theory of the previous theorem holds. Observe that,
since they are continuous operations on complex vector spaces, the exterior powers 
$\la^k$\index{exterior powers} can be applied fiberwise to give natural operations 
on vector bundles.

\begin{thm}
Let $H=H(E)$. Then $K(P(E))$ is the free $K(X)$-module on the basis 
$\sset{1,[H],\ldots\!,[H]^{n-1}}$, and the following formula holds:
$$\sum_{k=0}^n(-1)^k[H]^k[\la^kE]=0.$$
\end{thm}
\begin{proof}[Sketch proof] Suppose first that $E$ is the sum of $n$ line bundles. Using the fact 
that if $E$ is an $n$-plane bundle and $L$ is a line bundle, then $P(E)$ is canonically isomorphic 
to $P(E\ten L)$, one can reduce to the case when the last line bundle is trivial. One can then
argue by induction from the previous form of the Bott periodicity theorem. For a general bundle
$E$, one then deduces the structure of $K(P(E))$ as a $K(X)$-module by a patching argument from 
coordinate charts and the case of trivial bundles. This implies the splitting lemma and thus the
splitting principle in $K$-theory. It also implies that there must be some formula describing 
$[H]^n$ as a polynomial in the $[H]^k$ for $k<n$. One reason that the given formula holds will 
be indicated shortly.
\end{proof}

Projective bundles are closely related to Thom spaces.\index{Thom space} The inclusion of 
vector bundles
$\xi\subset \xi\oplus \epz$ induces an inclusion of projective bundles 
$P(E)\subset P(E\oplus\epz)$. We give $E$ a Hermitian metric and regard the Thom
space $T\xi$ as the quotient $D(E)/S(E)$ of the unit disk bundle by the unit 
sphere bundle. The total space of $\epz$ is $X\times \bC$ and we write $1_x=(x,1)$. 
Define a map $\et: D(E)\rtarr P(E\oplus \epz)$ 
by sending a point $e_x$ in the fiber over $x$ to the line generated by $e_x-(1-|e_x|^2)1_x$.
Then $\et$ maps $D(E)-S(E)$ homeomorphically onto $P(E\oplus\epz)-P(E)$ and maps
$S(E)$ onto $P(E)$ by the evident Hopf map. Therefore $\et$ induces a homeomorphism
$$T(\xi)\iso D(E)/S(E)\iso P(E\oplus \epz)/P(E).$$

Just as in ordinary cohomology, the Thom diagonal gives rise to a product
$$K(X)\ten \tilde K(T\xi)\rtarr \tilde K(T\xi).$$
The description of $K(P(E))$ and the exact sequence in $K$-theory induced by the cofibering
$$P(E)\rtarr P(E\oplus \epz)\rtarr T(\xi)$$
lead to the Thom isomorphism in $K$-theory. There is a natural way to associate elements 
of $K(X)$ to complexes of vector bundles over $X$, and the exterior algebra of the bundle
$E$ gives rise to an element $\la_E\in \tilde K(T\xi)$. This element restricts to a
generator of $\tilde K(S_x^n)$ for each $x\in X$, and these Thom classes\index{Thom class} are 
compatible with Whitney sum, in the sense that $\la_{E\oplus E'} = \la_E \cdot \la_{E'}$. Moreover,
the image of $\la_E$ in $K(P(E\oplus\epz))$ is $\sum_{k=0}^n(-1)^k[H]^k[\la^kE]$. Therefore this
element maps to zero in $K(P(E))$, and this gives the formula in the previous theorem.

\begin{thm}[Thom isomorphism theorem]\index{Thom isomorphism}
Define $\PH: K(X)\rtarr \tilde K(T(\xi))$ by $$\PH(x) = x\cdot\la_E.$$ Then $\PH$ is an
isomorphism. 
\end{thm} 

\section{The Chern character; almost complex structures on spheres}

We have seen above that ordinary cohomology and $K$-theory enjoy similar properties.
The splitting theorem implies a direct connection between them. Let $R$ be any
commutative ring and consider a formal power series $f(t) = \sum a_it^i\in R[[t]]$.
Given an element $x\in H^n(X;R)$, we let $f(x) = \sum a_i x^i\in H^{**}(X;R)$.
The sums will be finite in our applications of this formula. Via the splitting
principle, we can use $f$ to construct a natural homomorphism of Abelian monoids
$\hat{f}: Vect(X)\rtarr H^{**}(X;R)$, where $X$ is any compact space. For a line 
bundle over $X$, we set 
$$\hat{f}(L) = f(c_1(L)).$$ 
For a sum $E=L_1\oplus\cdots\oplus L_n$ of line bundles over $X$, we set 
$$\hat{f}(E) = \sum_{i=1}^n f(c_1(L_i)).$$
For a general $n$-plane bundle $E$ over $X$, we let $\hat{f}(E)$ be the 
unique element of $H^{**}(X;R)$ such that $p^*(\hat{f}(E))=\hat{f}(p^*(E))\in H^{**}(F(E))$. 
More explicitly, writing $p^*E=L_1\oplus\cdots\oplus L_n$, we see that $\hat{f}(p^*(E))$
is a symmetric polynomial in the $c_1(L_i)$ and can therefore be written  as a 
polynomial in the elementary symmetric polynomials $p^*(c_k(E))$. Application of
this polynomial to the $c_k(E)$ gives $\hat{f}(E)$. (For vector bundles $E$ over 
non-connected spaces $X$, we add the elements obtained by restricting $E$ to the
components of $X$.) By the universal property of $K(X)$, $\hat{f}$ extends to a 
homomorphism $\hat{f}: K(X) \rtarr H^{**}(X;R)$.

There is an analogous multiplicative extension $\bar{f}$ of $f$ that starts from the 
definition
$$\bar{f}(E) = \prod_{i=1}^n f(c_1(L_i))$$
on a sum $E=L_1\oplus\cdots\oplus L_n$ of line bundles $L_i$.

\begin{exmp} For any $R$, if $f(t)=1+t$, then $\bar{f}(E) = c(E)$ is the total 
Chern class\index{total Chern class} of $E$. 
\end{exmp}

The example we are interested in is the ``Chern character,'' which gives rise to an
isomorphism between rationalized $K$-theory and rational cohomology.

\begin{exmp} Taking $R=\bQ$, define the Chern character\index{Chern character} 
$ch(E)\in H^{**}(X;\bQ)$ by $ch(E)=\hat{f}(E)$, where $f(t) = e^t =\sum t^i/i!$.
\end{exmp}

For line bundles $L$ and $L'$, we have $c_1(L\ten L')=c_1(L)+c_1(L')$. One way
to see this is to recall that $BU(1)\htp K(\bZ,2)$ and that line bundles are classified 
by their Chern classes regarded as elements of 
$$[X_+,BU(1)]\iso H^2(X;\bZ).$$
The tensor product is represented by a product $\ph: BU(1)\times BU(1)\rtarr BU(1)$ that 
gives $BU(1)$ an $H$-space structure.  We may think of $\ph$ as an element of 
$$H^2(BU\times BU;\bZ)\iso H^2(BU;\bZ)\oplus H^2(BU;\bZ).$$
and this element is the sum of the Chern classes in the two copies of $H^2(BU;\bZ)$
(since a basepoint of $BU$ is a homotopy identity element for $\ph$). This has the 
following implication.

\begin{lem} The Chern character specifies a ring homomorphism 
$$ch: K(X)\rtarr H^{**}(X;Q).$$
\end{lem}
\begin{proof}
We must check that $ch(E\ten E)=ch(E)\cdot ch(E')$ for bundles $E$ and $E'$ over $X$.
It suffices to check this when $E$ and $E'$ are sums of line bundles, in which case
the result follows directly from the bilinearity of the tensor product and the
relation $e^{t+t'}=e^t e^{t'}$.
\end{proof}

This leads to the following calculation.

\begin{lem} For $n\geq 1$, the Chern character maps $\tilde K(S^{2n})$ isomorphically onto the image
of $H^{2n}(S^{2n};\bZ)$ in $H^{2n}(S^{2n};\bQ)$. Therefore 
$c_n:\tilde K(S^{2n}) \rtarr H^{2n}(S^{2n};\bZ)$ is a monomorphism with cokernel $\bZ_{(n-1)!}$.
\end{lem}
\begin{proof}
The first statement is clear for $n=1$, when $ch=c_1$, and follows by compatibility with 
external products for $n>1$. The definition of $ch$ implies that the component $ch_n$ of
$ch$ in degree $2n$ is $c_n/(n-1)!$ plus terms decomposable in terms of the $c_i$ for 
$i<n$, and the second statement follows.
\end{proof}

Together with some of the facts given in Chapter 23 \S7, this has a remarkable application 
to the study of almost complex structures on spheres. Recall that a smooth manifold of even
dimension admits an almost complex structure if its tangent bundle is the underlying real
vector bundle of a complex bundle. 

\begin{thm} $S^2$ and $S^6$ are the only spheres that admit an almost 
complex structure.\index{almost complex structure}
\end{thm}
\begin{proof} It is classical that $S^2$ and $S^6$ admit almost complex structures and that
$S^4$ does not. Assume that $S^{2n}$ admits an almost complex structure. We shall show
that $n\leq 3$. We are given that the tangent bundle $\ta$ is the realification of a 
complex bundle. Its $n$th Chern class is its Euler class: $c_n(\ta) =\ch(\ta)$. Since the 
Euler characteristic of $S^{2n}$ is $2$, $\ch(\ta) = 2 \io_{2n}$, where 
$\io_{2n}\in H^{2n}(S^{2n},\bZ)$ is the canonical generator. However, $c_n(\ta)$ must be 
divisible by $(n-1)!$. This can only happen if $n\leq 3$.
\end{proof}

Obviously the image of $ch$ lies in the sum of the even degree elements in $H^{**}(X;\bQ)$,
which we denote by $H^{even}(X;\bQ)$. We define $H^{odd}(X;\bQ)$ similarly, and we extend 
$ch$ to $\bZ_2$-graded reduced cohomology by defining $ch$ on $\tilde K^1(X)$ to be the composite 
$$\tilde K^1(X) \iso \tilde K(\SI X)\overto{ch} \tilde H^{even}(\SI X;\bQ) 
\iso \tilde H^{odd}(X;\bQ).$$
We then have the following basic result, which actually holds for general compact spaces
$X$ provided that we replace singular cohomology by \v Cech cohomology.

\begin{thm} For any finite based CW complex $X$, $ch$ induces an isomorphism 
$$\tilde K^*(X)\ten \bQ \rtarr \tilde H^{**}(X;\bQ).$$
\end{thm}
\begin{proof}[Sketch proof]
We think of both the source and target as $\bZ_2$-graded. The lemma above implies
the conclusion when $X=S^n$ for any $n$. One can check that the displayed maps for 
varying $X$ give a map of $\bZ_2$-graded cohomology theories. The conclusion then 
follows from the five lemma and induction on the number of cells of $X$. 
\end{proof}

\section{The Adams operations}

There are natural operations in $K$-theory, called the Adams operations, that are somewhat 
analogous to the Steenrod operations in mod $2$ cohomology. In fact, the analogy can be 
given content by establishing a precise relationship between the Adams and Steenrod
operations, but we will not go into that here.

\begin{thm}\index{Adams operations} For each non-zero integer $k$, there is a natural 
homomorphism of rings 
$\ps^k: K(X)\rtarr K(X)$. These operations satisfy the following properties. 
\begin{enumerate}
\item $\ps^1=\id$ and $\ps^{-1}$ is induced by complex conjugation of bundles.
\item $\ps^k\ps^{\ell} = \psi^{k\ell}= \ps^{\ell}\ps^{k}$.
\item $\ps^p(x)\equiv x^p$ mod $p$ for any prime $p$.
\item $\ps^k(\xi)=\xi^k$ if $\xi$ is a line bundle.
\item $\ps^k(x)= k^nx$ if $x\in \tilde{K}(S^{2n})$.
\end{enumerate}
\end{thm}

We explain the construction. By property 2, $\ps^{-k}= \ps^{k}\ps^{-1}$, hence by 
property 1 we can concentrate on the case $k > 1$. The exterior powers\index{exterior powers} of 
bundles satisfy the relation
$$\la^k(\xi\oplus \et)= \oplus_{i+j=k}\la^i(\xi)\ten \la^j(\et).$$
It follows formally that the $\la^k$ extend to operations $K(X)\rtarr K(X)$. Indeed,
form the group $G$ of power series with constant coefficient $1$ in the ring $K(X)[[t]]$
of formal power series in the variable $t$. We define a function from (equivalence 
classes of) vector bundles to this Abelian group by setting
$$\LA(\xi)=1 + \la^1(\xi)t +\cdots + \la^k(\xi)t^k +\cdots.$$
Visibly, this is a morphism of monoids,
$$\LA(\xi\oplus\et) = \LA(\xi)\LA(\et).$$
It therefore extend to a homomorphism of groups $\LA: K(X)\rtarr G$, and we let $\la^k(x)$
be the coefficient of $t^k$ in $\LA(x)$. 

We define the $\ps^k$ as suitable polynomials in the $\la^k$. Recall that the subring of
symmetric polynomials in the polynomial algebra $\bZ[x_1,\ldots\!,x_n]$ is the polynomial 
algebra $\bZ[\si_1,\ldots\!,\si_n]$, where $\si_i=x_1x_2\cdots x_i+\cdots$ is the
$i$th elementary symmetric function. We may write the power sum $\pi_k=x_1^k+\cdots+x_n^k$ 
as a polynomial
$$\pi_k = Q_k(\si_1,\ldots\!,\si_k)$$
in the first $k$ elementary symmetric functions. Provided $n\geq k$, $Q_k$ does not
depend on $n$. We define
$$\ps^k(x) = Q_k(\la^1(x),\ldots\!, \la^k(x)).$$
For example, $\pi_2=\si_1^2-2\si_2$, hence $\ps^2(x)=x^2-2\la^2(x)$. The naturality of
the $\ps^k$ is clear from the naturality of the $\la^k$. 

If $\xi$ is a line bundle, then $\la^1(\xi)=\xi$ and $\la^k(\xi)=0$ for $k\geq 2$. 
Clearly $\si^k_1 = \pi_k + \text{other terms}$ and $\pi_k$ does not occur as a 
summand of any other monomial in the $\si_i$. Therefore $Q_k \equiv \si_1^k$ modulo 
terms in the ideal generated by the $\si_i$ for $i>1$. This immediately implies 
property 4. Moreover, if $\xi_1,\ldots\!,\xi_n$ are line bundles, then 
\begin{eqnarray*}
\LA (\xi_1\oplus \cdots \oplus \xi_n) & = & (1+\xi_1t)\cdots(1+\xi_nt) \\
                        & = & 1+\si_1(\xi_1,\ldots\!,\xi_n)t+\si_2(\xi_1,\ldots\!,\xi_n)t^2 + \cdots.
\end{eqnarray*}
This implies the generalization of property 4 to sums of line bundles:
\begin{enumerate}
\item[$4'$] $\ps^k(\xi_1\oplus\cdots\oplus \xi_n)= \pi_k(\xi_1,\ldots\!, \xi_n)$ for 
line bundles $\xi_i$.
\end{enumerate}
Now, if $x$ and $y$ are sums of line bundles, the following formulas are immediate:
$$ \ps^k(x+y) = \ps^k(x)+\ps^k(y), \ \ \ps^k(xy) = \ps^k(x)\ps^k(y),\ \
\ps^k\ps^{\ell}(x)=\ps^{k\ell}(x) $$
$$\text{and}\ \  \ps^p(x)\equiv x^p\ \text{mod}\ p \ \ \text{for a prime}\ p. $$
For arbitrary bundles, these formulas follow directly from the splitting principle and 
naturality, and they then follow formally for arbitrary virtual bundles. This completes
the proof of all properties except 5. We have that $\tilde{K}(S^2)$ is generated by
$1-[H]$, where $(1-[H])^2=0$. 
Clearly $\ps^k(1-[H]) = 1-[H]^k$. By induction on $k$, $1-[H]^k=k(1-[H])$. 
Since $S^{2n}=S^2\sma\cdots\sma S^2$ and $\tilde{K}(S^{2n})$ is generated by the $k$-fold
external tensor power $(1-[H])\ten\cdots\ten(1-[H])$, property 5 follows
from the fact that $\ps^k$ preserves products.

\begin{rem} By the splitting principle, it is clear that the $\ps^k$ are the unique
natural and additive operations with the specified behavior on line bundles.
\end{rem}

Two further properties of the $\ps^k$ should be mentioned. The first is a direct 
consequence of the multiplicativity of the $\ps^k$ and their behavior on spheres.

\begin{prop}
The following diagram does not commute for based spaces $X$, where $\be$ is the periodicity 
isomorphism:
$$\diagram
\tilde K(X)\dto_{\ps^k} \rto^(0.43){\be} & \tilde K(\SI^2 X) \dto^{\ps^k}\\
\tilde K(X) \rto_(0.43){\be}  & \tilde K(\SI^2 X).\\
\enddiagram$$
Rather, $\ps^k\be=k\be \ps^k$. 
\end{prop}

Therefore the $\ps^k$ do not give stable operations on the $\bZ$-graded theory $K^*$.

\begin{prop}
Define $\ps^k_H$ on $H^{even}(X;\bZ)$ by letting $\ps^k_H(x) = k^rx$ for $x\in H^{2r}(X;\bZ)$.
Then the following diagram  commutes:
$$\diagram
K(X)\dto_{\ps^k} \rto^(0.35){ch} & H^{even}(X;\bQ)\dto^{\ps^k_H}\\
K(X) \rto_(0.35){ch} & H^{even}(X;\bQ).\\
\enddiagram$$
\end{prop}
\begin{proof}
It suffices to prove this on vector bundles $E$. By the splitting principle in $K$-theory 
and cohomology, we may assume that $E$ is a sum of line bundles. By additivity, we may
then assume that $E$ is a line bundle. Here $\ps^k(E)= E^k$ and $c_1(E^k) = kc_1(E)$.
The conclusion follows readily from the definition of $ch$ in terms of $e^t$.
\end{proof}

\begin{rem} The observant reader will have noticed that, by analogy with the definition
of the Stiefel-Whitney classes, we can define characteristic 
classes\index{characteristic classes!in $K$-theory} in $K$-theory
by use of the Adams operations and the Thom isomorphism, setting 
$\rh^k(E) = \PH^{-1}\ps^k\PH(1)$ for $n$-plane bundles $E$.
\end{rem}

\section{The Hopf invariant one problem and its applications}

We give one of the most beautiful and impressive illustrations of the philosophy
described in the first chapter. We define a numerical invariant, called the ``Hopf invariant,''
of maps $f: S^{2n-1}\rtarr S^n$ and show that it can only rarely take the value one. We then 
indicate several problems whose solution can be reduced to the question of when such maps 
$f$ take the value one. Adams' original solution to the Hopf invariant one problem used secondary 
cohomology operations in ordinary cohomology and was a critical starting point of modern algebraic 
topology. The later realization that a problem that required secondary operations in ordinary
cohomology could be solved much more simply using primary operations in $K$-theory had a
profound impact on the further development of the subject.

Take cohomology with integer coefficients unless otherwise specified.

\begin{defn} Let $X$ be the cofiber of a based map $f: S^{2n-1}\rtarr S^n$, where $n\geq 2$. 
Then $X$ is a CW complex with a single vertex, a single $n$-cell $i$, and a single $2n$-cell $j$.
The differential in the cellular chain complex of $X$ is zero for obvious dimensional 
reasons, hence $\tilde H^*(X)$ is free Abelian on generators $x=[i]$ and $y=[j]$.
Define an integer $h(f)$, the Hopf invariant\index{Hopf invariant} of $f$, by $x^2 = h(f) y$. 
We usually regard $h(f)$ as defined only up to sign (thus ignoring problems of orientations of 
cells). Note  that $h(f)$ depends only on the homotopy class of $f$.
\end{defn}

If $n$ is odd, then $2x^2 = 0$ and thus $x^2=0$. We assume from now on that $n$ is even.
Although not essential to the main point of this section, we record the following basic
properties of the Hopf invariant.

\begin{prop} The Hopf invariant enjoys the following properties.
\begin{enumerate}
\item If $g: S^{2n-1}\rtarr S^{2n-1}$ has degree $d$, then $h(f\com g) = dh(f)$.
\item If $e: S^n\rtarr S^n$ has degree $d$, then $h(e\com f)=d^2h(f)$.
\item The Hopf invariant defines a homomorphism $\pi_{2n-1}(S^n)\rtarr \bZ$.
\item There is a map $f: S^{2n-1}\rtarr S^n$ such that $h(f)=2$.
\end{enumerate}
\end{prop}
\begin{proof}
We leave the first three statements to the reader. For property 4, let 
$\pi: D^n\rtarr D^n/S^{n-1}\iso S^n$ be the quotient map and define
$$f: S^{2n-1}\iso (D^n\times S^{n-1})\cup (S^{n-1}\times D^n)\rtarr S^n$$
by $f(x,y)= \pi(x)$ and $f(y,x)=\pi(x)$ for $x\in D^n$ and $y\in S^{n-1}$.
We leave it to the reader to verify that $h(f)=2$.
\end{proof}

We have adopted the standard definition of $h(f)$, but we could just as well have 
defined it in terms of $K$-theory. To see this, consider the cofiber sequence
$$ S^{2n-1} \overto{f} S^n\overto{i} X \overto{\pi} S^{2n} \overto{\SI f} S^{n+1}.$$
Obviously $i^*: H^n(X)\rtarr H^n(S^n)$ and $\pi^*: H^{2n}(S^{2n})\rtarr H^{2n}(X)$
are isomorphisms. We have the commutative diagram with exact rows
$$\diagram
0 \rto & \tilde K(S^{2n}) \dto_{ch} \rto^{\pi^*} & \tilde K(X) \dto^{ch} \rto^{i^*}
& \tilde K(S^n) \dto^{ch}\rto & 0\\
0 \rto & \tilde H^{**}(S^{2n};\bQ)  \rto_{\pi^*} & \tilde H^{**}(X;\bQ) \rto_{i^*}
& \tilde H^{**}(S^n;\bQ) \rto & 0.\\
\enddiagram$$
Here the top row is exact since $\tilde K^1(S^n)=0$ and $\tilde K^1(S^{2n})=0$. The vertical
arrows are monomorphisms since they are rational isomorphisms. By a lemma in the previous
section, generators $i_n$ of $\tilde K(S^n)$ and $i_{2n}$ of $\tilde K(S^{2n})$ map under 
$ch$ to generators of $H^n(S^n)$ and $H^{2n}(S^{2n})$. Choose $a\in \tilde K(X)$ such that 
$i^*(a) = i_n$ and let $b=\pi^*(i_{2n})$. Then $\tilde K(X)$ is the free Abelian group on
the basis $\sset{a,b}$. Since $i_n^2=0$, we have $a^2=h'(f)b$ for some integer $h'(f)$.
The diagram implies that, up to sign, $ch(b)= y$ and $ch(a) = x+ qy$ for some rational 
number $q$. Since $ch$ is a ring homomorphism and since $y^2=0$ and $xy=0$, we conclude
that $h'(f)=h(f)$.

\begin{thm} If $h(f)=\pm 1$, then $n = 2$, $4$, or $8$.
\end{thm}
\begin{proof}
Write $n=2m$. Since $\ps^k(i_{2n})=k^{2m} i_{2n}$ and $\ps^k(i_n) = k^m i_n$, we have
$$\ps^k(b)=k^{2m} b \ \ \tand \ \ \ps^k(a) = k^m a + \mu_k b$$
for some integer $\mu_k$. Since $\ps^2(a)\equiv a^2\ \text{mod}\ 2$, $h(f)=\pm 1$ implies
that $\mu_2$ is odd. Now, for any odd $k$,
\begin{eqnarray*}
\ps^k\ps^2(a) & = & \ps^k(2^m a + \mu_2 b) \\
& = & k^m2^ma +(2^m\mu_k + k^{2m}\mu_2) b  
\end{eqnarray*}
while
\begin{eqnarray*}
\ps^2\ps^k(a) & = & \ps^2(k^m a + \mu_k b) \\
& = & 2^mk^ma + (k^m\mu_2 + 2^{2m} \mu_k) b.
\end{eqnarray*}
Since these must be equal, we find upon equating the coefficients of $b$ that
$$2^m(2^m-1)\mu_k = k^m(k^m-1)\mu_2.$$
If $\mu_2$ is odd, this implies that $2^m$ divides $k^m-1$. Already with $k=3$, an
elementary number theoretic argument shows that this implies $m=1$, $2$, or $4$.
\end{proof}

This allows us to determine which spheres can admit an $H$-space structure. Recall
from a problem in Chapter 18 that $S^{2m}$ cannot be an $H$-space. Clearly $S^n$ 
is an $H$-space for $n=0$, $1$, $3$, and $7$: view $S^n$ as the unit sphere in the 
space of real numbers, complex numbers, quaternions, or Cayley numbers.

\begin{thm}
If $S^{n-1}$ is an $H$-space,\index{Hspace@$H$-space} then $n=1$, $2$, $4$, or $8$.
\end{thm}

The strategy of proof is clear: given an $H$-space structure on $S^{n-1}$, we construct 
from it a map $f: S^{2n-1}\rtarr S^n$ of Hopf invariant one. The following construction
and lemma do this and more.

\begin{con}[Hopf construction]\index{Hopf construction} Let 
$\ph: S^{n-1}\times S^{n-1}\rtarr S^{n-1}$ be a map. 
Let $CX=(X\times I)/(X\times\sset{1})$ be the unreduced cone functor and note that we have
canonical homeomorphisms of pairs
$$(D^n,S^{n-1}) \iso(CS^{n-1},S^{n-1})$$
and
\begin{eqnarray*}
(D^{2n},S^{2n-1}) & \iso & (D^n\times D^n,(D^n\times S^{n-1})\cup (S^{n-1}\times D^n))\\
 & \iso & (CS^{n-1}\times CS^{n-1},(CS^{n-1}\times S^{n-1})\cup (S^{n-1}\times CS^{n-1})).
\end{eqnarray*}
Take $S^n$ to be the unreduced suspension of $S^{n-1}$, with the upper and lower hemispheres 
$D^n_+$ and $D^n_-$ corresponding to the points with suspension coordinate $1/2\leq t\leq 1$ 
and $0\leq t\leq 1/2$, respectively. Define 
$$f: S^{2n-1}\iso (CS^{n-1}\times S^{n-1})\cup (S^{n-1}\times CS^{n-1}) \rtarr S^n$$ 
as follows. Let $x,y\in S^{n-1}$ and $t\in I$.  On $CS^{n-1}\times S^{n-1}$, $f$ is the composite
$$CS^{n-1}\times S^{n-1} \overto{\al} C(S^{n-1}\times S^{n-1}) \overto{C \ph} C S^{n-1}
\overto{\be} D^n_-,$$
where $\al([x,t],y)=[(x,y),t]$ and $\be([x,t])=[x,(1-t)/2]$. On $S^{n-1}\times CS^{n-1}$, 
$f$ is the composite
$$S^{n-1}\times CS^{n-1} \overto{\al'} C(S^{n-1}\times S^{n-1}) \overto{C \ph} C S^{n-1}
\overto{\be'} D^n_+,$$
where $\al'(x,[y,t])=[(x,y),t]$ and $\be'([x,t])=[x,(1+t)/2]$. The map $f$, or 
rather the resulting  $2$-cell complex $X=S^n\cup_f D^{2n}$, is called the Hopf construction 
on $\ph$.
\end{con}

Giving $S^{n-1}$ a basepoint, we obtain inclusions of $S^{n-1}$ onto the first and second
copies of $S^{n-1}$ in $S^{n-1}\times S^{n-1}$. The bidegree\index{bidegree of a map} of a map 
$\ph: S^{n-1}\times S^{n-1}\rtarr S^{n-1}$ is the pair of integers given by the two 
resulting composite maps $S^{n-1}\rtarr S^{n-1}$. Thus $\ph$ gives $S^{n-1}$ an $H$-space
structure if its bidegree is $(1,1)$.

\begin{lem}
If the bidegree of $\ph: S^{n-1}\times S^{n-1}\rtarr S^{n-1}$ is $(d_1,d_2)$, then the Hopf
invariant of the Hopf construction on $\ph$ is $\pm d_1d_2$.
\end{lem}
\begin{proof}
Making free use of the homeomorphisms of pairs specified in the construction, we see that the
diagonal map of $X$, its top cell $j$, evident quotient maps, and projections $\pi_i$ onto
first and second coordinates give rise to a commutative diagram in which the maps marked
$\htp$ are homotopy equivalences and those marked $\iso$ are homeomorphisms:
$$\diagram
X \rto^{\DE} \dto & X\sma X \dto^{\htp} \\
X/S^n \rto^{\DE}  & X/D^n_+ \sma X/D^n_-  \\
S^{2n}\iso D^{2n}/S^{2n-1} \drto_{\iso} \rto^(0.3){\DE} \uto_j^{\iso}
& (D^n\times D^n)/(S^{n-1}\times D^n) \sma (D^n\times D^n)/(D^n\times S^{n-1})
 \uto_{j\sma j} \dto_{\htp}^{\pi_1\sma\pi_2} \\
 & D^n/S^{n-1} \sma D^n/S^{n-1}\iso S^n\sma S^n. \\
\enddiagram$$
The cup square of $x\in H^n(X)$ is the image under $\DE^*$ of the external product of $x$ with
itself. The maps on the left induce isomorphisms on $H^{2n}$. The inclusions of $D^n$
in the $i$th factor of $D^n\times D^n$ induce homotopy inverses
$$\io_1: D^n/S^{n-1}\rtarr (D^n\times D^n)/(S^{n-1}\times D^n)$$
and
$$\io_2: D^n/S^{n-1} \rtarr (D^n\times D^n)/(D^n\times S^{n-1})$$ 
to the projections $\pi_i$ in the diagram, and it suffices to prove that, up to sign, the 
composites
$$j\com \io_1: D^n/S^{n-1}\rtarr X/D^n_+ \tand j\com\io_2: D^n/S^{n-1}\rtarr X/D^n_-$$
induce multiplication by $d_1$ and by $d_2$ on $H^n$. However, by construction, these maps 
factor as composites
$$D^n/S^{n-1}\overto{\ga_1}S^n/D^n_+\rtarr X/D^n_+ \tand 
D^n/S^{n-1}\overto{\ga_2}S^n/D^n_-\rtarr X/D^n_-,$$ 
where, up to signs and identifications of spheres, $\ga_1$ and $\ga_2$ are the suspensions
of the restrictions of $\ph$ to the two copies of $S^{n-1}$ in $S^{n-1}\times S^{n-1}$.
\end{proof}

The determination of which spheres are $H$-spaces has the following implications.

\begin{thm}\index{products on $\bR^n$}
Let $\om: \bR^n\times \bR^n\rtarr \bR^n$ be a map with a two-sided identity element 
$e\neq 0$ and no zero divisors. Then $n=1$, $2$, $4$, or $8$.
\end{thm}
\begin{proof}
The product restricts to give $\bR^n-\sset{0}$ an $H$-space structure. Since $S^{n-1}$
is homotopy equivalent to $\bR^n-\sset{0}$, it inherits an $H$-space structure.
Explicitly, we may assume that $e\in S^{n-1}$, by rescaling the metric, and we give
$S^{n-1}$ the product $\ph: S^{n-1}\times S^{n-1}\rtarr S^{n-1}$ specified by 
$\ph(x,y)=\om(x,y)/|\om(x,y)|$.
\end{proof}

Note that $\om$ need not be bilinear, just continuous. Also, it need not have a strict
unit; all that is required is that $e$ be a two-sided unit up to homotopy for the 
restriction of $\om$ to $\bR^n-\sset{0}$.

\begin{thm}
If $S^{n}$ is parallelizable,\index{parallelizable spheres} then $n=0$, $1$, $3$, or $7$.
\end{thm}
\begin{proof}
Exclude the trivial case $n=0$ and suppose that $S^{n}$ is parallelizable, so that its 
tangent bundle $\ta$ is trivial. We will show that $S^{n}$ is an $H$-space. Define
a map $\mu: \ta \rtarr S^{n}$ as follows. Think of the tangent plane $\ta_x$ as 
affinely embedded in $\bR^{n+1}$ with origin at $x$. We have a parallel translate of 
this plane to an affine plane with origin at $-x$. Define $\mu$ by sending a tangent 
vector $y\in \ta_x$ to the intersection with $S^{n}$ of the line from $x$ to the 
translate of $y$. Composing with a trivialization $S^{n}\times\bR^n\iso \ta$, this
gives a map $\mu: S^{n}\times\bR^n\rtarr S^n$. Let $S^n_{\infty}$ be the one-point
compactification of $\bR^n$. Extend $\mu$ to a map $\ph: S^n\times S^n_{\infty}\rtarr S^n$
by letting $\ph(x,\infty)=x$; $\ph$ is continuous since $\mu(x,y)$ approaches $x$ as
$y$ approaches $\infty$. By construction, $\infty$ is a right unit for this product.
For a fixed $x$, $y\rtarr \ph(x,y)$ is a degree one homeomorphism 
$S^n_{\infty}\rtarr S^n_{\infty}$. The conclusion follows.
\end{proof}

\chapter{An introduction to cobordism}

Cobordism theories were introduced shortly after $K$-theory, and their use pervades
modern algebraic topology. We shall describe the cobordism of smooth closed manifolds,
but this is in fact a particularly elementary example. Other examples include smooth
closed manifolds with extra structure on their stable normal bundles: orientation, complex
structure, Spin structure, or symplectic structure for example. All of these except the 
symplectic case have been computed completely. The complex case is particularly important 
since complex cobordism and theories constructed from it have been of central importance in 
algebraic topology for the last few decades, quite apart from their geometric origins 
in the classification of manifolds. The area is pervaded by insights from algebraic 
topology that are quite mysterious geometrically. For example, the complex cobordism groups 
turn out to be concentrated in even degrees: every smooth closed manifold of odd dimension with
a complex structure on its stable normal bundle is the boundary of a compact manifold (with  
compatible bundle information). However, there is no geometric understanding of why this 
should be the case. The analogue with ``complex'' replaced by ``symplectic'' is false. 

\section{The cobordism groups of smooth closed manifolds}

We consider the problem of classifying smooth closed $n$-manifolds $M$. One's first
thought is to try to classify them up to diffeomorphism, but that problem is in principle
unsolvable. Thom's discovery that one can classify such manifolds up to the weaker 
equivalence relation of ``cobordism''\index{cobordism} is one of the most beautiful advances of 
twentieth
century mathematics. We say that two smooth closed $n$-manifolds $M$ and $N$ are 
cobordant\index{cobordant manifolds}
if there is a smooth compact manifold $W$ whose boundary is the disjoint union of $M$ and $N$, 
$\pa W = M\amalg N$. We write $\sN_n$\index{Naa@$\sN_n$} for the set of cobordism classes of 
smooth closed $n$-manifolds. It is convenient to allow the empty set $\emptyset$ as an $n$-manifold
for every $n$. Disjoint union gives an addition on the set $\sN_n$. This operation
is clearly associative and commutative and it has $\emptyset$ as a zero element. Since 
$$\pa(M\times I) =M\amalg M,$$
$M\amalg M$ is cobordant to $\emptyset$. Thus $M=-M$ and $\sN_n$ is a vector space over $\bZ_2$.
Cartesian product of manifolds defines a multiplication $\sN_m\times \sN_n\rtarr \sN_{m+n}$.
This operation is bilinear, associative, and commutative, and the zero dimensional manifold with a 
single point provides an identity element. We conclude that $\sN_*$ is a graded $\bZ_2$-algebra.

\begin{thm}[Thom]\index{Thom cobordism theorem} $\sN_*$ is a polynomial algebra over 
$\bZ_2$ on generators $u_i$ of dimension
$i$ for $i > 1$ and not of the form $2^r-1$. 
\end{thm}

As already stated in our discussion of Stiefel-Whitney numbers, it follows from the proof of 
the theorem that a manifold is a boundary if and only if its normal Stiefel-Whitney numbers
are zero. We can restate this as follows.\index{Stiefel-Whitney numbers}
\index{Stiefel-Whitney numbers!tangential}\index{Stiefel-Whitney numbers!normal}

\begin{thm} Two smooth closed $n$-manifolds are cobordant if and only if their normal
Stiefel-Whitney numbers, or equivalently their tangential Stiefel-Whitney numbers, are equal.
\end{thm}

Explicit generators $u_i$ are known. Write $[M]$ for the cobordism class of a manifold $M$. 
Then we can take $u_{2i}=[\bR P^{2i}]$. We have seen that the Stiefel-Whitney numbers of
$\bR P^{2i-1}$ are zero, so we need different generators in odd dimensions. For $m<n$, define
$H_{n,m}$\index{hypersurface $H_{n,m}$} to be the hypersurface in $\bR P^n\times \bR P^m$ consisting 
of those pairs
$([x_0,\ldots\!,x_n],[y_0,\ldots\!,y_m])$ such that $x_0y_0+\cdots + x_my_m = 0$; here 
$(x_0,\ldots\!, x_n)\in S^n$ and $[x_0,\ldots\!,x_n]$ denotes its image in $\bR P^n$. We may
write an odd number $i$ not of the form $2^r-1$ in the form $i=2^p(2q+1)-1=2^{p+1}q+2^p-1$, 
where $p\geq 1$ and $q\geq 1$. Then we can take $u_i=[H_{2^{p+1}q,2^p}]$.

The strategy for the proof of Thom's theorem is to describe $\sN_n$ as a homotopy group
of a certain Thom space. The homotopy group is a stable one, and it turns out to be 
computable by the methods of generalized homology theory. 

Consider the universal
$q$-plane bundle $\ga_q: E_q\rtarr G_q(\bR^{\infty}) = BO(q)$. Let $TO(q)$\index{TO(q)@$TO(q)$} 
be its Thom space\index{Thom space}. Recall that we have maps $i_q: BO(q)\rtarr BO(q+1)$ such that 
$i_q^*(\ga_{q+1})=\ga_q\oplus \epz$. The Thom space $T(\ga_q\oplus \epz)$ is canonically
homeomorphic to the suspension $\SI TO(q)$, and the bundle map $\ga_q\oplus \epz \rtarr \ga_{q+1}$
induces a map $\si_q: \SI TO(q)\rtarr TO(q+1)$. Thus the spaces $TO(q)$ and maps $\si_q$ 
constitute a prespectrum $TO$\index{Thom prespectrum}.\index{TO@$TO$} By definition, the homotopy 
groups of a \index{prespectrum!homotopy groups of} prespectrum $T=\sset{T_q}$ are 
$$\pi_n(T) = \colim\,\pi_{n+q}(T_q),$$
where the colimit is taken over the maps
$$ \pi_{n+q}(T_q) \overto{\SI} \pi_{n+q+1}(\SI T_q)\overto{{\si_q}_*} \pi_{n+q+1}(T_{q+1}).$$
In the case of $TO$, it turns out that these maps are isomorphisms if $q$ is sufficiently large, 
and we have 
the following translation of our problem in manifold theory to a problem in homotopy theory. We
shall sketch the proof in the next section, where we shall also explain the ring structure on
$\pi_*(TO)$ that makes it a $\bZ_2$-algebra.

\begin{thm}[Thom]\index{Thom cobordism theorem}
For sufficiently large $q$, $\sN_n$ is isomorphic to $\pi_{n+q}(TO(q))$. Therefore
$$\sN_n\iso \pi_n(TO).$$
Moreover, $\sN_*$ and $\pi_*(TO)$ are isomorphic as $\bZ_2$-algebras.
\end{thm}

\section{Sketch proof that $\sN_*$ is isomorphic to $\pi_*(TO)$}

Given a smooth closed $n$-manifold $M$, we may embed it in $\bR^{n+q}$ for $q$ sufficiently 
large, and we let $\nu$ be the normal bundle of the embedding. (By the Whitney embedding 
theorem, $q=n$ suffices, but the precise estimate is not important to us.)  Embed $M$ as the 
zero section of the total space $E(\nu)$. Then a standard result in differential topology 
known as the tubular neighborhood theorem\index{tubular neighborhood theorem} implies that 
the identity map of $M$ extends to an 
embedding of $E(\nu)$ onto an open neighborhood $U$ of $M$ in $\bR^{n+q}$. 

Think of $S^{n+q}$ as the one-point compactification of $\bR^{n+q}$. The ``Pontryagin-Thom
construction''\index{Pontryagin-Thom construction} associates a map $t: S^{n+q}\rtarr T(\nu)$ 
to our tubular neighborhood $U$.
Observing that $T\nu-\sset{\infty} = E(\nu)$, we let $t$ restrict on $U$ to the identification 
$U\iso E(\nu)$ and let $t$ send all points of $\bR^{n+q}-U$ to the point at infinity. 
The Thom space was tailor made to allow this construction. For $q$ large enough, any two 
embeddings of $M$ in $\bR^{n+q}$ are isotopic,
and the homotopy class of $t$ is independent of the choice of the embedding of $M$ in $\bR^{n+q}$.  
Now choose a classifying map $f: M\rtarr BO(q)$ for $\nu$. The composite 
$Tf\com t: S^{n+q} \rtarr TO(q)$ represents an element of $\pi_{n+q}(TO(q))$. 

As the reader should think through, it is intuitively plausible that cobordant manifolds induce 
homotopic maps  $S^{n+q} \rtarr TO(q)$, so that this construction gives a well defined
function $\al: \sN_n\rtarr \pi_{n+q}(TO(q))$. However, technically, one can arrange the argument
so that this fact drops out without explicit verification. Given two $n$-manifolds, we can embed 
them and their tubular neighborhoods disjointly in $\bR^{n+q}$, and it follows easily that $\al$
is a homomorphism. 

We construct an inverse $\be$ to $\al$. Any map $g: S^{n+q}\rtarr TO(q)$
has image contained in $T(\ga_q^{r})$ for a sufficiently large $r>q$, where $\ga_q^r$ is
the restriction of the universal bundle $\ga_q$ to the compact manifold $G_q(\bR^r)$.
By an implication of Sard's theorem known as the transversality\index{transversality} theorem, 
we can deform the
restriction of $g$ to $g^{-1}(T\ga_q^r-\sset{\infty})=g^{-1}(E(\ga_q^r))$ so as to obtain a
homotopic map that is smooth and transverse to the zero section. This use of transversality 
is the crux of the proof of the theorem. It follows that the inverse image
$g^{-1}(G_q(\bR^r))$ is a smooth closed $n$-manifold embedded in $\bR^{n+q}=S^{n+q}-\sset{\infty}$.
It is intuitively plausible that homotopic maps $g_i: S^{n+q}\rtarr TO(q)$, $i=0,\, 1$, give rise 
to cobordant $n$-manifolds by this construction. Indeed, with the $g_i$ smooth and transverse to
the zero section, we can approximate a homotopy between them by a homotopy $h$ which is smooth on
$h^{-1}(T(\ga_q^r)-\sset{\infty})$ and transverse to the zero section. Then
$h^{-1}(G_q(\bR^{r}))$ is a manifold whose boundary 
is $g_0^{-1}(G_q(\bR^r))\amalg g_1^{-1}(G_q(\bR^r))$.
It is easy to verify that the resulting function $\be: \pi_{n+q}(TO(q))\rtarr \sN_n$ is a 
homomorphism.

If we start with a manifold $M$ embedded in $\bR^{n+q}$ and construct the classifying
map $f$ for its normal bundle to be the Gauss map described in our sketch proof of
the classification theorem in Chapter 23 \S1, then the composite $Tf\com t$ is smooth and
transverse to the zero section, and the inverse image of the zero section is exactly $M$.
This proves that $\be$ is an epimorphism. To complete the proof, it suffices to show that 
$\be$ is a monomorphism. It will follow formally that $\al$ is well defined and inverse 
to $\be$. 

Thus suppose given $g: S^{n+q}\rtarr T\ga_q^r$ such that $g^{-1}(E(\ga_q^r))$
is smooth and transverse to the zero section and suppose that  $M=g^{-1}(G_q(\bR^r))$ is 
a boundary, say $M=\pa W$. The inclusion of $M$ in $S^{n+q}$ extends to a embedding of $W$ 
in $D^{n+q+1}$, by the Whitney embedding theorem for manifolds with boundary (assuming as 
always that $q$ is sufficiently large).  We may assume that $U=g^{-1}(T\ga_q^r-\sset{\infty})$ 
is a tubular neighborhood and that $g: U \rtarr E(\ga_q^r)$ is a map of vector bundles. 
A relative version of the tubular neighborhood theorem then shows that $U$ can be extended to a 
tubular neighborhood $V$ of $W$ in $D^{n+q+1}$ and that $g$ extends to a map of vector bundles 
$h: V\rtarr E(\ga_q^r)$. We can then extend $h$ to a map $D^{n+q+1}\rtarr T(\ga_q^r)$ by mapping 
$D^{n+q+1}-V$ to $\infty$. This extension of $g$ to the disk implies that $g$ is null homotopic.

We must still define the ring structure on $\pi_*(TO)$ and prove that we have an isomorphism
of rings and therefore of $\bZ_2$-algebras. Recall that we have maps 
$p_{m,n}: BO(m)\times BO(n)\rtarr BO(m+n)$ 
such that $p_{m,n}^*(\ga_{m+n})=\ga_m\times \ga_n$. The Thom space $T(\ga_m\times\ga_n)$ is 
canonically homeomorphic to the smash product $TO(m)\sma TO(n)$, and the bundle map 
$\ga_m\times \ga_n \rtarr \ga_{m+n}$ induces a map $\ph_{m,n}: TO(m)\sma TO(n)\rtarr TO(m+n)$.
If we have maps $f: S^{m+q}\rtarr TO(m)$ and $g: S^{n+q}\rtarr TO(n)$, then we can compose
their smash product with $\ph_{m,n}$ to obtain a composite map
$$ S^{m+n+q+r}\iso S^{m+q}\sma S^{n+r} \overto{f\sma g} TO(m)\sma TO(n) 
\overto{\ph_{m,n}} TO(m+n).$$
We can relate the maps $\ph_{m,n}$ to the maps $\si_n$. In fact, $TO$ is a commutative and
associative ring prespectrum in the sense of the following definition.

\begin{defn} Let $T$ be a prespectrum. Then $T$ is a ring prespectrum\index{ring prespectrum}
\index{prespectrum!ring} if there are maps
$\et: S^0 \rtarr T_0$ and $\ph_{m,n}: T_m\sma T_n\rtarr T_{m+n}$ such that the
following diagrams are homotopy commutative:
$$\diagram
T_m\sma\SI T_n \ddouble \rto^{\id\sma\si_n} & T_m\sma T_{n+1} \drto^{\ph_{m,n+1}} & \\
\SI(T_m\sma T_n) \dto_{(-1)^n} \rto^(0.56){\SI\ph_{m,n}}& \SI T_{m+n} \rto^(0.44){\si_{m+n}} & T_{m+n+1} \\
(\SI T_m)\sma T_n \rto_(0.5){\si_m\sma \id} & T_{m+1}\sma T_n \urto_{\ph_{m+1,n}} & \\
\enddiagram$$

\vspace{.1in}

$$\diagram
S^0\sma T_n \rto^{\et\sma\id} \drto_{\iso} 
& T_0\sma T_n \dto^{\ph_{0,n}} \\
& T_n\\
\enddiagram
\ \ \ \ \text{and} \ \ \ \
\diagram
T_n\sma T_0 \dto_{\ph_{n,0}} & T_n\sma S^0; \lto_{\id\sma \et} \dlto^{\iso}\\
T_{n} & \\
\enddiagram$$
$T$ is associative if the following diagrams are homotopy commutative:
$$\diagram
T_m\sma T_n\sma T_p \dto_{\id\sma\ph_{n,p}}\rto^(0.53){\ph_{m,n}\sma\id} 
& T_{m+n}\sma T_p \dto^{\ph_{m+n,p}} \\
T_m\sma T_{n+p} \rto_{\ph_{m,n+p}} & T_{m+n+p}; \\
\enddiagram$$
$T$ is commutative if there are equivalences $(-1)^{mn}: T_{m+n}\rtarr T_{m+n}$ that suspend
to $(-1)^{mn}$ on $\SI T_{m+n}$ and if the following diagrams are homotopy commutative:
$$\diagram
T_m\sma T_n \dto_{\ph_{m,n}}\rto^{t} & T_n\sma T_m \dto^{\ph_{n,m}} \\
T_{m+n} \rto_{(-1)^{mn}} & T_{m+n}.\\
\enddiagram$$
When $T$ is an $\OM$-prespectrum, we can restate this as $\ph_{m,n}\htp (-1)^{mn} \ph_{n,m}t$.
\end{defn}

For example, the Eilenberg-Mac\,Lane $\OM$-prespectrum of a commutative ring $R$ is an 
associative and commutative ring prespectrum by the arguments in Chapter 22 \S3. It is 
denoted $HR$ or sometimes, by abuse, $K(R,0)$. Similarly, the $K$-theory $\OM$-prespectrum 
is an associative and commutative ring prespectrum. The sphere prespectrum, whose $n$th space 
is $S^n$, is another example. For $TO$, the required maps $(-1)^{mn}: TO(m+n)\rtarr TO(m+n)$ 
are obtained by passage to Thom complexes from a map $\ga_{m+n}\rtarr \ga_{m+n}$ of universal 
bundles given on the domains of coordinate charts by the evident interchange isomorphism 
$\bR^{m+n}\rtarr \bR^{m+n}$. The following lemma is immediate by passage to colimits.

\begin{lem}
If $T$ is an associative ring prespectrum, then $\pi_*(T)$ is a graded ring. If $T$ is
commutative, then $\pi_*(T)$ is commutative in the graded sense.
\end{lem}

Returning to the case at hand, we show that the maps $\al$ for varying $n$ transport
products of manifolds to products in $\pi_*(TO)$. Thus let $M$ be an $m$-manifold embedded in 
$\bR^{m+q}$ with tubular neighborhood $U\iso E(\nu_M)$ and $N$ be an 
$n$-manifold embedded in $\bR^{n+r}$ with tubular neighborhood $V\iso E(\nu_N)$. Then $M\times N$ 
is embedded in $\bR^{m+q+n+r}$ with tubular neighborhood $U\times V\iso E(\nu_{M\times N})$. 
Identifying 
$S^{m+q+n+r}$ with $S^{m+q}\sma S^{n+r}$, we find that the Pontryagin-Thom construction
for $M\times N$ is the smash product of the Pontryagin-Thom constructions for $M$ and $N$. That is,
the left square in the following diagram commutes. The right square commutes up to homotopy by the
definition of $\ph_{q,r}$.
$$\diagram
S^{m+q}\sma S^{n+r} \rto^{t\sma t} \ddouble & T\nu_m\sma T\nu_N \dto^{\iso} \rto 
& TO(q)\sma TO(r) \dto^{\ph_{q,r}} \\
S^{m+q+n+r} \rto_{t} & T(\nu_{M\times N}) \rto & TO(q+r).\\
\enddiagram$$
This implies the claimed multiplicativity of the maps $\al$.

\section{Prespectra and the algebra $H_*(TO;\bZ_2)$}

Calculation of the homotopy groups $\pi_*(TO)$ proceeds by first computing the 
homology groups $H_*(TO;\bZ_2)$ and then showing that the stable Hurewicz homomorphism
maps $\pi_*(TO)$ monomorphically onto an identifiable part of $H_*(TO;\bZ_2)$. 
We explain the calculation of homology groups in this section and the next, connect the 
calculation with Stiefel-Whitney numbers in \S5, and describe how to complete the 
desired calculation of homotopy groups in \S6.

We must first define the homology groups of prespectra and the stable Hurewicz homomorphism.
Just as we defined the homotopy groups of a prespectrum $T$ by the formula
$$\pi_n(T)=\colim \pi_{n+q}(T_q),$$
we define the homology and cohomology groups\index{prespectrum!homology groups of}
\index{prespectrum!cohomology groups of} of $T$ with respect to a 
homology theory $k_*$ and cohomology theory $k^*$ on spaces by the formulas
$$k_n(T)=\colim \tilde{k}_{n+q}(T_q),$$
where the colimit is taken over the maps
$$ \tilde{k}_{n+q}(T_q) \overto{\SI_*} \tilde{k}_{n+q+1}(\SI T_q)
\overto{{\si_q}_*} \tilde{k}_{n+q+1}(T_{q+1}),$$
and 
$$k^n(T) = \lim \tilde{k}^{n+q}(T_q),$$
where the limit is taken over the maps
$$\tilde{k}^{n+q+1}(T_{q+1})\overto{\si^*_q} \tilde{k}^{n+q+1}(\SI T_q) 
\overto{\SI^{-1}} \tilde{k}^{n+q}(T_q).$$
In fact, this definition of cohomology is inappropriate in general, differing from
the appropriate definition by a ${\lim}^1$ error term. However, the definition is
correct when $k^*$ is ordinary cohomology with coefficients in a field $R$ and each
$\tilde{H}^{n+q}(T_q;R)$ is a finite dimensional vector space over $R$. This is the 
only case that we will need in the work of this chapter. In this case, it is clear
that $H^n(T;R)$ is the vector space dual of $H_n(T;R)$, a fact that we shall use
repeatedly.  

Observe that there is no cup product in $H^*(T;R)$: the maps in the 
limit system factor through the reduced cohomologies of suspensions, in which
cup products are identically zero (see Problem 5 at the end of Chapter 19).
However, if $T$ is an associative and commutative ring prespectrum, then the 
homology groups $H_*(T;R)$ form a graded commutative $R$-algebra.

The Hurewicz homomorphisms $\pi_{n+q}(T_q)\rtarr \tilde{H}_{n+q}(T_q;Z)$ pass to 
colimits to give the stable Hurewicz homomorphism\index{Hurewicz homomorphism!stable}
$$h: \pi_n(T)\rtarr H_n(T;\bZ).$$
We may compose this with the map $H_n(T;\bZ)\rtarr H_n(T;R)$ induced by the unit of
a ring $R$, and we continue to denote the composite by $h$. If $T$ is an associative
and commutative ring prespectrum, then $h: \pi_*(T)\rtarr H_*(T;R)$ is a map of graded
commutative rings. 

We shall write $H_*$ and $H^*$ for homology and cohomology with coefficients in $\bZ_2$
throughout \S\S3--6, and we tacitly assume that all homology and cohomology groups 
in sight are finite dimensional $\bZ_2$-vector spaces. Recall that we have Thom isomorphisms 
$$\PH_q: H^n(BO(q))\rtarr \tilde{H}^{n+q}(TO(q))$$
obtained by cupping with the Thom class $\mu_q\in \tilde{H}^q(TO(q))$.
Naturality of the Thom diagonal applied to the map of bundles $\ga_q\oplus\epz \rtarr \ga_{q+1}$
gives the commutative diagram
$$\diagram 
\SI TO(q) \rto^(0.4){\DE} \dto_{\si_q} & BO(q)_+\sma \SI TO(q) \dto^{i_q\sma \si_q}\\
TO(q+1) \rto_(0.35){\DE} & BO(q+1)_+\sma TO(q+1).\\
\enddiagram$$
This implies that the following diagram is commutative:
$$\diagram 
H^n(BO(q+1))\rrto^{i_q^*} \dto_{\PH_{q+1}} & & H^n(BO(q)) \dto^{\PH_q}\\
\tilde{H}^{n+q+1}(TO(q+1))\rto_{\si^*_q} & \tilde{H}^{n+q+1}(\SI TO(q)) 
\rto_{\SI^{-1}} & \tilde{H}^{n+q}(TO(q)).\\
\enddiagram$$
We therefore obtain a ``stable Thom isomorphism''\index{Thom isomorphism!stable}
$$\PH: H^n(BO)\rtarr H^n(TO)$$
on passage to limits. We have dual homology Thom isomorphisms
$$\PH_n: \tilde{H}_{n+q}(TO(q))\rtarr H_n(BO(q))$$
that pass to colimits to give a stable Thom isomorphism
$$\PH: H_n(T) \rtarr H_n(BO).$$

Naturality of the Thom diagonal applied to the map of bundles $\ga_q\oplus\ga_r \rtarr \ga_{q+r}$
gives the commutative diagram
$$\diagram
TO(q)\sma TO(r) \ddto_{\ph_{q,r}} \rto^(0.33){\DE\sma\DE} 
& BO(q)_+\sma TO(q)\sma BO(r)_+ \sma TO(r) \dto^{\id\sma t\sma \id}\\
& (BO(q)\times BO(r))_+\sma TO(q)\sma TO(r) \dto^{(p_{q,r})_+\sma \ph_{q,r}} \\
TO(q+r) \rto_(0.38){\DE} & BO(q+r)_+\sma TO(q+r). \\
\enddiagram$$
As we observed for $BU$ in the previous chapter, the maps $p_{q,r}$ pass to colimits to give
$BO$ an $H$-space structure, and it follows that $H_*(BO)$ is a $\bZ_2$-algebra. On passage
to homology and colimits, these diagrams imply the following conclusion.

\begin{prop} The Thom isomorphism $\PH: H_*(TO)\rtarr H_*(BO)$ is an isomorphism of 
$\bZ_2$-algebras.
\end{prop}

The description of the $H^*(BO(n))$ and the maps $i_q^*$ in Chapter 23 \S2 implies that 
$$H^*(BO)=\bZ_2[w_i|i\geq 1]$$
as an algebra. However, we are more interested in its ``coalgebra''\index{coalgebra} structure, 
which is given by the vector space dual 
$$\ps: H^*(BO)\rtarr H^*(BO)\ten H^*(BO)$$
of its product in homology. It is clear from the description of the $p_{q,r}^*$ that 
$$\ps(w_k)=\sum_{i+j=k} w_i\ten w_j.$$
From here, determination of $H_*(BO)$ and therefore $H_*(TO)$ as an algebra is a purely algebraic,
but non-trivial, problem in dualization. Let $i: \bR P^{\infty}=BO(1)\rtarr BO$ be the inclusion. 
Let $x_i\in H_i(\bR P^{\infty})$ be the unique non-zero element and let $b_i=i_*(x_i)$. 
Then the solution of our dualization problem takes the following form.

\begin{thm} $H_*(BO)$ is the polynomial algebra $\bZ_2[b_i|i\geq 1]$.
\end{thm}

Let $a_i\in H_i(TO)$ be the element characterized by $\PH(a_i) = b_i$.

\begin{cor}
$H_*(TO)$ is the polynomial algebra $\bZ_2[a_i|i\geq 1]$.
\end{cor}

Using the compatibility of the Thom isomorphisms for $BO(1)$ and $BO$, we see that the
$a_i$ come from $H_*(TO(1))$. Remember that elements of $H_{i+1}(TO(1))$ map to elements 
of $H_i(TO)$ in the colimit; in particular, the non-zero element of $H_1(TO(1))$ maps to
the identity element $1\in H_0(TO)$. Recall from Chapter 23 \S6 that we have a homotopy equivalence 
$j: \bR P^{\infty}\rtarr TO(1)$.

\begin{cor} For $i\geq 0$, $j_*(x_{i+1})$ maps to $a_i$ in $H_*(TO)$, where $a_0=1$.
\end{cor}

\section{The Steenrod algebra and its coaction on $H_*(TO)$}

Since the Steenrod operations are stable and natural, they pass to limits to define
natural operations\index{prespectrum!Steenrod operations of} 
$Sq^i: H^n(T)\rtarr H^{n+i}(T)$ for $i\geq 0$ and prespectra $T$. Here 
$Sq^0=\id$, but it is not true that $Sq^i(x)=0$ for $i>\deg\,x$. For example, we have the 
``stable Thom class''\index{Thom class!stable} $\PH(1)=\mu\in H^0(TO)$, and it is immediate 
from the definition of the 
Stiefel-Whitney classes that $\PH(w_i)=Sq^i(\mu)$. Of course, $Sq^i(1)=0$ for 
$i>0$, so that $\PH$ does not commute with Steenrod operations. The homology and
cohomology of $TO$ are built up from $\pi_*(TO)$ and Steenrod operations. We need
to make this statement algebraically precise to determine $\pi_*(TO)$, and we need
to assemble the Steenrod operations into an algebra to do this.

\begin{defn} The mod $2$ Steenrod algebra\index{Steenrod algebra} $A$ is the quotient 
of the free associative
$\bZ_2$-algebra generated by elements $Sq^i$, $i\geq 1$, by the ideal generated by the 
Adem relations (which are stated in Chapter 22 \S5).
\end{defn}

The following lemmas should be clear.

\begin{lem} For spaces $X$, $H^*(X)$ has a natural $A$-module structure.  
\end{lem}

\begin{lem} For prespectra $T$, $H^*(T)$ has a natural $A$-module structure.
\end{lem}

The elements of $A$ are stable mod $2$ cohomology operations, and our description of the cohomology 
of $K(\bZ_2,q)$s in Chapter 22 \S5 implies that $A$ is in fact the algebra of all stable mod $2$ cohomology
operations, with multiplication given by composition. Passage to limits over $q$ leads to the
following lemma. Alternatively, with the more formal general definitions of the next section, 
it will become yet another application of the Yoneda lemma. Recall 
that $H\bZ_2$ denotes the Eilenberg-Mac\,Lane $\OM$-prespectrum $\sset{K(\bZ_2,q)}$.

\begin{lem} As a vector space, $A$ is isomorphic to $H^*(H\bZ_2)$.
\end{lem}

We shall see how to describe the composition in $A$ homotopically in the next section. 
What is more important at the moment is that the lemma allows us to read off a basis for $A$.

\begin{thm} $A$ has a basis consisting of the operations $Sq^I = Sq^{i_1}\cdots Sq^{i_j}$, 
where $I$ runs over the sequences $\sset{i_1,\ldots\!,i_j}$ of positive integers such that 
$i_{r}\geq 2 i_{r+1}$ for $1\leq r < j$. 
\end{thm}

What is still more important to us is that $A$ not only has the composition product
$A\ten A\rtarr A$, it also has a coproduct $\ps: A\rtarr A\ten A$. Giving $A\ten A$
its natural structure as an algebra, $\ps$ is the unique map of algebras specified
on generators by $\ps(Sq^k) = \sum_{i+j=k} Sq^i\ten Sq^j$. The fact that $\ps$ is a
well defined map of algebras is a formal consequence of the Cartan formula. Algebraic
structures like this, with compatible products and coproducts, are called 
``Hopf algebras.''\index{Hopf algebra}

We write $A_*$ for the vector space dual of $A$, and we give it the dual basis to
the basis just specified on $A$. While $A_*$ is again a Hopf algebra, we are
only interested in its algebra structure at the moment. In contrast with $A$, the algebra
$A_*$ is commutative, as is apparent from the form of the coproduct on the generators of $A$. 
Recall that $H\bZ_2$ is an associative and commutative ring prespectrum, so that $H_*(H\bZ_2)$ 
is a commutative $\bZ_2$-algebra. The definition of the product on $H\bZ_2$ (in Chapter 22 \S3) and 
the Cartan formula directly imply the following observation.

\begin{lem} $A_*$ is isomorphic as an algebra to $H_*(H\bZ_2)$.
\end{lem}

We need an explicit description of this algebra. In principle, this is a matter of pure
algebra from the results already stated, but the algebraic work is non-trivial.

\begin{thm} For $r\geq 1$, define $I_r=(2^{r-1}, 2^{r-2},\ldots\!, 2, 1)$ and define $\xi_r$
to be the basis element of $A_*$ dual to $Sq^{I_r}$. Then $A_*$ is the polynomial algebra
$\bZ_2[\xi_r|r\geq 1]$. 
\end{thm}

We need a bit of space level motivation for the particular relevance of the elements $\xi_r$. 
We left the computation of the Steenrod operations in $H^*(\bR P^{\infty})$ as an exercise,
and the reader should follow up by proving the following result.

\begin{lem} In $H^*(\bR P^{\infty})=\bZ_2[\al]$, $Sq^{I_r}(\al)=\al^{2^r}$ for $r\geq 1$
and $Sq^{I}(\al)=0$ for all other basis elements $Sq^I$ of $A$.
\end{lem}

The $A$-module structure maps 
$$A\ten H^*(X)\rtarr H^*(X) \ \ \tand \ \ A\ten H^*(T)\rtarr H^*(T)$$
for spaces $X$ and prespectra $T$ dualize to give ``$A_*$-comodule''\index{comodule} structure maps
$$\ga: H_*(X)\rtarr A_*\ten H_*(X) \ \ \tand \ \ \ga: H_*(T)\rtarr A_*\ten H_*(T).$$
We remind the reader that we are implicitly assuming that all homology and cohomology groups 
in sight are finitely generated $\bZ_2$-vector spaces, although these ``coactions'' can in fact be 
defined without this assumption. 

Formally, the notion of a comodule $N$ over a coalgebra $C$ 
is defined by reversing the direction of arrows in a diagrammatic definition of a module over 
an algebra. For example, for any vector space $V$, $C\ten V$ is a comodule with action 
$$\ps\ten\id: C\ten V\rtarr C\ten C\ten V.$$ 
Note that, dualizing the unit of an algebra, a $\bZ_2$-coalgebra is
required to have a counit $\epz: C\rtarr \bZ_2$. We understand all of these algebraic structures
to be graded, and we say that a coalgebra is connected if $C_i=0$ 
for $i<0$ and $\epz: C_0\rtarr \bZ_2$ is an isomorphism. When considering the Hurewicz homomorphism 
of $\pi_*(TO)$, we shall need the following observation.

\begin{lem} Let $C$ be a connected coalgebra and $V$ be a vector space. 
An element $y\in C\ten V$ satisfies $(\ps\ten\id)(y) = 1\ten y$ if and 
only if $y\in C_0\ten V\iso V$. 
\end{lem}

If $V$ is a $C$-comodule with coaction $\nu: V\rtarr C\ten V$,
then $\nu$ is a morphism of $C$-comodules. Therefore the coaction maps $\ga$ above are maps of
$A_*$-comodules for any space $X$ or prespectrum $T$. We also need the following observation, 
which is implied by the Cartan formula.

\begin{lem} If $T$ is an associative ring prespectrum, then $\ga: H_*(T)\rtarr A_*\ten H_*(T)$
is a homomorphism of algebras.
\end{lem}

The lemma above on Steenrod operations in $H^*(\bR P^{\infty})$ dualizes as follows.

\begin{lem} Write the coaction $\ga: H_*(\bR P^{\infty})\rtarr A_*\ten H_*(\bR P^{\infty})$
in the form $\ga(x_i) = \sum_j a_{i,j}\ten x_j$. Then 
$$ a_{i,1}= \left\{ \begin{array}{ll}
\xi_r  &  \mbox{if $i=2^r$ for some $r\geq 1$}\\
0      &  \mbox{otherwise.}
\end{array} \right. $$
\end{lem}

Note that $a_{i,i}=1$, dualizing $Sq^0(\al^i)=\al^i$.

Armed with this information, we return to the study of the algebra $H_*(TO)$. 
We know that it is isomorphic to $H_*(BO)$, but the crux of the matter is to
redescribe it in terms of $A_*$. 

\begin{thm} Let $N_*$ be the algebra defined abstractly by 
$$N_*=\bZ_2[u_i|i>1  \tand i\neq 2^r-1],$$
where $\deg u_i = i$. Define a homomorphism of algebras $f: H_*(TO)\rtarr N_*$ by
$$ f(a_i)= \left\{ \begin{array}{ll}
u_i  &  \mbox{if $i$ is not of the form $2^r-1$}\\
0      &  \mbox{if $i=2^r-1$.}
\end{array} \right. $$
Then the composite
$$g: H_*(TO)\overto{\ga} A_*\ten H_*(TO) \overto{\id\ten f} A_*\ten N_*$$
is an isomorphism of both $A$-comodules and $\bZ_2$-algebras.
\end{thm}
\begin{proof} It is clear from things already stated that $g$ is a map of both $A$-comodules 
and $\bZ_2$-algebras. We must prove that it is an isomorphism. Its source and target are both 
polynomial algebras with one generator of degree $i$ for each $i\geq 1$, hence it suffices to
show that $g$ takes generators to generators. Recall that $a_i=j_*(x_{i+1})$. This allows
us to compute $\ga(a_i)$. Modulo terms that are decomposable in the algebra $A_*\ten H_*(TO)$,
we find 
$$ \ga(a_i)\equiv  \left\{ \begin{array}{ll}
1\ten a_i  &  \mbox{if $i$ is not of the form $2^r-1$}\\
\xi_r\ten 1 + 1\ten a_{2^r-1}      &  \mbox{if $i=2^r-1$.}
\end{array} \right. $$
Applying $\id\ten f$ to these elements, we obtain $1\ten u_i$ in the first case and 
$\xi_r\ten 1$ in the second case.
\end{proof}

Now consider the Hurewicz homomorphism $h: \pi_*(T)\rtarr H_*(T)$ of a prespectrum $T$.
We have the following observation, which is a direct consequence of the definition of 
the Hurewicz homomorphism and the fact that $Sq^i = 0$ for $i>0$ in the cohomology of spheres.

\begin{lem} For $x\in\pi_*(T)$, $\ga(h(x))=1\ten h(x)$. 
\end{lem}

Therefore, identifying $N_*$ as the subalgebra $\bZ_2\ten N_*$ of $A_*\ten N_*$, we see that 
$g\com h$ maps $\pi_*(TO)$ to $N_*$.  We shall prove the following result in \S6 
and so complete the proof of Thom's theorem.

\begin{thm} $h: \pi_*(TO)\rtarr H_*(TO)$ is a monomorphism and $g\com h$ maps $\pi_*(TO)$
isomorphically onto $N_*$.
\end{thm}

\section{The relationship to Stiefel-Whitney numbers}

We shall prove that a smooth closed $n$-manifold $M$ is a boundary if and only if all
of its normal Stiefel-Whitney numbers\index{Stiefel-Whitney numbers!normal} are zero. Polynomials 
in the Stiefel-Whitney 
classes are elements of $H^*(BO)$. We have seen that the normal Stiefel-Whitney numbers 
of a boundary are zero, and it follows that cobordant manifolds have the same normal 
Stiefel-Whitney numbers. The assignment of Stiefel-Whitney numbers to
corbordism classes of $n$-manifolds specifies a homomorphism
$$\#: H^n(BO)\ten \sN_n \rtarr \bZ_2.$$
We claim that the following diagram is commutative:
$$\diagram
H^n(BO)\ten \sN_n \rto^(0.45){\id\ten\al} \dto_{\#} & H^n(BO)\ten \pi_n(TO) \rto^{\id\ten h} 
& H^n(BO)\ten H_n(TO) \dto^{\id\ten \PH} \\
\bZ_2 & & H^n(BO)\ten H_n(BO). \llto_{\langle \, , \, \rangle}\\
\enddiagram$$
To say that all normal Stiefel-Whitney numbers of $M$ are zero is to say that $w\#[M]=0$ 
for all $w\in H^n(BO)$. Granted the commutativity of the diagram, this is the same as to say that
$\langle w,(\PH\com h\com \al)([M])\rangle = 0$ for all $w\in H^n(BO)$. Since 
$\langle \, , \, \rangle$ is the evaluation pairing of dual vector spaces, this implies that 
$(\PH\com h\com \al)([M])=0$. Since $\PH$ and $\al$ are isomorphisms and $h$ is a monomorphism, 
this implies that $[M]=0$ and thus that $M$ is a boundary. 

Thus we need only prove that the diagram is commutative. Embed $M$ in $\bR^{n+q}$ with normal
bundle $\nu$ and let $f: M\rtarr BO(q)$ classify $\nu$. Then $\al([M])$ is 
represented by the composite $S^{n+q}\overto{t} T\nu\overto{Tf} TO(q)$. In homology, we have
the commutative diagram
$$\diagram
\tilde{H}_{n+q}(S^{n+q}) \rto^{t_*} 
& \tilde{H}_{n+q}(T\nu) \rto^{(Tf)_*} \dto^{\PH} & \tilde{H}_{n+q}(TO(q)) \dto^{\PH} \\
& H_n(M) \rto_{f_*} & H_n(BO(q)).\\
\enddiagram$$
Let $i_{n+q}\in \tilde{H}_{n+q}(S^{n+q})$ be the fundamental class. By the diagram and the
definitions of $\al$ and the Hurewicz homomorphism,
$$ (f_*\com \PH\com t_*)(i_{n+q}) = 
(\PH\com (Tf)_*\com t_*)(i_{n+q}) = (\PH\com h\com \al)([M]) \in H_n(BO(q)).$$ 
Let $z=(\PH\com t_*)(i_{n+q})\in H_n(M)$. We claim that $z$ is the fundamental class. 
Granting the claim, it follows immediately that, for $w\in H^n(BO(q))$,
\begin{eqnarray*}
w\# [M] = \langle w(\nu), z\rangle   & = & \langle (f^*w(\ga_q)),(\PH\com t_*)(i_{n+q}) \rangle \\
& = & \langle w(\ga_q), (f_*\com \PH\com t_*)(i_{n+q})\rangle \\
& = & \langle w(\ga_q), (\PH\com h\com \al)([M])\rangle.
\end{eqnarray*}

Thus we are reduced to proving the claim. It suffices to show
that $z$ maps to a generator of $H_n(M,M-x)$ for each $x\in M$. Since we must deal with pairs,
it is convenient to use the homeomorphism between $T\nu$ and the quotient $D(\nu)/S(\nu)$ 
of the unit disk bundle by the unit sphere bundle. Recall that we have a relative cap 
product
$$\cap: H^q(D(\nu),S(\nu))\ten H_{i+q}(D(\nu),S(\nu))\rtarr H_i(D(\nu)).$$
Letting $p:D(\nu)\rtarr M$ be the projection, which of course is a homotopy equivalence, we 
find that the homology Thom isomorphism 
$$\PH: H_{i+q}(D(\nu),S(\nu))\rtarr H_i(M)$$
is given by the explicit formula
$$ \PH(a) = p_*(\mu \cap a).$$
Let $x\in U\subset M$, where $U\iso \bR^n$. Let $D(U)$ and $S(U)$ be the inverse images in
$U$ of the unit disk and unit sphere in $\bR^n$ and let $V=D(U)-S(U)$. Since $D(U)$ is 
contractible, $\nu|_{D(U)}$ is trivial and thus isomorphic to $D(U)\times D^q$. Write
$$\pa(D(U)\times D^q) = (D(U)\times S^{q-1})\cup (S(U)\times D^q)$$
and observe that we obtain a homotopy equivalence
$$t: S^{n+q} \rtarr (D(U)\times D^q)/\pa (D(U)\times D^q)\iso S^{n+q}$$
by letting $t$ be the quotient map on the restriction of the tubular neighborhood of $\nu$
to $D(\nu|_{D(U)})$ and letting $t$ send the complement of this restriction to the basepoint.
Interpreting $t: S^{n+q}\rtarr D(\nu)/S(\nu)$ similarly, we obtain the following commutative 
diagram:
\begin{small}
$$\diagram
\tilde{H}_{n+q}(S^{n+q}) \rto^(0.3){t_*}_(0.3){\iso} \ddto_{t_*} 
& H_{n+q}(D(U)\times D^q,\pa(D(U)\times D^q)) \rto^(0.63){\PH}_(0.63){\iso} \dto 
& H_n(D(U),S(U)) \dto^{\iso} \\
& H_{n+q}(D(\nu),S(\nu)\cup D(\nu|_{M-V}))
\rto^(0.6){\PH} & H_n(M,M-V) \dto^{\iso} \\
H_{n+q}(D(\nu),S(\nu))\urto \rto_(0.55){\PH} & H_n(M) \urto \rto & H_n(M,M-x).\\
\enddiagram$$
\end{small}
The unlabeled arrows are induced by inclusions, and the right vertical arrows are 
excision isomorphisms. The maps $\PH$ are of the general form $\PH(a)=p_*(\mu\cap a)$.
For the top map $\PH$, $\mu\in H_{n+q}(D(\nu|_{D(U)}),S(\nu|_{D(U)}))\iso H_{n+q}(S^{n+q})$,
and, up to evident isomorphisms, $\PH$ is just the inverse of the suspension isomorphism 
$\tilde{H}_n(S^n) \rtarr \tilde{H}_{n+q}(S^{n+q})$. The diagram shows that $z$ maps to
a generator of $H_n(M,M-x)$, as claimed.

\section{Spectra and the computation of $\pi_*(TO) =\pi_*(MO)$}

We must still prove that $h:\pi_*(TO)\rtarr H_*(TO)$ is a monomorphism and
that $g\com h$ maps $\pi_*(TO)$ isomorphically onto $N_*$. Write $N$ for
the dual vector space of $N_*$. (Of course, $N$ is a coalgebra, but that
is not important for this part of our work.) Remember that the Steenrod
algebra $A$ is dual to $A_*$ and that $A\iso H^*(H\bZ_2)$. The dual of 
$g:H_*(TO)\rtarr A_*\ten N_*$ is an isomorphism of $A$-modules (and of 
coalgebras) $g^*: A\ten N\rtarr H^*(TO)$. Thus, if we choose a basis $\sset{y_i}$
for $N$, where $\deg\,y_i = n_i$ say, then $H^*(TO)$ is the free graded $A$-module
on the basis $\sset{y_i}$.

At this point, we engage in a conceptual thought exercise. We think of prespectra
as ``stable objects''\index{stable objects} that have associated homotopy, homology, and 
cohomology groups. Imagine that we have a good category of stable objects, analogous to the
category of based spaces, that is equipped with all of the constructions that we 
have on based spaces: wedges (= coproducts), colimits, products, limits, suspensions, 
loops, homotopies, cofiber sequences, fiber sequences, smash products, function 
objects, and so forth. Let us call the stable objects in our imagined category 
``spectra''\index{spectrum} and call the category of such objects $\sS$.\index{S@$\sS$} 
We have in mind an analogy with the notions of presheaf and sheaf.

Whatever spectra are, there must be a way of constructing a spectrum from a 
prespectrum without changing its homotopy, homology, and cohomology groups. 
In turn, a based space $X$ determines the prespectrum $\SI^{\infty} X=\sset{\SI^nX}$.
The homology and cohomology groups of $\SI^{\infty} X$ are the (reduced) homology and cohomology
groups of $X$; the homotopy groups of $\SI^{\infty} X$ are the stable homotopy groups of $X$.

Because homotopy groups, homology groups, and cohomology groups on based spaces satisfy the 
weak equivalence axiom, the real domain of definition of these invariants is the category 
$\bar{h}\sT$ that is obtained from the homotopy category $h\sT$ of based spaces by adjoining 
inverses to the weak equivalences. This category is equivalent to the homotopy 
category $h\sC$ of based CW complexes. Explicitly, the morphisms from $X$ to $Y$ in 
$\bar{h}\sT$ can be defined to be the based homotopy classes of maps $\GA X\rtarr \GA Y$, 
where $\GA X$ and $\GA Y$ are CW approximations of $X$ and $Y$. Composition is defined 
in the evident way.

Continuing our thought exercise, we can form the homotopy category $h\sS$ of spectra and
can define homotopy groups in terms of homotopy classes of maps from sphere spectra to
spectra. Reflection on the periodic nature of $K$-theory suggests that we should define
sphere spectra of negative dimension and define homotopy groups $\pi_q(X)$ for all integers $q$. 
We say that a map of spectra is a weak equivalence if it induces an isomorphism on homotopy 
groups. We can form the ``stable category''\index{stable category} $\bar{h}\sS$ from $h\sS$ 
exactly as we formed the 
category  $\bar{h}\sT$ from $h\sT$. That is, we develop a theory of CW spectra using sphere
spectra as the domains of attaching maps. The Whitehead and cellular approximation theorems
hold, and every spectrum $X$ admits a CW approximation  $\GA X\rtarr X$. We define the set
$[X,Y]$ of morphisms $X\rtarr Y$ in $\bar{h}\sS$ to be the set of homotopy classes of maps
$\GA X\rtarr \GA Y$. This is a {\em stable} category in the sense that the functor 
$\SI: \bar{h}\sS \rtarr \bar{h}\sS$ is an equivalence of categories. More explicitly, the 
natural maps $X\rtarr \OM\SI X$ and $\SI\OM X\rtarr X$ are isomorphisms in $\bar{h}\sS$. 

In particular, up to isomorphism, 
every object in the category $\bar{h}\sS$ is a suspension, hence a double suspension. This 
implies that each $[X,Y]$ is an Abelian group and composition is bilinear. Moreover, for 
any map $f: X\rtarr Y$, the canonical map $Ff\rtarr \OM Cf$ and its adjoint $\SI Ff\rtarr Cf$
(see Chapter 8 \S7) are also isomorphisms in $\bar{h}\sS$, so that cofiber sequences and 
fiber sequences are equivalent. Therefore cofiber sequences give rise to long exact sequences
of homotopy groups.

The homotopy groups of wedges and products of spectra are given by
$$\pi_*(\textstyle{\bigvee}_i\, X_i) = \textstyle{\sum}_i\, \pi_*(X_i) 
\tand \pi_*(\textstyle{\prod}_i\, X_i)=\textstyle{\prod}_i\, \pi_*(X_i).$$
Therefore, if only finitely many $\pi_q(X_i)$ are non-zero for each $q$, then the natural 
map $\bigvee_i\, X_i\rtarr \prod_i\, X_i$ is an isomorphism. 

We have homology groups and cohomology groups defined on $\bar{h}\sS$. A spectrum $E$ 
represents a homology theory\index{homology theory!on spectra} $E_*$ and a cohomology 
theory\index{cohomology theory!on spectra} $E^*$ specified in terms 
of smash products and function spectra by 
$$E_q(X) =\pi_q(X\sma E) \ \tand \ E^q(X) = \pi_{-q}F(X,E) \iso [X,\SI^qE].$$ 
Verifications of the exactness, suspension, additivity, and weak equivalence axioms are
immediate from the properties of the category $\bar{h}\sS$. Moreover,
every homology or cohomology theory on $\bar{h}\sS$ is so represented by some spectrum $E$. 

As will become clear later, $\OM$-prespectra are more like spectra than general prespectra, 
and we continue to write $H\pi$ for the ``Eilenberg-Mac\,Lane spectrum''
\index{Eilenberg-Mac\,Lane spectrum} that represents 
ordinary cohomology with coefficients in $\pi$. Its only non-zero homotopy group is 
$\pi_0(H\pi)=\pi$, and the Hurewicz homomorphism maps this group isomorphically onto 
$H_0(H\pi;\bZ)$. When $\pi=\bZ_2$, the natural map $H_0(H\bZ_2;\bZ)\rtarr H_0(H\bZ_2;\bZ_2)$ 
is also an isomorphism.

Returning to our motivating example, we write $MO$\index{MO@$MO$} for the 
``Thom spectrum''\index{Thom spectrum}
that arises from the Thom prespectrum $TO$. The reader may sympathize 
with a student who claimed that 
$MO$ stands for ``Mythical Object.''\index{Mythical Object} 

We may choose a map $\bar{y}_i: MO \rtarr \SI^{n_i}H\bZ_2$ that represents the
element $y_i$. Define $K(N_*)$ to be the wedge of a copy of $\SI^{n_i}H\bZ_2$ for 
each basis element $y_i$ and note that $K(N_*)$ is isomorphic in $\bar{h}\sS$ to
the product of a copy of $\SI^{n_i}H\bZ_2$ for each $y_i$. We think of $K(N_*)$ as a 
``generalized Eilenberg-Mac\,Lane spectrum.'' It satisfies $\pi_*(K(N_*))\iso N_*$ 
(as Abelian groups and so as $\bZ_2$-vector spaces), and the mod $2$ Hurewicz 
homomorphism $h: \pi_*(K(N_*))\rtarr H_*(K(N_*))$ is a monomorphism. Using the 
$\bar{y}_i$ as coordinates, we obtain a map
$$\om: MO\rtarr \textstyle{\prod}_i\, \SI^{n_i} H\bZ_2 \htp K(N_*).$$

The induced map $\om^*$ on mod $2$ cohomology is an isomorphism of $A$-modules: $H^*(MO)$ 
and $H^*(K(N_*))$ are free $A$-modules, and we have defined $\om$  
so that $\om^*$ sends basis elements to basis elements. Therefore the induced map on
homology groups is an isomorphism. Here we are using mod $2$ homology, but it 
can be deduced from the fact that both $\pi_*(MO)$ and $\pi_*(K(N_*))$
are $\bZ_2$-vector spaces that $\om$ induces an isomorphism on integral homology groups.
Therefore the integral homology groups of $C\om$ are zero. By the Hurewicz theorem in
$\bar{h}\sS$, the homotopy groups of $C\om$ are also zero. Therefore $\om$ induces
an isomorphism of homotopy groups. That is, $\om$ is an isomorphism in $\bar{h}\sS$. 
Therefore $\pi_*(MO)\iso N_*$ and the Hurewicz homomorphism $h:\pi_*(MO)\rtarr H_*(MO)$ 
is a monomorphism. It follows that $g\com h:\pi_*(MO)\rtarr N_*$ is an isomorphism since 
it is a monomorphism between vector spaces of the same finite dimension in each degree.

\section{An introduction to the stable category}

To give content to the argument just sketched, we should construct a good category of spectra. 
In fact, no such category was available when Thom first proved his theorem in 1960. With 
motivation from the introduction of $K$-theory and cobordism, a good stable category was 
constructed by Boardman (unpublished) around 1964 and an exposition of his category was 
given by Adams soon after. However, these early constructions were far more primitive than 
our outline suggests. While they gave a satisfactory stable category, the underlying category
of spectra did not have products, limits, and function objects, and its smash product was
not associative, commutative, or unital. In fact, a fully satisfactory category of spectra
was not constructed until 1995.

We give a few definitions to indicate what is involved. 

\begin{defn}
A spectrum\index{spectrum} $E$ is a prespectrum
such that the adjoints $\tilde{\si}: E_n\rtarr \OM E_{n+1}$ of the structure maps
$\si: \SI E_n \rtarr E_{n+1}$ are {\em homeomorphisms}. A map $f: T\rtarr T'$ of prespectra
is a sequence of maps $f_n: T_n\rtarr T'_n$ such that $\si'_n\com \SI f_n = f_{n+1}\com \si_n$
for all $n$. A map $f:E\rtarr E'$ of spectra is a map between $E$ and $E'$ regarded as
prespectra. 
\end{defn}

We have a forgetful functor from the category $\sS$\index{S@$\sS$} of spectra to the 
category $\sP$\index{P@$\sP$} of 
prespectra. It has a left adjoint $L:\sP\rtarr \sS$. In $\sP$, we define wedges, colimits, 
products, and limits spacewise. For example, $(T\wed T')_n = T_n\wed T'_n$, with the 
evident structure maps. We define wedges and colimits of spectra by first performing the
construction on the prespectrum level and then applying the functor $L$. If we start with 
spectra and construct products or limits spacewise, then the result is again a spectrum;
that is, limits of spectra are the limits of their underlying prespectra. Thus the category
$\sS$ is complete and cocomplete.

Similarly, we define the smash product $T\sma X$ and function prespectrum $F(X,T)$ of a 
based space $X$ and a prespectrum $T$ spacewise.  For a spectrum $E$, we define $E\sma X$ 
by applying $L$ to the prespectrum level construction; the prespectrum $F(X,E)$ is already
a spectrum. We now have cylinders $E\sma I_+$ and thus can define homotopies between maps
of spectra. Similarly we have cones $CE=E\sma I$ (where $I$ has basepoint $1$), suspensions 
$\SI E=E\sma S^1$, path spectra $F(I,E)$ (where $I$ has basepoint $0$), and loop spectra
$\OM E =F(S^1,E)$. The development of cofiber and fiber sequences proceeds exactly as for
based spaces. 

The left adjoint $L$ can easily be described explicitly on those prespectra $T$ whose
adjoint structure maps $\tilde{\si}_n: T_n\rtarr \OM T_{n+1}$ are inclusions:
we define $(LT)_n$ to be the union of the expanding sequence 
$$T_n \overto{\tilde{\si}_n} \OM T_{n+1} \overto{\OM\tilde{\si}_{n+1}} \OM^2 T_{n+2} \rtarr \cdots.$$
We then have
$$\OM (LT)_{n+1} =\OM(\bigcup \OM^qT_{n+1+q}) \iso \bigcup \OM^{q+1}T_{n+q+1} \iso (LT)_n.$$

We have an evident map of prespectra $\la: T\rtarr LT$, and a comparison of colimits shows
(by a cofinality argument) that $\la$ induces isomorphisms on homotopy and homology groups.
The essential point is that homotopy and homology commute with colimits. It is not true 
that cohomology converts colimits to limits in general, because of ${\lim}^1$ error terms, and
this is one reason that our definition of the cohomology of prespectra via limits is 
inappropriate except under restrictions that guarantee the vanishing of ${\lim}^1$ terms. 
Observe that there is no problem in the case of $\OM$-prespectra, for which $\la$ is a 
spacewise weak equivalence.

For a based space $X$, we define the suspension spectrum\index{suspension spectrum} 
$\SI^{\infty}X$ by applying $L$ to the suspension prespectrum $\SI^{\infty} X =\sset{\SI^nX}$. 
The inclusion condition is satisfied in this case. We define $QX=\cup \OM^q\SI^q X$,\index{QX@$QX$} 
and we find that the $n$th space of $\SI^{\infty} X$ is $Q\SI^n X$. It should be apparent that 
the homotopy groups of the space $QX$ are the stable homotopy groups of $X$.

The adjoint structure maps of the Thom prespectrum $TO$ are also inclusions, and our mythical 
object is $MO=L TO$.\index{MO@$MO$}

In general, for a prespectrum $T$, we can apply an iterated mapping cylinder construction to
define a spacewise equivalent prespectrum $KT$ whose adjoint structure maps are inclusions.
The prespectrum level homotopy, homology, and cohomology groups of $KT$ are isomorphic to 
those of $T$. Thus, if we have a prespectrum $T$ whose invariants we are interested in, such as
an Eilenberg-Mac\,Lane $\OM$-prespectrum or the $K$-theory $\OM$-prespectrum, then we can 
construct a spectrum $LKT$ that has the same invariants. 

For a based space $X$ and $q\geq 0$, we construct a prespectrum $\SI^{\infty}_qX$ whose
$n$th space is a point for $n<q$ and is $\SI^{n-q}X$ for $n\geq q$; its structure maps
for $n\geq q$ are identity maps. We continue to write $\SI^{\infty}_qX$ for the spectrum
obtained by applying $L$ to this prespectrum. We then define sphere spectra $S^q$ for all 
integers $q$ by letting $S^q = \SI^{\infty} S^q$ for $q\geq 0$
and $S^{-q} = \SI^{\infty}_qS^0$ for $q>0$. The definition is appropriate since 
$\SI S^q \iso S^{q+1}$ for all integers $q$. We can now define homotopy groups in the
obvious way. For example, the homotopy groups of the $K$-theory spectrum are $\bZ$ for every
even integer and zero for every odd integer. 

From here, we can go on to define CW spectra in very much the same way that we defined
CW complexes, and we can fill in the rest of the outline in the previous section. The real
work involves the smash product of spectra, but this does not belong in our rapid course.
While there is a good deal of foundational work involved, there is also considerable payoff 
in explicit concrete calculations, as the computation of $\pi_*(MO)$ well illustrates. 

With the hope that this glimpse into the world of stable homotopy theory has whetted the 
reader's appetite for more, we will end at this starting point.

\clearpage

\thispagestyle{empty}

\chapter*{Suggestions for further reading}

\setcounter{section}{0}

Rather than attempt a complete bibliography, I will give a number of basic references.
I will begin with historical references and textbooks. I will then give references for 
specific topics, more or less in the order in which topics appear in the text. Where 
material has been collected in one or another book, I have often referred to such books 
rather than to original articles. However, the importance and quality of exposition of 
some of the original sources often make them still to be preferred today. The subject in 
its earlier days was blessed with some of the finest expositors of mathematics, for example 
Steenrod, Serre, Milnor, and Adams. Some of the references are intended to give historical 
perspective, some are classical papers in the subject, some are follow-ups to material in 
the text, and some give an idea of the current state of the subject. In fact, 
many major parts of algebraic topology are nowhere mentioned in any of the existing 
textbooks, although several were well established by the mid-1970s. I will indicate 
particularly accessible references for some of them; the reader can find more of the
original references in the sources given.

\section{A classic book and historical references}

The axioms for homology and cohomology theories were set out in the classic:

\vspace{1mm}

\noindent
{\em S. Eilenberg and N. Steenrod. Foundations of algebraic topology.}
Princeton University Press. 1952.

\vspace{1.3mm}

I believe the only historical monograph on the subject is:

\vspace{1mm}

\noindent
{\em J. Dieudonn\'e. A history of algebraic and differential topology, 1900--1960.}
Birk\-h\"auser. 1989.

\vspace{1.3mm}

A large collection of historical essays will appear soon:

\vspace{1mm}

\noindent
{\em I.M. James, editor. The history of topology.} Elsevier Science. To appear.

\vspace{1.3mm}

Among the contributions, I will advertise one of my own, available on the web:

\vspace{1mm}

\noindent
{\em J.P. May. Stable algebraic topology, 1945--1966.} http://hopf.math.purdue.edu

\section{Textbooks in algebraic topology and homotopy theory}

These are ordered roughly chronologically (although this is obscured by the fact that 
the most recent editions or versions are cited). I have included only those texts that 
I have looked at myself, that are at least at the level of the more elementary chapters 
here, and that offer significant individuality of treatment. There are many other textbooks in 
algebraic topology.

\vspace{1mm}

Two classic early textbooks:

\vspace{1.3mm}

\noindent
{\em P.J. Hilton and S. Wylie. Homology theory.} Cambridge University Press. 1960.

\vspace{1mm}

\noindent
{\em E. Spanier. Algebraic topology.} McGraw-Hill. 1966.

\vspace{1.3mm}

An idiosyncratic pre-homology level book giving much material about groupoids:

\vspace{1mm}

\noindent
{\em R. Brown. Topology. A geometric account of general topology, homotopy types,
and the fundamental groupoid.} Second edition. Ellis Horwood. 1988.

\vspace{1.3mm}

A homotopical introduction close to the spirit of this book:

\vspace{1mm}

\noindent
{\em B. Gray. Homotopy theory, an introduction to algebraic topology.} Academic Press. 1975.

\vspace{1.3mm}

The standard current textbooks in basic algebraic topology:

\vspace{1mm}

\noindent
{\em M.J. Greenberg and J. R. Harper. Algebraic topology, a first course.} 
Benjamin/\linebreak
Cummings. 1981.

\vspace{1mm}

\noindent
{\em W.S. Massey. A basic course in algebraic topology.} Springer-Verlag. 1991.

\vspace{1mm}

\noindent
{\em A. Dold. Lectures on algebraic topology.} Reprint of the 1972 edition. 
Springer-Verlag. 1995.

\vspace{1mm}

\noindent
{\em J.W. Vick. Homology theory; an introduction to algebraic topology.}
Second edition. Springer-Verlag. 1994.

\vspace{1mm}

\noindent
{\em J.R. Munkres. Elements of algebraic topology.} Addison Wesley. 1984.

\vspace{1mm}

\noindent
{\em J.J. Rotman. An introduction to algebraic topology.} Springer-Verlag. 1986.

\vspace{1mm}

\noindent
{\em G.E. Bredon. Topology and geometry.} Springer-Verlag. 1993.

\vspace{1.3mm}

Sadly, the following are still the only more advanced textbooks in the subject:

\vspace{1mm}

\noindent
{\em R.M. Switzer. Algebraic topology. Homotopy and homology.} Springer-Verlag. 1975.

\vspace{1mm}

\noindent
{\em G.\!W. Whitehead. Elements of homotopy theory.} Springer-Verlag. 1978.

\section{Books on CW complexes}

Two books giving more detailed studies of CW complexes than are found in textbooks
(the second giving a little of the theory of compactly generated spaces):

\vspace{1mm}

\noindent
{\em A.T. Lundell and S. Weingram The topology of CW complexes.} 
Van Nostrand Reinhold. 1969.

\vspace{1mm}

\noindent
{\em R. Fritsch and R.A. Piccinini. Cellular structures in topology.} 
Cambridge University Press. 1990.

\section{Differential forms and Morse theory}

Two introductions to algebraic topology starting from de Rham cohomology:

\vspace{1mm}

\noindent
{\em R. Bott and L.\!W. Tu. Differential forms in algebraic topology.} Springer-Verlag. 1982.

\vspace{1mm}

\noindent
{\em I. Madsen and J. Tornehave. From calculus to cohomology. de Rham cohomology and
characteristic classes.} Cambridge University Press. 1997. 

\vspace{1.3mm} 

The classic reference on Morse theory, with an exposition of the Bott periodicity theorem:

\vspace{1mm}

\noindent
{\em J. Milnor. Morse theory.} Annals of Math. Studies No. 51. Princeton University Press. 1963.

\vspace{1.3mm}

A modern use of Morse theory for the analytic construction of homology:

\vspace{1mm}

\noindent
{\em M. Schwarz. Morse homology.} Progress in Math. Vol. 111. Birkh\"auser. 1993.

%R. Bott. An application of the Morse theory to the topology of Lie-groups.
%Bull. Soc. Math. France 84(1956), 251-281.
%R. Bott. The stable homotopy of the classical groups. Annals of Math. 70(1959), 313-337.
%R. Bott. Quelques remarques sur les th\'eor\`emes de periodicit\'e de topology. Bull. Soc.
%Math. France 87(1959), 293-310. 
%M.F. Atiyah and R. Bott. On the periodicity theorem for complex vector bundles.
%Acta Math. 112(1964), 229-247.

\section{Equivariant algebraic topology}

Two good basic references on equivariant algebraic topology, classically 
called the theory of transformation groups (see also \S\S16, 21 below):

\vspace{1mm}

\noindent
{\em G. Bredon. Introduction to compact transformation groups.} Academic Press. 1972.

\vspace{1mm}

\noindent
{\em T. tom Dieck. Transformation groups.} Walter de Gruyter. 1987.

\vspace{1.3mm}

A more advanced book, a precursor to much recent work in the area:

\vspace{1mm}

\noindent
{\em T. tom Dieck. Transformation groups and representation theory.}
Lecture Notes in Mathematics Vol. 766. Springer-Verlag. 1979.

\section{Category theory and homological algebra}

A revision of the following classic on basic category theory is in preparation:

\vspace{1mm}

\noindent
{\em S. Mac\,Lane. Categories for the working mathematician.} Springer-Verlag. 1971.

\vspace{1.3mm}

Two classical treatments and a good modern treatment of homological algebra:

\vspace{1mm}

\noindent
{\em H. Cartan and S. Eilenberg. Homological algebra.} Princeton University Press. 1956.

\vspace{1mm}

\noindent
{\em S. MacLane. Homology.} Springer-Verlag. 1963.

\vspace{1mm}

\noindent
{\em C.A. Weibel. An introduction to homological algebra.} Cambridge University Press. 1994.

\section{Simplicial sets in algebraic topology} 

Two older treatments and a comprehensive modern treatment:

\vspace{1mm}

\noindent
{\em P. Gabriel and M. Zisman. Calculus of fractions and homotopy theory.} Springer-Verlag. 1967.

\vspace{1mm}

\noindent
{\em J.P. May. Simplicial objects in algebraic topology.}  D. Van Nostrand 1967; 
reprinted by the University of Chicago Press 1982 and 1992.

\vspace{1mm}

\noindent
{\em P.G. Goerss and J.F. Jardine. Simplicial homotopy theory.} Birkh\"auser. To appear.

\section{The Serre spectral sequence and Serre class theory}

Two classic papers of Serre:

\vspace{1mm}

\noindent
{\em J.-P. Serre. Homologie singuli\'ere des espaces fibr\'es. Applications.} Annals
of Math. (2)54(1951), 425--505.

\vspace{1mm}

\noindent
{\em J.-P. Serre. Groupes d'homotopie et classes de groupes ab\'eliens.} Annals of 
Math. (2)58(1953), 198--232.

\vspace{1.3mm}

A nice exposition of some basic homotopy theory and of Serre's work:

\vspace{1mm}

\noindent
{\em S.-T. Hu. Homotopy theory.} Academic Press. 1959.

\vspace{1.3mm}

Many of the textbooks cited in \S2 also treat the Serre spectral sequence.

\section{The Eilenberg-Moore spectral sequence}

There are other important spectral sequences in the context of fibrations, 
mainly due to Eilenberg and Moore. Three references:

\vspace{1mm}

\noindent
{\em S. Eilenberg and J.C. Moore. Homology and fibrations, I.} Comm. Math. Helv.
40(1966), 199--236.

\vspace{1mm}

\noindent
{\em L. Smith. Homological algebra and the Eilenberg-Moore spectral sequences.}
Trans. Amer.  Math.  Soc. 129(1967), 58--93. 

\vspace{1mm}

\noindent
{\em V.K.A.M. Gugenheim and J.P. May. On the theory and applications of differential
torsion products.} Memoirs Amer. Math. Soc. No. 142. 1974.

\vspace{1.3mm}

There is a useful guidebook to spectral sequences:

\vspace{1mm}

\noindent
{\em J. McCleary. User's guide to spectral sequences.} Publish or Perish. 1985.

\section{Cohomology operations}

A compendium of the work of Steenrod and others on the construction and analysis 
of the Steenrod operations:

\vspace{1mm}

\noindent
{\em N.E. Steenrod and D.B.A. Epstein. Cohomology operations.} Annals of Math. Studies No. 50.
Princeton University Press. 1962.

\vspace{1.3mm}

A classic paper that first formalized cohomology operations, among other things:

\vspace{1mm}

\noindent
{\em J.-P. Serre. Cohomologie modulo $2$ des complexes d'Eilenberg-Mac\,Lane.} 
Comm. Math. Helv. 27(1953), 198--232.

\vspace{1.3mm}

A general treatment of Steenrod-like operations:

\vspace{1mm}

\noindent
{\em J.P. May. A general algebraic approach to Steenrod operations.} In Lecture Notes 
in Mathematics Vol. 168, 153--231. Springer-Verlag. 1970.

\vspace{1.3mm}

A nice book on mod $2$ Steenrod operations and the Adams spectral sequence:

\vspace{1mm}

\noindent
{\em R. Mosher and M. Tangora. Cohomology operations and applications in homotopy theory.}
Harper and Row. 1968.

\section{Vector bundles}

A classic and a more recent standard treatment that includes $K$-theory:

\vspace{1mm}

\noindent
{\em N.E. Steenrod. Topology of fibre bundles.} Princeton University Press. 
1951. Fifth printing, 1965.

\vspace{1mm}

\noindent
{\em D. Husemoller. Fibre bundles.} Springer-Verlag. 1966. Third edition, 1994.

\vspace{1.3mm}

A general treatment of classification theorems for bundles and fibrations:

\vspace{1mm}

\noindent
{\em J.P. May. Classifying spaces and fibrations.} Memoirs Amer. Math. Soc. No. 155. 1975.

\section{Characteristic classes}

The classic introduction to characteristic classes:

\vspace{1mm}

\noindent
{\em J. Milnor and J.D. Stasheff. Characteristic classes.} Annals of Math. Studies No. 76. 
Princeton University Press. 1974.

\vspace{1.3mm}

A good reference for the basic calculations of characteristic classes: 

\vspace{1mm}

\noindent
{\em A. Borel. Topology of Lie groups and characteristic classes.} Bull. Amer. Math. Soc.
61(1955), 297--432. 

\vspace{1.3mm}

Two proofs of the Bott periodicity theorem that only use standard techniques of algebraic 
topology, starting from characteristic class calculations:

\vspace{1mm}

\noindent
{\em H. Cartan et al. P\'eriodicit\'e des groupes d'homotopie stables des groupes 
classiques, d'apr\`es Bott.} S\'eminaire Henri Cartan, 1959/60. Ecole Normale Sup\'erieure. Paris.

\vspace{1mm}

\noindent
{\em E. Dyer and R.K. Lashof. A topological proof of the Bott periodicity theorems.}
Ann. Mat. Pure Appl. (4)54(1961), 231--254.

\section{$K$-theory}

%M.F. Atiyah and F. Hirzebruch. Vector bundles and homogeneous spaces, in Differential 
%Geometry. Amer. Math. Soc. Proc. Symp. Pure Math 3(1961), 7--38.

Two classical lecture notes on $K$-theory: 

\vspace{1mm}

\noindent
{\em R. Bott. Lectures on $K(X)$.} W.A. Benjamin. 1969.

\vspace{1mm}

This includes a reprint of perhaps the most accessible proof of the complex
case of the Bott periodicity theorem, namely:

\vspace{1mm}

\noindent
{\em M.F. Atiyah and R. Bott. On the periodicity theorem for complex vector bundles.}
Acta Math. 112(1994), 229--247.

\vspace{1.3mm}

\noindent
{\em M.F. Atiyah. $K$-theory.} Notes by D.W. Anderson. Second Edition.
Addison-Wesley. 1967.

\vspace{1mm}

This includes reprints of two classic papers of Atiyah, one that relates Adams 
operations in $K$-theory to Steenrod operations in cohomology and another that 
sheds insight on the relationship between real and complex $K$-theory:

\vspace{1mm}

\noindent
{\em M.F. Atiyah. Power operations in $K$-theory.} Quart. J. Math. (Oxford) (2)17(1966),
165--193.

\vspace{1mm}

\noindent
{\em M.F. Atiyah. $K$-theory and reality.} Quart. J. Math. (Oxford) (2)17(1966), 367--386. 

\vspace{1.3mm}

Another classic paper that greatly illuminates real $K$-theory:

\vspace{1mm}

\noindent
{\em M.F. Atiyah, R. Bott, and A. Shapiro. Clifford algebras.} Topology 
3(1964), suppl. 1, 3--38.

\vspace{1.3mm}

A more recent book on $K$-theory:

\noindent
{\em M. Karoubi. $K$-theory.} Springer-Verlag. 1978.

\vspace{1.3mm}

Some basic papers of Adams and Adams and Atiyah giving applications of $K$-theory:

\vspace{1mm}

\noindent
{\em J.F. Adams. Vector fields on spheres.} Annals of Math. 75(1962), 603--632.

\vspace{1mm}

\noindent
{\em J.F. Adams. On the groups $J(X)$ I, II, III, and IV.} Topology 2(1963), 181--195;
3(1965), 137-171 and 193--222; 5(1966), 21--71. 

\vspace{1mm}

\noindent
{\em J.F. Adams and M.F. Atiyah. $K$-theory and the Hopf invariant.} Quart. J. Math. (Oxford)
(2)17(1966), 31--38.

\section{Hopf algebras; the Steenrod algebra, Adams spectral sequence}

The basic source for the structure theory of (connected) Hopf algebras:

\vspace{1mm}

\noindent
{\em J. Milnor and J.C. Moore. On the structure of Hopf algebras.} Annals of Math. 81(1965),
211--264. 

\vspace{1.3mm}

The classic analysis of the structure of the Steenrod algebra as a Hopf algebra:

\vspace{1mm}

\noindent
{\em J. Milnor. The Steenrod algebra and its dual.} Annals of Math. 67(1958), 150--171.

\vspace{1.3mm}

Two classic papers of Adams; the first constructs the Adams spectral sequence 
relating the Steenrod algebra to stable homotopy groups and the second uses
secondary cohomology operations to solve the Hopf invariant one problem:

\vspace{1mm}

\noindent
{\em J.F. Adams. On the structure and applications of the Steenrod algebra.}
Comm. Math. Helv. 32(1958), 180--214.

\vspace{1mm}

\noindent
{\em J.F. Adams. On the non-existence of elements of Hopf invariant one.} 
Annals of Math. 72(1960), 20--104. 

\section{Cobordism}

The beautiful classic paper of Thom is still highly recommended:

\vspace{1mm}

\noindent
{\em R. Thom. Quelques propri\'et\'es globals des vari\'et\'es diff\'erentiables.} 
Comm. Math. Helv. 28(1954), 17--86.

\vspace{1.3mm}

Thom computed unoriented cobordism. Oriented and complex cobordism
came later. In simplest form, the calculations use the Adams spectral 
sequence:

\vspace{1mm}

\noindent
{\em J. Milnor. On the cobordism ring $\Omega^*$ and a complex analogue.} Amer. J. 
Math. 82(1960), 505--521. 

\vspace{1mm}

\noindent
{\em C.T.C. Wall. A characterization of simple modules over the Steenrod algebra 
mod $2$.} Topology 1(1962), 249--254. 

\vspace{1mm}

\noindent
{\em A. Liulevicius. A proof of Thom's theorem.} Comm. Math. Helv. 37(1962), 121--131. 

\vspace{1mm}

\noindent
{\em A. Liulevicius. Notes on homotopy of Thom spectra.} Amer. J. Math. 86(1964), 1--16. 

\vspace{1.3mm}

A very useful compendium of calculations of cobordism groups:

\vspace{1mm}

\noindent
{\em R. Stong. Notes on cobordism theory.} Princeton University Press. 1968. 

\section{Generalized homology theory and stable homotopy theory}

Two classical references, the second of which also gives detailed information about complex
cobordism that is of fundamental importance to the subject.

\vspace{1mm}

\noindent
{\em G.W. Whitehead. Generalized homology theories.} Trans. Amer. Math. Soc. 102(1962), 227--283.

\vspace{1mm}

\noindent
{\em J.F. Adams. Stable homotopy and generalised homology.} Chicago Lectures in Mathematics.
University of Chicago Press. 1974. Reprinted in 1995.

\vspace{1.3mm} 

An often overlooked but interesting book on the subject: 

\vspace{1mm}

\noindent
{\em H.R. Margolis. Spectra and the Steenrod algebra. Modules over the Steenrod
algebra and the stable homotopy category.} North-Holland. 1983.

\vspace{1.3mm}

Foundations for equivariant stable homotopy theory are established in:

\vspace{1mm}

\noindent
{\em L.G. Lewis, Jr., J.P. May, and M.Steinberger (with contributions by 
J.E. McClure). Equivariant stable homotopy theory.}  Lecture Notes in
Mathematics Vol. 1213. Springer-Verlag. 1986.

\section{Quillen model categories}

In the introduction, I alluded to axiomatic treatments of ``homotopy theory.''
Here are the original and two more recent references:

\vspace{1mm}

\noindent
{\em D.G. Quillen. Homotopical algebra.} Lecture Notes in Mathematics
Vol. 43. Springer-Verlag. 1967. 

\vspace{1mm}

\noindent
{\em W.G. Dwyer and J. Spalinski. Homotopy theories and model categories}.
In A handbook of algebraic topology, edited by I.M. James. North-Holland. 1995.

\vspace{1.3mm}

The cited ``{\em Handbook}'' (over 1300 pages) contains an uneven but very interesting 
collection of expository articles on a wide variety of topics in algebraic topology. 

\vspace{1.3mm}

\noindent
{\em M. Hovey. Model categories.} Amer. Math. Soc. Surveys and Monographs No. 63. 1998.

\section{Localization and completion; rational homotopy theory}

Since the early 1970s, it has been standard practice in algebraic topology to localize 
and complete topological spaces, and not just their algebraic invariants, at sets of primes
and then to study the subject one prime at a time, or rationally. Two of the basic original
references are:

\vspace{1mm}

\noindent
{\em D. Sullivan. The genetics of homotopy theory and the Adams conjecture.}
Annals of Math. 100(1974), 1--79.

\vspace{1mm}

\noindent
{\em A.K. Bousfield and D.M. Kan. Homotopy limits, completions, and localizations.}
Lecture Notes in Mathematics Vol. 304. Springer-Verlag. 1972.

\vspace{1.3mm}

A more accessible introduction to localization and a readable recent 
paper on completion are:

\vspace{1mm}

\noindent
{\em P. Hilton, G. Mislin, and J. Roitberg. Localization of nilpotent groups
and spaces.} North-Holland. 1975.

\vspace{1mm}

\noindent
{\em F. Morel. Quelques remarques sur la cohomologie modulo $p$ continue des 
pro-$p$-espaces et les resultats de J. Lannes concernent les espaces fonctionnel 
Hom$(BV,X)$.} Ann. Sci. Ecole Norm. Sup. (4)26(1993), 309--360. 

\vspace{1.3mm}

When spaces are rationalized, there is a completely algebraic description of the
result. The main original reference and a more accessible source are:

\vspace{1mm}

\noindent
{\em D. Sullivan. Infinitesimal computations in topology.} Publ. Math. 
IHES 47(1978), 269--332.

\vspace{1mm}

\noindent
{\em A.K. Bousfield and V.K.A.M. Gugenheim. On PL de Rham theory and rational homotopy
type.} Memoirs Amer. Math. Soc. No. 179. 1976.

\section{Infinite loop space theory}

Another area well established by the mid-1970s. The following book is a 
delightful read, with capsule introductions of many topics other than infinite 
loop space theory, a very pleasant starting place for learning modern 
algebraic topology:

\vspace{1mm}

\noindent
{\em J.F. Adams. Infinite loop spaces.} Annals of Math. Studies No. 90. Princeton
University Press. 1978.

\vspace{1.3mm}

The following survey article is less easy going, but gives an indication of
the applications to high dimensional geometric topology and to algebraic $K$-theory:

\vspace{1mm}

\noindent
{\em J.P. May. Infinite loop space theory.} Bull. Amer. Math. Soc. 83(1977), 
456--494.

\vspace{1.3mm}

Five monographs, each containing a good deal of expository material, 
that give a variety of theoretical and calculational developments and applications 
in this area:

\vspace{1mm}

\noindent
{\em J.P. May. The geometry of iterated loop spaces.}  Lecture Notes in Mathematics 
Vol. 271.  Springer-Verlag. 1972.

\vspace{1mm}

\noindent
{\em J.M. Boardman and R.M. Vogt. Homotopy invariant algebraic structures on topological
spaces.} Lecture Notes in Mathematics Vol. 347. Springer-Verlag. 1973.

\vspace{1mm}

\noindent
{\em F.R. Cohen, T.J. Lada, and J.P. May.  The homology of iterated loop spaces.}
Lecture Notes in Mathematics Vol. 533. Springer-Verlag. 1976.

\vspace{1mm}

\noindent
{\em J.P. May (with contributions by F. Quinn, N. Ray, and J. Tornehave). 
$E_{\infty}$ ring spaces and $E_{\infty}$ ring spectra.}  Lecture Notes in 
Mathematics Vol. 577.  Springer-Verlag. 1977.

\vspace{1mm}

\noindent
{\em R. Bruner, J.P. May, J.E. McClure, and M. Steinberger. $H_{\infty}$ ring    
spectra and their applications.}  Lecture Notes in Mathematics  Vol. 1176.  
Springer-Verlag. 1986.

\section{Complex cobordism and stable homotopy theory}

Adams' book cited in \S16 gives a spectral sequence for the computation of stable
homotopy groups in terms of generalized cohomology theories. Starting from complex cobordism
and related theories, its use has been central to two waves of major developments in stable 
homotopy theory.  A good exposition for the first wave:

\vspace{1mm}

\noindent
{\em D.C. Ravenel. Complex cobordism and stable homotopy groups of spheres.} Academic 
Press. 1986.

\vspace{1.3mm}

The essential original paper and a very nice survey article on the second wave:

\vspace{1mm}

\noindent
{\em E. Devinatz, M.J. Hopkins, and J.H. Smith. Nilpotence and stable homotopy theory.}
Annals of Math. 128(1988), 207--242.

\vspace{1mm}

\noindent
{\em M.J. Hopkins. Global methods in homotopy theory.} In Proceedings of the 1985 LMS 
Symposium on homotopy theory, edited by J.D.S. Jones and E. Rees. 
London Mathematical Society. 1987.

\vspace{1.3mm}

The cited {\em Proceedings} contain good introductory survey articles on several other
topics in algebraic topology. A larger scale exposition of the second wave is:

\vspace{1mm}

\noindent
{\em D.C. Ravenel. Nilpotence and periodicity in stable homotopy theory.}
Annals of Math. Studies No. 128. Princeton University Press. 1992.

\section{Follow-ups to this book}

There is a leap from the level of this introductory book to that of the most 
recent work in the subject. One recent book that helps fill the gap is:

\vspace{1mm}

{\em P. Selick. Introduction to homotopy theory.} Fields Institute Monographs No. 9. 
American Mathematical Society. 1997.

\vspace{1.3mm}

There is a recent expository book for the reader who would like to jump 
right in and see the current state of algebraic topology; although it focuses on 
equivariant theory, it contains introductions and discussions of many non-equivariant 
topics:

\vspace{1mm}

\noindent
{\em J.P. May et al. Equivariant homotopy and cohomology theory.} 
NSF-CBMS Regional Conference Monograph. 1996.

\vspace{1.3mm}

For the reader of the last section of this book whose appetite has been whetted for 
more stable homotopy theory, there is an expository article that motivates and explains 
the properties that a satisfactory category of spectra should have:

\vspace{1mm}

\noindent
{\em J.P. May. Stable algebraic topology and stable topological algebra.}
Bulletin London Math. Soc. 30(1998), 225--234.

\vspace{1.3mm}

The following monograph gives such a category, with many applications; more
readable accounts appear in the {\em Handbook} cited in \S17 and in the book just 
cited:

\vspace{1mm}

\noindent
{\em A. Elmendorf, I. Kriz, M.A. Mandell, and J.P. May, with an appendix by M. Cole.
Rings, modules, and algebras in stable homotopy theory.} Amer. Math. Soc. Surveys and 
Monographs No. 47. 1997.

%\input{ConciseRevised.ind}

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