michael/Spectral
1
0
Fork 0
Code Issues Pull requests Projects Releases Packages Wiki Activity Actions

work on notes

Browse source
This commit is contained in:
Floris van Doorn 2017-03-09 17:34:30 -05:00
parent b781de8473
commit b61fb7e685
1 changed files with 71 additions and 26 deletions
Show stats Download patch file Download diff file Expand all files Collapse all files

97
Notes/smash.tex
View file

@ -12,6 +12,10 @@
\renewcommand{\epsilon}{\varepsilon} \renewcommand{\epsilon}{\varepsilon}
\newcommand{\tr}{\cdot} \newcommand{\tr}{\cdot}
\renewcommand{\o}{\ensuremath{\circ}} \renewcommand{\o}{\ensuremath{\circ}}
\newcommand{\auxl}{\mathsf{auxl}}
\newcommand{\auxr}{\mathsf{auxr}}
\newcommand{\gluel}{\mathsf{gluel}}
\newcommand{\gluer}{\mathsf{gluer}}
\begin{document} \begin{document}
@ -43,9 +47,7 @@
started with. In diagrams, we will denote pointed homotopies by equalities, but we always mean started with. In diagrams, we will denote pointed homotopies by equalities, but we always mean
pointed homotopies. pointed homotopies.
\item The type $A\to B$ of pointed maps from $A$ to $B$ is itself pointed, with as basepoint the \item The type $A\to B$ of pointed maps from $A$ to $B$ is itself pointed, with as basepoint the
constant map $0\equiv0_{A,B}:A\to B$ which has as underlying function $\lam{a:A}b_0$. In these constant map $0\equiv0_{A,B}:A\to B$ which has as underlying function $\lam{a:A}b_0$. We have $0\o g \sim 0$ and $f \o 0 \sim 0$.
notes we will not use $0$ for the empty type (since that is not pointed, we will not use the empty
type).
\end{rmk} \end{rmk}
\begin{lem} \begin{lem}
@ -65,17 +67,35 @@
\section{Smash Product} \section{Smash Product}
\begin{lem}\mbox{}\label{lem:smash-general}
\begin{lem}\mbox{}
\begin{itemize} \begin{itemize}
\item The smash of $A$ and $B$ is the HIT generated by point constructor
$({-},{-}):A\to B \to A\smash B$ and two auxilliary points $\auxl,\auxr:A\smash B$ and path
constructors $\gluel:\prd{a:A}(a,b_0)=\auxl$ and $\gluer:\prd{b:B}(a_0,b)=\auxr$ (the
constructors are given as unpointed maps). $A\smash B$ is pointed with point $(a_0,b_0)$.
\item The smash is functorial: if $f:A\pmap A'$ and $g:B\pmap B'$ then \item The smash is functorial: if $f:A\pmap A'$ and $g:B\pmap B'$ then
$f\smash g:A\smash B\pmap A'\smash B'$. We write $A\smash g$ or $f\smash B$ if one of the $f\smash g:A\smash B\pmap A'\smash B'$. We write $A\smash g$ or $f\smash B$ if one of the
functions is the identity function. functions is the identity function.
\item Smash preserves composition, which gives rise to the interchange law: \item Smash preserves composition, which gives rise to the interchange law:
$(f' \o f)\smash (g' \o g) \sim f' \smash g' \o f \smash g$ $i:(f' \o f)\smash (g' \o g) \sim f' \smash g' \o f \smash g$
\item If $p:f\sim f'$ and $q:g\sim g'$ then $p\smash q:f\smash g\sim f'\smash g'$. This operation \item If $p:f\sim f'$ and $q:g\sim g'$ then $p\smash q:f\smash g\sim f'\smash g'$. This operation
preserves reflexivities, symmetries and transitivies. preserves reflexivities, symmetries and transitivies.
\item $f\smash0\sim0$ and $0\smash g\sim 0$. \item There are homotopies $f\smash0\sim0$ and $0\smash g\sim 0$ such that the following diagrams
commute for given homotopies $p : f\sim f'$ and $q : g\sim g'$.
\begin{center}
\begin{tikzcd}
f\smash 0 \arrow[rr, equals,"p\smash1"]\arrow[dr,equals] & &
f'\smash 0\arrow[dl,equals] \\
& 0 &
\end{tikzcd}
\qquad
\begin{tikzcd}
0\smash g\arrow[rr, equals,"1\smash q"]\arrow[dr,equals] & &
0\smash g'\arrow[dl,equals] \\
& 0 &
\end{tikzcd}
\end{center}
\end{itemize} \end{itemize}
\end{lem} \end{lem}
@ -87,8 +107,8 @@
\begin{center} \begin{center}
\begin{tikzcd} \begin{tikzcd}
(f_2 \o f_1)\smash (g_2 \o 0) \arrow[r, equals]\arrow[dd,equals] & (f_2 \o f_1)\smash (g_2 \o 0) \arrow[r, equals]\arrow[dd,equals] &
(f_2 \smash g_1)\o (f_1 \smash 0)\arrow[d,equals] \\ (f_2 \smash g_2)\o (f_1 \smash 0)\arrow[d,equals] \\
& (f_2 \smash g_1)\o 0\arrow[d,equals] \\ & (f_2 \smash g_2)\o 0\arrow[d,equals] \\
(f_2 \o f_1)\smash 0 \arrow[r,equals] & (f_2 \o f_1)\smash 0 \arrow[r,equals] &
0 0
\end{tikzcd} \end{tikzcd}
@ -104,8 +124,24 @@
\end{lem} \end{lem}
\begin{proof} \begin{proof}
We will only do the case where $g_1\jdeq 0$, the other case is similar (but slightly easier). In this case, we need to show that the following diagram commutes. We will only do the case where $g_1\jdeq 0$, i.e. fill the diagram on the left. The other case is
Proof to do. similar (but slightly easier). First apply induction on the paths that $f_2$, $f_1$ and $g_2$
respect the basepoint. In this case $g_2\o0$ is definitionally equal to $0$, and the canonical
proof that $g_2\o 0~0$ is (definitionally) equal to reflexivity. This means that the homotopy
$(f_2 \o f_1)\smash (g_2 \o 0)~(f_2 \o f_1)\smash 0$ is also equal to reflexivity.
For the underlying homotopy, take $x : A_1\smash B_1$. We apply induction on $x$.
Suppose $x\equiv(a,b)$ for $a:A_1$ and $b:B_1$. Then we have to fill the square (denote the basepoints of $A_i$ and $B_i$ by $a_i$ and $b_i$)
\begin{center}
\begin{tikzcd}
(f_2(f_1(a)),g_2(b_2))\arrow[r, equals,"1"]\arrow[dd,equals,"1"] &
(f_2(f_1(a)),g_2(b_2))\arrow[d,equals] \\
& (f_2(a_2),g_2(b_2))\o (f_1 \smash g_1)\arrow[d,equals] \\
(f_2(f_1(a)),y_0) \arrow[r,equals,"1"] &
(x_0,y_0)
\end{tikzcd}
\end{center}
\end{proof} \end{proof}
\begin{thm}\label{thm:smash-functor-right} \begin{thm}\label{thm:smash-functor-right}
@ -138,8 +174,9 @@ which is natural in $A$, $B$ and $C$. (note: $(A\smash C\pmap B\smash C)$ is bot
\end{tikzcd} \end{tikzcd}
\end{center} \end{center}
Let $k:A\pmap B$. Then as homotopy the naturality in $A$ becomes Let $k:A\pmap B$. Then as homotopy the naturality in $A$ becomes
$(k\o f)\smash C=k\smash C\o f\smash C$. To prove an equality between pointed maps, we need to give a pointed homotopy, which is given by interchange. To show that this homotopy $(k\o f)\smash C=k\smash C\o f\smash C$. To prove an equality between pointed maps, we need to give
is pointed, we need to fill the following square, which follows from \autoref{lem:smash-coh}. a pointed homotopy, which is given by interchange. To show that this homotopy is pointed, we need to
fill the following square (after reducing out the applications of function extensinality), which follows from \autoref{lem:smash-coh}.
\begin{center} \begin{center}
\begin{tikzcd} \begin{tikzcd}
(0 \o f)\smash C \arrow[r, equals]\arrow[dd,equals] & (0 \o f)\smash C \arrow[r, equals]\arrow[dd,equals] &
@ -161,20 +198,28 @@ square, which follows from \autoref{lem:smash-coh}.
0 0
\end{tikzcd} \end{tikzcd}
\end{center} \end{center}
The naturality in $C$ is harder. For the underlying homotopy we need to show The naturality in $C$ is a bit harder. For the underlying homotopy we need to show
$B\smash h\o g\smash C=g\smash C'\o A\smash h$. This follows by applying interchange twice: $B\smash h\o k\smash C=k\smash C'\o A\smash h$. This follows by applying interchange twice:
$$B\smash h\o g\smash C\sim(\idfunc[B]\o g)\smash(h\o\idfunc[C])\sim(g\o\idfunc[A])\smash(\idfunc[C']\o h)\sim g\smash C'\o A\smash h.$$ $$B\smash h\o k\smash C\sim(\idfunc[B]\o k)\smash(h\o\idfunc[C])\sim(k\o\idfunc[A])\smash(\idfunc[C']\o h)\sim k\smash C'\o A\smash h.$$
To show that this homotopy is pointed, we need to fill the following square: To show that this homotopy is pointed, we need to fill the following square:
% \begin{center} \begin{center}
% \begin{tikzcd} \begin{tikzcd}
% (B\smash h\o g\smash C) \arrow[r, equals]\arrow[dd,equals] & B\smash h\o 0\smash C \arrow[r, equals]\arrow[d,equals] &
% (g \smash C)\o (0 \smash C)\arrow[d,equals] \\ (\idfunc[B]\o 0)\smash(h\o\idfunc[C]) \arrow[r, equals]\arrow[d,equals] &
% & (g\smash C) \o 0\arrow[d,equals] \\ (0\o\idfunc[A])\smash(\idfunc[C']\o h)\arrow[r, equals]\arrow[d,equals] &
% 0\smash C \arrow[r,equals] & 0\smash C'\o A\smash h\arrow[d,equals] \\
% 0 B\smash h\o 0 \arrow[d,equals] &
% \end{tikzcd} 0\smash(h\o\idfunc[C]) \arrow[r, equals]\arrow[d,equals] &
% \end{center} 0\smash(\idfunc[C']\o h) \arrow[d,equals] &
0\o A\smash h\arrow[d,equals] \\
B\smash h\o 0 \arrow[r, equals] &
0 \arrow[r, equals] &
0 \arrow[r, equals] &
0
\end{tikzcd}
\end{center}
The left and the right squares are filled by \autoref{lem:smash-coh}. The squares in the middle
are filled by (corollaries of) \autoref{lem:smash-general}.
\end{proof} \end{proof}
\section{Adjunction} \section{Adjunction}