work on notes
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Floris van Doorn
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1 changed files with 71 additions and 26 deletions
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@ -12,6 +12,10 @@
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\renewcommand{\epsilon}{\varepsilon}
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\newcommand{\tr}{\cdot}
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\renewcommand{\o}{\ensuremath{\circ}}
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\newcommand{\auxl}{\mathsf{auxl}}
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\newcommand{\auxr}{\mathsf{auxr}}
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\newcommand{\gluel}{\mathsf{gluel}}
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\newcommand{\gluer}{\mathsf{gluer}}
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\begin{document}
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@ -43,9 +47,7 @@
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started with. In diagrams, we will denote pointed homotopies by equalities, but we always mean
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pointed homotopies.
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\item The type $A\to B$ of pointed maps from $A$ to $B$ is itself pointed, with as basepoint the
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constant map $0\equiv0_{A,B}:A\to B$ which has as underlying function $\lam{a:A}b_0$. In these
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notes we will not use $0$ for the empty type (since that is not pointed, we will not use the empty
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type).
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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 $f \o 0 \sim 0$.
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\end{rmk}
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\begin{lem}
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@ -65,17 +67,35 @@
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\section{Smash Product}
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\begin{lem}\mbox{}
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\begin{lem}\mbox{}\label{lem:smash-general}
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\begin{itemize}
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\item The smash of $A$ and $B$ is the HIT generated by point constructor
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$({-},{-}):A\to B \to A\smash B$ and two auxilliary points $\auxl,\auxr:A\smash B$ and path
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constructors $\gluel:\prd{a:A}(a,b_0)=\auxl$ and $\gluer:\prd{b:B}(a_0,b)=\auxr$ (the
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constructors are given as unpointed maps). $A\smash B$ is pointed with point $(a_0,b_0)$.
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\item The smash is functorial: if $f:A\pmap A'$ and $g:B\pmap B'$ then
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$f\smash g:A\smash B\pmap A'\smash B'$. We write $A\smash g$ or $f\smash B$ if one of the
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functions is the identity function.
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\item Smash preserves composition, which gives rise to the interchange law:
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$(f' \o f)\smash (g' \o g) \sim f' \smash g' \o f \smash g$
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$i:(f' \o f)\smash (g' \o g) \sim f' \smash g' \o f \smash g$
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\item If $p:f\sim f'$ and $q:g\sim g'$ then $p\smash q:f\smash g\sim f'\smash g'$. This operation
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preserves reflexivities, symmetries and transitivies.
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\item $f\smash0\sim0$ and $0\smash g\sim 0$.
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\item There are homotopies $f\smash0\sim0$ and $0\smash g\sim 0$ such that the following diagrams
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commute for given homotopies $p : f\sim f'$ and $q : g\sim g'$.
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\begin{center}
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\begin{tikzcd}
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f\smash 0 \arrow[rr, equals,"p\smash1"]\arrow[dr,equals] & &
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f'\smash 0\arrow[dl,equals] \\
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& 0 &
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\end{tikzcd}
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\qquad
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\begin{tikzcd}
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0\smash g\arrow[rr, equals,"1\smash q"]\arrow[dr,equals] & &
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0\smash g'\arrow[dl,equals] \\
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& 0 &
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\end{tikzcd}
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\end{center}
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\end{itemize}
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\end{lem}
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@ -87,8 +107,8 @@
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\begin{center}
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\begin{tikzcd}
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(f_2 \o f_1)\smash (g_2 \o 0) \arrow[r, equals]\arrow[dd,equals] &
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(f_2 \smash g_1)\o (f_1 \smash 0)\arrow[d,equals] \\
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& (f_2 \smash g_1)\o 0\arrow[d,equals] \\
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(f_2 \smash g_2)\o (f_1 \smash 0)\arrow[d,equals] \\
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& (f_2 \smash g_2)\o 0\arrow[d,equals] \\
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(f_2 \o f_1)\smash 0 \arrow[r,equals] &
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0
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\end{tikzcd}
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@ -104,8 +124,24 @@
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\end{lem}
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\begin{proof}
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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.
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Proof to do.
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We will only do the case where $g_1\jdeq 0$, i.e. fill the diagram on the left. The other case is
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similar (but slightly easier). First apply induction on the paths that $f_2$, $f_1$ and $g_2$
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respect the basepoint. In this case $g_2\o0$ is definitionally equal to $0$, and the canonical
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proof that $g_2\o 0~0$ is (definitionally) equal to reflexivity. This means that the homotopy
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$(f_2 \o f_1)\smash (g_2 \o 0)~(f_2 \o f_1)\smash 0$ is also equal to reflexivity.
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For the underlying homotopy, take $x : A_1\smash B_1$. We apply induction on $x$.
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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 $b_i$)
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\begin{center}
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\begin{tikzcd}
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(f_2(f_1(a)),g_2(b_2))\arrow[r, equals,"1"]\arrow[dd,equals,"1"] &
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(f_2(f_1(a)),g_2(b_2))\arrow[d,equals] \\
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& (f_2(a_2),g_2(b_2))\o (f_1 \smash g_1)\arrow[d,equals] \\
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(f_2(f_1(a)),y_0) \arrow[r,equals,"1"] &
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(x_0,y_0)
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\end{tikzcd}
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\end{center}
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\end{proof}
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\begin{thm}\label{thm:smash-functor-right}
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@ -138,8 +174,9 @@ which is natural in $A$, $B$ and $C$. (note: $(A\smash C\pmap B\smash C)$ is bot
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\end{tikzcd}
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\end{center}
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Let $k:A\pmap B$. Then as homotopy the naturality in $A$ becomes
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$(k\o f)\smash C=k\smash C\o f\smash C$. To prove an equality between pointed maps, we need to give a pointed homotopy, which is given by interchange. To show that this homotopy
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is pointed, we need to fill the following square, which follows from \autoref{lem:smash-coh}.
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$(k\o f)\smash C=k\smash C\o f\smash C$. To prove an equality between pointed maps, we need to give
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a pointed homotopy, which is given by interchange. To show that this homotopy is pointed, we need to
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fill the following square (after reducing out the applications of function extensinality), which follows from \autoref{lem:smash-coh}.
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\begin{center}
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\begin{tikzcd}
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(0 \o f)\smash C \arrow[r, equals]\arrow[dd,equals] &
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@ -161,20 +198,28 @@ square, which follows from \autoref{lem:smash-coh}.
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0
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\end{tikzcd}
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\end{center}
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The naturality in $C$ is harder. For the underlying homotopy we need to show
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$B\smash h\o g\smash C=g\smash C'\o A\smash h$. This follows by applying interchange twice:
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$$B\smash h\o g\smash C\sim(\idfunc[B]\o g)\smash(h\o\idfunc[C])\sim(g\o\idfunc[A])\smash(\idfunc[C']\o h)\sim g\smash C'\o A\smash h.$$
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The naturality in $C$ is a bit harder. For the underlying homotopy we need to show
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$B\smash h\o k\smash C=k\smash C'\o A\smash h$. This follows by applying interchange twice:
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$$B\smash h\o k\smash C\sim(\idfunc[B]\o k)\smash(h\o\idfunc[C])\sim(k\o\idfunc[A])\smash(\idfunc[C']\o h)\sim k\smash C'\o A\smash h.$$
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To show that this homotopy is pointed, we need to fill the following square:
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% \begin{center}
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% \begin{tikzcd}
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% (B\smash h\o g\smash C) \arrow[r, equals]\arrow[dd,equals] &
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% (g \smash C)\o (0 \smash C)\arrow[d,equals] \\
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% & (g\smash C) \o 0\arrow[d,equals] \\
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% 0\smash C \arrow[r,equals] &
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% 0
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% \end{tikzcd}
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% \end{center}
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\begin{center}
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\begin{tikzcd}
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B\smash h\o 0\smash C \arrow[r, equals]\arrow[d,equals] &
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(\idfunc[B]\o 0)\smash(h\o\idfunc[C]) \arrow[r, equals]\arrow[d,equals] &
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(0\o\idfunc[A])\smash(\idfunc[C']\o h)\arrow[r, equals]\arrow[d,equals] &
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0\smash C'\o A\smash h\arrow[d,equals] \\
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B\smash h\o 0 \arrow[d,equals] &
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0\smash(h\o\idfunc[C]) \arrow[r, equals]\arrow[d,equals] &
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0\smash(\idfunc[C']\o h) \arrow[d,equals] &
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0\o A\smash h\arrow[d,equals] \\
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B\smash h\o 0 \arrow[r, equals] &
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0 \arrow[r, equals] &
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0 \arrow[r, equals] &
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0
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\end{tikzcd}
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\end{center}
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The left and the right squares are filled by \autoref{lem:smash-coh}. The squares in the middle
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are filled by (corollaries of) \autoref{lem:smash-general}.
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\end{proof}
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\section{Adjunction}
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