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Make the proof that reflectors commute with pushouts a bit more rigorous

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commit 7ae4ffe0324a1ffcd49477e238840b92ca487708 1 parent 112a50d
Guillaume Brunerie authored
Showing with 320 additions and 83 deletions.
  1. +320 −83 homotopy.tex
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403 homotopy.tex
@@ -4,8 +4,6 @@
\newcommand{\typele}[1]{\ensuremath{\type_{\le #1}}\xspace}
% Natural numbers
\newcommand{\N}{\mathbb{N}}
-% Type of truncation levels (integers starting at -2)
-\newcommand{\Nt}{\N_{-2}}
% Function definition with domain and codomain
\newcommand{\function}[4]{\left\{\begin{array}{rcl}#1 &
\longrightarrow & #2 \\ #3 & \longmapsto & #4 \end{array}\right.}
@@ -36,7 +34,7 @@
\newcommand{\ext}{\mathsf{ext}}
% Projection and extension for truncations
\newcommand{\tproj}{\mathsf{proj}}
-\newcommand{\extendsmb}{\mathsf{extend}}
+\newcommand{\extendsmb}{\mathsf{ext}}
\newcommand{\extend}[1]{\extendsmb(#1)}
% Apply a function to a cocone
\newcommand{\composecocone}[2]{#1\circ#2}
@@ -60,7 +58,7 @@ \subsection{Definition}
Let’s consider \P a subuniverse of \type. More precisely we have a map
$P:\type\to\prop$ and we use the notation $A:\P$ to mean that $A:\type$ and
$P(A)$ holds. For instance \P could be \prop, \set, or more generally
-$\typele{n}$ for any $n\in\Nt$.
+$\typele{n}$ for any $n\ge-2$.
\begin{defn}
A \emph{pushout diagram} in $\P$ is 5-uple $\Ddiag=(A,B,C,f,g)$ with
@@ -391,16 +389,21 @@ \section{Homotopy pullbacks}
\section{Reflective subuniverses of $\type$}
+Reflective subuniverses are particularly nice subuniverses in the sense that
+every type $A:\type$ can be reflected down to a type $\reflect(A)$ in the
+subuniverse which is (in a precise way) the free type generated by $A$ in the
+subuniverse.
+
\begin{defn}
- A subuniverse $\P$ of $\type$ is said to be a \emph{reflective subuniverse} of
- $\type$ if for every $A:\type$ we have a type $\reflect(A):\P$ and a map
+ A subuniverse $\P$ of $\type$ is a \emph{reflective subuniverse of $\type$} if
+ for every $A:\type$ we have a type $\reflect(A):\P$ and a map
$\project_A:A\to\reflect(A)$ satisfying the following universal property:
For every $A:\type$ and $B:\P$, the following map is an equivalence
\[\function{(\reflect(A)\to{}B)}{(A\to{}B)}{f}{f\circ\project_A}\]
- In other words, for every $g:A\to{}B$ in the following diagram there is a
- unique map $\ext(g):\reflect(A)\to{}B$ making the diagram commute.
+ In other words, for every $B:\P$ and $g:A\to{}B$ there is a unique map
+ $\ext(g):\reflect(A)\to{}B$ making the following diagram commute.
\[\uppercurveobject{{ }}\lowercurveobject{{ }}\twocellhead{{ }}
\xymatrix{A \ar^{\project_A}[r] \ar_g[rd] \druppertwocell{=} & \reflect(A)
@@ -408,50 +411,50 @@ \section{Reflective subuniverses of $\type$}
& B}\]
\end{defn}
-% If $\P$ is a reflective subuniverse of $\type$, then we can construct pushouts
-% in $\P$ by reflecting the corresponding pushout in $\type$.
-
-The reflector $\reflect$ is a map $\type\to\P$ and using the universal property
-we should be able to prove that $\reflect$ is $(\infty,1)$-functorial. But as
-before we don’t know how to express this property, so we will only express a few
-bits of it.
+The universal property means that $\reflect$ is left adjoint to the inclusion
+$\iota:\P\to\type$ with $\project_A$ as the unit, because the last $B$ in the
+universal property can be replaced by $\iota(B)$ given that $B$ is in $\P$.
-\begin{defn}
- If $f:A\to{}B$, there is a map $\reflect(f):\reflect(A)\to\reflect(B)$ and
- this operation satisfy the following functoriality conditions:
- \begin{align*}
- \reflect(\idfunc[A])=\idfunc[\reflect(A)]\\
- \reflect(f\circ{}g)=\reflect(f)\circ\reflect(g)
- \end{align*}
-\end{defn}
+\begin{lem}
+ Let \P be a subuniverse of \type. The fact that \P is reflective is \anhprop.
+\end{lem}
\begin{proof}
- The map $\reflect(f)$ is defined to be the unique map
- $\reflect(A)\to\reflect(B)$ such that
- $\reflect(f)\circ\project_A=\project_B\circ{}f$ (using the universal property
- of $\reflect$), or in other words $\reflect(f)=\ext(\project_B\circ{}f)$.
+ Let’s assume that \P is reflective in two different ways
+ $(\reflect,\project,\ext)$ and $(\reflect',\project',\ext')$. We need to
+ construct an equivalence between $\reflect(A)$ and $\reflect'(A)$ for every
+ $A:\type$ and we need to prove that the following diagram commutes:
\[\uppercurveobject{{ }}\lowercurveobject{{ }}\twocellhead{{ }}
- \xymatrix{A \ar^-{\project_A}[r] \ar_-f[d] \drtwocell{=} & \reflect(A)
- \ar@{-->}^-{\reflect(f)}[d]
- \\ B \ar_-{\project_B}[r] & \reflect(B)}\]
- In order to prove that $\reflect(\idfunc[A])=\idfunc[\reflect(A)]$, we only
- need to prove that
- $\reflect(\idfunc[A])\circ\project_A=\idfunc[\reflect(A)]\circ\project_A$. But
- both sides are equal to $\project_A$.
-
- Similarly, to prove that $\reflect(f\circ{}g)=\reflect(f)\circ\reflect(g)$ we
- only need to prove that $\reflect(f\circ{}g)\circ\project_A=
- \reflect(f)\circ\reflect(g)\circ\project_A$ and both sides are equal to
- $\project_C\circ{}f\circ{}g$.
+ \xymatrix{A \ar^{\project_A}[r] \ar_{\project'_A}[rd] \druppertwocell{=} &
+ \reflect(A) \ar@{->}^\sim[d] \\
+ & \reflect'(A)}\]
+
+ The type $\reflect'(A)$ is in \P, so we can define the map
+ \[\ext(\project'_A):\reflect(A)\to\reflect'(A)\]
+ which is exactly the map making the previous diagram commute.
+
+ We can also define
+ \[\ext'(\project_A):\reflect'(A)\to\reflect(A)\]
+
+ In order to prove that the composite is the identity, we only need to prove
+ that $\ext'(\project_A)\circ\ext(\project'_A)\circ\project_A=\project_A$
+ which is the case:
+ \[\uppercurveobject{{ }}\lowercurveobject{{ }}\twocellhead{{ }}
+ \xymatrix{A \ar^{\project_A}[r] \ar_{\project'_A}[rd]
+ \ar@/_5mm/_{\project_A}[rdd] &
+ \reflect(A) \ar@{->}^{\ext(\project'_A)}[d] \\
+ & \reflect'(A) \ar@{->}^{\ext'(\project_A)}[d] \\
+ & \reflect(A)}\]
\end{proof}
+For the rest of this section, we assume that $\P$ is a reflective subuniverse of
+$\type$.
+
+The following lemma says that the counit of the adjunction is an equivalence.
\begin{lem}
\label{reflectPequiv}
If $A:\P$, then the map $\project_A:A\to\reflect(A)$ is an equivalence.
-
- Moreover, if $f:A\to{}B$ with $B:\P$, then
- $\project_B^{-1}\circ\reflect(f)\circ\project_A = f$.
\end{lem}
\begin{proof}
Given that $A$ is in $\P$, we can define $\ext(\idfunc[A]):\reflect(A)\to{}A$.
@@ -460,27 +463,118 @@ \section{Reflective subuniverses of $\type$}
definition, and in order to prove that
$\project_A\circ\ext(\idfunc[A])=\idfunc[\reflect(A)]$ we only need to prove
that $\project_A\circ\ext(\idfunc[A])\circ\project_A=
- \idfunc[\reflect(A)]\circ\project_A$ which is true.
+ \idfunc[\reflect(A)]\circ\project_A$ which is again true.
- For the second point is just a consequence of the fact that
- $\reflect(f)\circ\project_A=\project_B\circ{}f$.
+ \[\xymatrix{
+ A \ar^{\project_A}[r] \ar_{\idfunc[A]}[rd] &
+ \reflect(A) \ar^>>>{\ext(\idfunc[A])}[d] \ar@/^40pt/^{\idfunc[\reflect(A)]}[dd] \\
+ & A \ar_{\project_A}[d] \\
+ & \reflect(A)}\]
\end{proof}
-\begin{lem}
- For every $A,B,C:\type$, $f:A\to{}B$ and $i:B\to{}C$ we have
+The reflector $\reflect$ is a map $\type\to\P$ and using the universal property
+we should be able to prove that $\reflect$ is $(\infty,1)$-functorial. But we
+don’t know how to express $(\infty,1)$-functoriality so we will only prove a
+few bits of it.
+
+\begin{defn}
+ If $f:A\to{}B$, there is a map $\reflect(f):\reflect(A)\to\reflect(B)$ defined
+ by
+ \[\reflect(f)\circ\project_A=\project_B\circ{}f\]
+ (or in other words $\reflect(f)=\ext(\project_B\circ{}f)$).
+
+ \[\uppercurveobject{{ }}\lowercurveobject{{ }}\twocellhead{{ }}
+ \xymatrix{A \ar^-{\project_A}[r] \ar_-f[d] \drtwocell{=} & \reflect(A)
+ \ar@{-->}^-{\reflect(f)}[d]
+ \\ B \ar_-{\project_B}[r] & \reflect(B)}\]
+
+ This operation satisfies the following functoriality conditions:
+ \begin{align*}
+ \reflect(\idfunc[A])=\idfunc[\reflect(A)]\\
+ \reflect(f\circ{}g)=\reflect(f)\circ\reflect(g)
+ \end{align*}
+
+ In order to define these equalities, we only need to define them after
+ composition by $\project(A)$ because of the universal property.
+
+ The first one is defined by
+
+ \[\xymatrix{
+ \reflect(\idfunc[A])\circ\project_A \ar@{==}[r] \ar@{=}[d] &
+ \idfunc[\reflect(A)]\circ\project_A \ar@{=}[d] \\
+ \project_A \ar@{=}[r] & \project_A
+ }\]
+
+ The second one is defined by
+
+ \[\xymatrix{
+ \reflect(f\circ g)\circ\project_A \ar@{==}[r] \ar@{=}[d] &
+ \reflect(f)\circ\reflect(g)\circ\project_A \ar@{=}[d] \\
+ \project_C\circ f\circ g \ar@{=}[r] & \reflect(f)\circ\project_B\circ g
+ }\]
+\end{defn}
+
+% \begin{proof}
+% The map $\reflect(f)$ is defined to be the unique map
+% $\reflect(A)\to\reflect(B)$ such that
+% $\reflect(f)\circ\project_A=\project_B\circ{}f$ (using the universal property
+% of $\reflect$), or in other words $\reflect(f)=\ext(\project_B\circ{}f)$.
+
+% \[\uppercurveobject{{ }}\lowercurveobject{{ }}\twocellhead{{ }}
+% \xymatrix{A \ar^-{\project_A}[r] \ar_-f[d] \drtwocell{=} & \reflect(A)
+% \ar@{-->}^-{\reflect(f)}[d]
+% \\ B \ar_-{\project_B}[r] & \reflect(B)}\]
+% In order to prove that $\reflect(\idfunc[A])=\idfunc[\reflect(A)]$, we only
+% need to prove that
+% $\reflect(\idfunc[A])\circ\project_A=\idfunc[\reflect(A)]\circ\project_A$. But
+% both sides are equal to $\project_A$.
+
+% Similarly, to prove that $\reflect(f\circ{}g)=\reflect(f)\circ\reflect(g)$ we
+% only need to prove that $\reflect(f\circ{}g)\circ\project_A=
+% \reflect(f)\circ\reflect(g)\circ\project_A$ and both sides are equal to
+% $\project_C\circ{}f\circ{}g$.
+
+% Note that by definition, the following diagram commutes:
+
+% \[\xymatrix{
+% \reflect(f\circ g)\circ\project_A \ar@{=}[r] \ar@{=}[d] &
+% \reflect(f)\circ\reflect(g)\circ\project_A \ar@{=}[d] \\
+% \project_C\circ f\circ g \ar@{=}[r] & \reflect(f)\circ\project_B\circ g
+% }\]
+% \end{proof}
+
+\begin{defn}
+ For every $A,B:\type$, $C:\P$, $f:A\to{}B$ and $i:B\to{}C$ we have
\[\ext(i\circ{}f)=\ext(i)\circ\reflect(f)\]
-\end{lem}
-\begin{proof}
- \[\xymatrix{A \ar^{\project_A}[r] \ar_f[d] & \reflect(A) \ar^{\reflect(f)}[d]
- \\
- B \ar^{\project_B}[r] \ar_i[d] & \reflect(B) \ar^{\ext(i)}[ld] \\
- C &}\]
+ \[\xymatrix{
+ A \ar^{\project_A}[r] \ar_f[d] & \reflect(A) \ar^{\reflect(f)}[d]
+ \ar@/^15mm/^{\ext(i\circ f)}[dd] \\
+ B \ar^{\project_B}[r] \ar_i[rd] & \reflect(B) \ar^{\ext(i)}[d] \\
+ & C}\]
+
+ This equality is defined by
+ \[\xymatrix{
+ \ext(i\circ f)\circ\project_A \ar@{==}[r] \ar@{=}[d] &
+ \ext(i)\circ\reflect(f)\circ\project_A \ar@{=}[d] \\
+ i\circ f \ar@{=}[r] & \ext(i)\circ\project_B\circ f
+ }\]
- We only have to prove that
- \[\ext(i\circ{}f)\circ\project_A=\ext(i)\circ\reflect(f)\circ\project_A\]
- which is the case because they are both equal to $i\circ{}f$.
-\end{proof}
+\end{defn}
+
+% \begin{proof}
+% \[\xymatrix{A \ar^{\project_A}[r] \ar_f[d] & \reflect(A) \ar^{\reflect(f)}[d]
+% \\
+% B \ar^{\project_B}[r] \ar_i[d] & \reflect(B) \ar^{\ext(i)}[ld] \\
+% C &}\]
+
+% We only have to prove that
+% \[\ext(i\circ{}f)\circ\project_A=\ext(i)\circ\reflect(f)\circ\project_A\]
+% which is the case because they are both equal to $i\circ{}f$.
+% \end{proof}
+
+We now want to prove that pushouts in \P are related to pushouts in \type. We
+first need to explain how to reflect pushout diagrams and cocones.
\begin{defn}
Let
@@ -489,9 +583,11 @@ \section{Reflective subuniverses of $\type$}
pushout diagram in $\P$:
\[\reflect(\Ddiag)=\xymatrix{\reflect(C) \ar^{\reflect(g)}[r]
\ar_{\reflect(f)}[d] & \reflect(B) \\ \reflect(A) & }\]
+\end{defn}
- We can also reflect cocones. If $D:\P$ and $c=(i,j,h):\cocone{\Ddiag}{D}$ we
- define
+\begin{defn}
+ Let $D:\type$ and $c=(i,j,h):\cocone{\Ddiag}{D}$.
+ We define
\[\reflect(c)=(\reflect(i),\reflect(j),\reflect(h)):
\cocone{\reflect(\Ddiag)}{\reflect(D)}\]
where
@@ -505,7 +601,7 @@ \section{Reflective subuniverses of $\type$}
\[\mapfunc{\reflect}(\funext(h)):\reflect(i\circ{}f)=\reflect(j\circ{}g)\]
Now we can compose with the fact that $\reflect$ commutes with composition and
we get the following (the proofs that $\reflect$ commutes with composition are
- not written in order to keep terms readable, and we will see that they don’t
+ not written in order to keep terms readable, we will later see that they don’t
get in the way):
\[\mapfunc{\reflect}(\funext(h)):
\reflect(i)\circ\reflect(f)=\reflect(j)\circ\reflect(g)\]
@@ -538,7 +634,8 @@ \section{Reflective subuniverses of $\type$}
proving that it’s a composite of five maps which are already known to be
equivalences.
- The first map comes from the universal property of $\reflect$, we have
+ The first map comes from the universal property of $\reflect$ and the fact
+ that $E$ is in \P, we have
\[f_1(t)=t\circ\project_D\]
The second map comes from the universal property of $(D,c)$, as a pushout of
@@ -562,18 +659,25 @@ \section{Reflective subuniverses of $\type$}
The fifth map comes from \autoref{coneispb}, we have
\[f_5(a,b,q)=(a,b,\happly(q))\]
- Putting everything together we get
+ We now need to prove that the diagram commutes because it’s not enough to
+ know that $(\reflect(D)\to{}E)$ is equivalent to
+ $\cocone{\reflect(\Ddiag)}{E}$, we need to know that this is the map
+ $t\mapsto{}t\circ\reflect(c)$ which is an equivalence.
+
+ We have
\begin{align*}
f_5(f_4(f_3(f_2(f_1(t))))) &= f_5(f_4(f_3(f_2(t\circ\project_D)))) \\
&= f_5(f_4(f_3(t\circ\project_D\circ{}i,t\circ\project_D\circ{}j,
\mapfunc{(t\circ\project_D)}\circ{}h))) \\
&=f_5(f_4(t\circ\project_D\circ{}i,t\circ\project_D\circ{}j,
\funext(\mapfunc{(t\circ\project_D)}\circ{}h))) \\
- &=f_5(\ext(t\circ\project_D\circ{}i),\ext(t\circ\project_D\circ{}j),
- \mapfunc{\ext}(\funext(\mapfunc{(t\circ\project_D)}\circ{}h))) \\
+ &=f_5(\ext(t\circ\project_D\circ{}i),\ext(t\circ\project_D\circ{}j),\\
+ &\qquad\qquad\mapfunc{\ext}(\funext(\mapfunc{(t\circ\project_D)}\circ{}h)))
+ \\
&=f_5(\ext(t\circ\project_D)\circ\reflect(i),
- \ext(t\circ\project_D)\circ\reflect(j),
- \mapfunc{\ext}(\funext(\mapfunc{(t\circ\project_D)}\circ{}h))) \\
+ \ext(t\circ\project_D)\circ\reflect(j),\\
+ &\qquad\qquad\mapfunc{\ext}(\funext(\mapfunc{(t\circ\project_D)}\circ{}h)))
+ \\
&=f_5(t\circ\reflect(i),
t\circ\reflect(j),
\mapfunc{\ext}(\funext(\mapfunc{(t\circ\project_D)}\circ{}h))) \\
@@ -581,6 +685,19 @@ \section{Reflective subuniverses of $\type$}
\happly(\mapfunc{\ext}(\funext(\mapfunc{(t\circ\project_D)}\circ{}h))))
\end{align*}
+ Note that in the two steps before the last equality we are using the fact
+ that if $a=a'$ and $b=b'$, then $(a,b,q)=(a',b',q)$. This is actually not
+ even well typed given that the type of $q$ depends on $a$ and $b$, so we
+ have to transport $q$ along the proofs $p:a=a'$ and $q:b=b'$. In the present
+ case we have $q:a\circ\reflect(f)=b\circ\reflect(g)$ hence the correct
+ statement is
+ \[(a,b,q)=(a',b',q')\] where
+ \[q'\defeq\map{(\lambda{}u.\,u\circ\reflect(f))}{\rev p} \ct q \ct
+ \map{(\lambda{}u.\,u\circ\reflect(g))}{p'})\] Again we will leave this
+ implicit and take care of it only when it will be needed.
+
+ \bigskip
+
In order to prove that the diagram commutes, we now only need to prove that
\begin{align*}
\happly(\mapfunc{\ext}(\funext(\mapfunc{(t\circ\project_D)}\circ{}h))) &=
@@ -611,38 +728,158 @@ \section{Reflective subuniverses of $\type$}
\[\map{\ext}{\map{(\lambda{}u.\,t\circ\project_D\circ{}u)}{\funext(h)}}=
\map{(\lambda{}u.\,t\circ{}u)}{\map{\reflect}{\funext(h)}}\]
- This is the point where we need to check that the implicit equalities around
- $\mapfunc{\ext}$ and $\mapfunc{\reflect}$ cancel (TODO, this seems true but
- awfully complicated)
+ The idea is now to do an induction on the path $\funext(h):i\circ f = j\circ
+ g$ which should be allowed because nothing in the previous expression seems to
+ depend on the fact that $\funext(h)$ goes from a composition to another
+ composition. But let’s not forget that there are implicit equalities around
+ both sides of the equation and they do use the fact that $i\circ f$ and
+ $j\circ g$ are compositionss, so we first need to get rid of that.
- After that, using functoriality of $f\mapsto\mapfunc{f}$ it will be enough to
- prove the following:
+ % After that, using the fact that for every $k:C\to{}D$ we have
- \[\ext\circ(\lambda{}u.\,t\circ\project_D\circ{}u)=
- (\lambda{}u.\,t\circ{}u)\circ\reflect\]
+ % \begin{align*}
+ % \ext(t\circ\project_D\circ{}k) &= \ext(t\circ\project_D)\circ\reflect(k) \\
+ % &= t\circ\reflect(k)
+ % \end{align*}
- We apply function extensionality, so for every $u:C\to{}D$ we have to prove
+ % We will have
- \[\ext(t\circ\project_D\circ{}u) = t\circ\reflect(u) : \reflect(C)\to E\]
+ % \[\ext\circ(\lambda{}u.\,t\circ\project_D\circ{}u)=
+ % (\lambda{}u.\,t\circ{}u)\circ\reflect\]
- But we have
- $\ext(t\circ\project_D\circ{}u)=\ext(t\circ\project_D)\circ\reflect(u)=t\circ\reflect(u)$
- hence the equality holds.
+ % Hence the result, using the functoriality of $f\mapsto{}f_*$.
+
+ \bigskip
+
+ % Let’s prove that the implicit equalities around $\mapfunc{r}$ and
+ % $\mapfunc{\ext}$ cancel.
+ We have the following diagram:
+
+ \[\xymatrix{
+ t\circ\reflect(i\circ f) \ar@{=}[r] \ar@{=}[d] &
+ t\circ\reflect(j\circ g) \ar@{=}[d] \\
+ t\circ\reflect(i)\circ\reflect(f) \ar@{=}[d] &
+ t\circ\reflect(j)\circ\reflect(g) \ar@{=}[d] \\
+ \extend{t\circ\project_D}\circ\reflect(i)\circ\reflect(f) \ar@{=}[d] &
+ \extend{t\circ\project_D}\circ\reflect(j)\circ\reflect(g) \ar@{=}[d] \\
+ \extend{t\circ\project_D\circ i}\circ\reflect(f) \ar@{=}[d] &
+ \extend{t\circ\project_D\circ j}\circ\reflect(g) \ar@{=}[d] \\
+ \extend{t\circ\project_D\circ i\circ f} \ar@{=}[r] &
+ \extend{t\circ\project_D\circ j\circ g}\\
+ }\]
+
+ \begin{itemize}
+ \item The top horizontal equality is
+ $\map{(\lambda{}u.\,t\circ{}u)}{\map{\reflect}{\funext(h)}}$ without the
+ implicit equalities around
+ \item The first line of vertical equalities are the implicit equalities around
+ the term $\map{(\lambda{}u.\,t\circ{}u)}{\map{\reflect}{\funext(h)}}$ (the
+ right hand side of the equality we want to prove)
+ \item The next two lines of vertical equalities are around the left hand side
+ of the equality we want to prove and come from the long computation a few
+ pages ago
+ \item The last line of vertical equalities are the implicit equalities around
+ $\mapfunc{\extendsmb}$
+ \item The bottom horizontal equality is
+ $\map{\ext}{\map{(\lambda{}u.\,t\circ\project_D\circ{}u)}{\funext(h)}}$
+ without the implicit equalities around
+ \end{itemize}
+
+ The commutation of the diagram is exactly the equality we want to prove and
+ the fact that the compositions $i\circ f$ and $j\circ g$ are being split in
+ the middle of the diagram is the reason why we cannot directly induct on
+ $\funext(h)$.
+
+ So we will prove that the following diagram (where the vertical part is the
+ right hand side of the previous diagram) commutes. Note that in the right hand
+ side of this diagram, we do not use anymore the fact that $j\circ g$ is a
+ composition of two functions, so induction on it will be allowed.
+
+ \[\xymatrix{
+ t\circ\reflect(j\circ g) \ar@{=}[d] & \\
+ t\circ\reflect(j)\circ\reflect(g) \ar@{=}[d] & \\
+ \extend{t\circ\project_D}\circ\reflect(j)\circ\reflect(g) \ar@{=}[d] &
+ \extend{t\circ\project_D}\circ\reflect(j\circ g) \ar@{=}[l] \ar@{=}[luu]
+ \ar@{=}[ldd] \\
+ \extend{t\circ\project_D\circ j}\circ\reflect(g) \ar@{=}[d] & \\
+ \extend{t\circ\project_D\circ j\circ g} & \\
+ }\]
+
+ The top square commutes because it’s the concatenation of the two equalities
+ $t=\extend{t\circ\project_D}$ and $\reflect(j)\circ\reflect(g)=\reflect(j\circ
+ g)$ in two different orders (interchange law). In order to prove that the
+ bottom square commute, we only have to prove that it commutes after
+ composition with $\project_C$ to the right. Let’s write $u=t\circ\project_D$
+ and let’s consider the following diagram.
+
+ \[\xymatrix@C=-30pt{
+ \extend{u}\circ\reflect(j)\circ\reflect(g)\circ\project_C
+ \ar@{=}[rrrr] \ar@{=}[ddd] \ar@{=}[rd] & & & &
+ \extend{u}\circ\reflect(j\circ g)\circ\project_C
+ \ar@{=}[ld] \ar@{=}[ddd] \\ &
+ \extend{u}\circ\reflect(j)\circ\project_B\circ g
+ \ar@{=}[rr] \ar@{=}[d] & {\phantom{thisisanuglyhack}} &
+ \extend{u}\circ\project_D\circ j\circ g
+ \ar@{=}[d] & \\ &
+ \extend{u\circ j}\circ\project_B\circ g
+ \ar@{=}[rr] \ar@{=}[ld] & &
+ u\circ j\circ g
+ \ar@{=}[rd] & \\
+ \extend{u\circ j}\circ\reflect(g)\circ\project_C
+ \ar@{=}[rrrr] & & & &
+ \extend{u\circ j\circ g}\circ\project_C
+ }\]
+
+ We need to prove that the exterior square commutes.
+
+ \begin{itemize}
+ \item The left square commutes because of the interchange law
+ \item The top square commutes because of the definition of the equality
+ $\reflect(j\circ g)=\reflect(j)\circ\reflect(g)$
+ \item The bottom, right and interior squares commute because of the definition
+ of the equality $\extend{a}\circ\reflect(b)=\extend{a\circ b}$
+ \end{itemize}
+
+ This proves that the exterior square commutes, hence the previous triangular
+ diagram commutes. We can do the same for the left part of the first
+ rectangular diagram so we now only have to prove that the following diagram
+ commutes:
+
+ \[\xymatrix{
+ t\circ\reflect(i\circ f) \ar@{=}[r] \ar@{=}[d] &
+ t\circ\reflect(j\circ g) \ar@{=}[d] \\
+ \extend{t\circ\project_D}\circ\reflect(i\circ f) \ar@{=}[d] &
+ \extend{t\circ\project_D}\circ\reflect(j\circ g) \ar@{=}[d] \\
+ \extend{t\circ\project_D\circ i\circ f} \ar@{=}[r] &
+ \extend{t\circ\project_D\circ j\circ g}\\
+ }\]
+
+ And now we are finally allowed to induct on $\funext(h)$ because the diagram
+ is not relying anymore on the fact that $i\circ f$ and $j\circ g$ are
+ compositions.
+
+ \bigskip
This proves that the diagram commutes, hence the map
$t\mapsto{}\composecocone{t}\reflect(c)$ is an equivalence which proves that
the reflector commutes with pushouts.
\end{proof}
+This proof was rather tedious, but we can hope that at some point we will
+understand better what $(\infty,1)$-functorial means in homotopy type theory and
+that we will be able to rigorously omit the equalities I left implicit and just
+say that every coherence condition is always satisfied.
+
\begin{cor}
Every pushout diagram $\Ddiag$ in $\P$ has a pushout in $\P$.
\end{cor}
\begin{proof}
- According to \autoref{reflectPequiv}, the diagram $\reflect(\Ddiag)$ is
- equivalent to $\Ddiag$. But we just proved that $\reflect(\Ddiag)$ has a
- pushout, namely the reflection of the pushout in \type of $\Ddiag$, hence
- $\Ddiag$ has a pushout in \P.
+ According to \autoref{reflectPequiv} and to the diagram defining the action of
+ $\reflect$ on functions, the diagram $\reflect(\Ddiag)$ is equivalent to
+ $\Ddiag$. But we just proved that $\reflect(\Ddiag)$ has a pushout, namely the
+ reflection of the pushout in \type of $\Ddiag$, hence $\Ddiag$ has a pushout
+ in \P.
\end{proof}
\section{Truncations}
@@ -747,7 +984,7 @@ \section{Connectedness of suspensions}
connectedness.
\begin{thm}
- Let $n:\Nt$ and $A:\type$.
+ Let $n\ge-2$ and $A:\type$.
If $A$ is $n$-connected then the suspension of $A$ is $(n+1)$-connected.
\end{thm}
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