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Update univalence proof to use our standard macros

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1 parent a1b20a3 commit 18e160858ae1b1d5d90149e415e2b8ae605723d4 @mikeshulman mikeshulman committed Nov 20, 2012
Showing with 179 additions and 190 deletions.
  1. +1 −0 macros.tex
  2. +0 −17 pa-macros.tex
  3. +44 −44 ua-append.tex
  4. +68 −66 ua-fe.tex
  5. +66 −63 ua.tex
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@@ -77,6 +77,7 @@
\defthm{rmk}{Remark}
\defthm{eg}{Example}
\defthm{egs}{Examples}
+\defthm{notes}{Notes}
% Number exercises within chapters, with their own counter.
\newtheorem{ex}{Exercise}[chapter]
\def\exautorefname{Exercise}
View
@@ -1,21 +1,4 @@
\newcommand{\comment}[1]{} %%%to comment out blocks of text
-\newcommand{\be}{\begin}
-\newcommand{\en}{\end}
-
-%\newtheorem{thm}{Theorem:}[section]
-%\newtheorem{cor}[thm]{Corollary:}
-\newtheorem{propp}[thm]{Proposition:}
-%\newtheorem{lem}[thm]{Lemma:}
-\newtheorem{deff}[thm]{Definition:}
-\newtheorem{exe}[thm]{Exercise:}
-\newtheorem{cla}[thm]{Claim:}
-\newtheorem{rem}[thm]{Remark:}
-\newtheorem{examples}[thm]{Examples:}
-\newtheorem{dfn}[thm]{Definition:}
-\newtheorem{rmr}[thm]{Remark:}
-\newtheorem{pro}[thm]{Proposition:}
-\newtheorem{conj}[thm]{Conjecture:}
-\newtheorem{notes}[thm]{Notes:}
\newcommand{\ra}{\rightarrow}
View
@@ -8,52 +8,52 @@ \subsection{Adjoint Isomorphisms}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
In order to define the notion of an adjoint isomorphism we need the following definition.
-\be{deff} $\;$
+\begin{defn} $\;$
If $f:A\ra B$ define, by Id-induction, $fz:fx=fx'$ for $x,x':A, z:x=x'$ such that $f\refl{x} = \refl{fx}$ for $x:A$.
-\en{deff}
+\end{defn}
-\be{deff} A function $f:A\ra B$ is an {\em adjoint isomorphism} if there are $g:B\ra A$, $\eta:\forall_{x:A}\; [x=g(fx)]$ and $\epsilon:\forall_{y:B}\; [f(gy)=y]$ such that
+\begin{defn} A function $f:A\ra B$ is an {\em adjoint isomorphism} if there are $g:B\ra A$, $\eta:\forall_{x:A}\; [x=g(fx)]$ and $\epsilon:\forall_{y:B}\; [f(gy)=y]$ such that
\[\forall_{x:A}\;[(f(\eta x)\; @\;\epsilon(fx))=1_{fx}].\]
-\en{deff}
-\be{prop}\label{prop:6.3} The following are logically equivalent for $f:A\ra B$.
-\be{enumerate}
+\end{defn}
+\begin{prop}\label{prop:6.3} The following are logically equivalent for $f:A\ra B$.
+\begin{enumerate}
\item $f$ is an equivalence.
\item $f$ is an isomorphism
\item $f$ is an adjoint isomorphism
-\en{enumerate}
-\en{prop}
-\proof $1\ra 2$ is Proposition~\ref{prop:3.4}. The other logical implications might best be left to another chapter.
-\qed
+\end{enumerate}
+\end{prop}
+\begin{proof} $1\ra 2$ is \autoref{prop:3.4}. The other logical implications might best be left to another chapter.
+\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The Main Lemma - version 2}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Let $\bbU$ be a universe. In the following, when we write $Z:X=X'$ it will be implicit that $X,X':\bbU$. We use transport to define, for $Z:X=X'$, $Z_0$ and also $Z_B$, for each type $B$.
-\be{deff} $\;$\label{def:6.4}
-\be{enumerate}
+\begin{defn} $\;$\label{def:6.4}
+\begin{enumerate}
\item For $Z:X=X'$ let $Z_0:= Z^*_{P_0}$ where $P_0[X]:=X$ for $X:\bbU$.
So $Z_0:X\ra X'$ is an equivalence for $Z:X=X'$.
\item Let $B$ be a type. For $Z:X=X'$ let $Z_B:=(Z^{-1})^*_{P_B}$ where
$P_B[X]:=(X\ra B)$ for $X:\bbU$. So $Z_B: (X'\ra B)\ra(X\ra B)$ is an equivalence for $Z:X=X'$.
-\en{enumerate}
-\en{deff}
+\end{enumerate}
+\end{defn}
-\be{lem} $\;$\label{lem:6.5}
-\be{enumerate}
+\begin{lem} $\;$\label{lem:6.5}
+\begin{enumerate}
\item For $Z:X=X'$
\[ Z_Bg = g\circ Z_0 \mbox{ for $g:X'\ra B$}.\]
\item For $Z:X=X'$
\[ Z_0 = \pi_{X,X'}(E_{X,X'}Z),\]
where $\pi_{X,X'}:(X\simeq X')\ra (X\ra X')$ is defined by
\[ \pi_{X,X'}(f,q):=f\mbox{ for } f:X\ra X', q:EQ(f).\]
-\en{enumerate}
-\en{lem}
-\proof
-\be{enumerate}
+\end{enumerate}
+\end{lem}
+\begin{proof}
+\begin{enumerate}
\item By Id-induction on $Z:X=X'$, for $X,X':\bbU$, it suffices to show that, for $X:\bbU$ and $g:X\ra B$,
\[ (\refl{X})_Bg = g\circ(\refl{X})_0.\]
But
- \[\be{array}{ll}
+ \[\begin{array}{ll}
(\refl{X})_Bg
&= (\refl{X}^{-1})_{P_B}^*g\\
&= (\refl{X})_{P_B}^*g\\
@@ -62,58 +62,58 @@ \subsection{The Main Lemma - version 2}
&= \lambda_{x:X}gx\\
&= g\circ 1_X\\
&= g\circ (\refl{X})_0
-\en{array}\]
+\end{array}\]
\item Observe that
\[(\refl{X})_0=1_X=\pi_{X,X}(1_X,s_X)=\pi_{X,X}(E_{X,X}\refl{X})\]
for $X:\bbU$, where $s_X:EQ(1_X)$. So, by Id-induction on $Z:X=X'$, for
$X,X':\bbU$,
\[ Z_0 = \pi_{X,X'}(E_{X,X'}Z).\]
-\en{enumerate}
-\qed
+\end{enumerate}
+\end{proof}
-\be{lem}[Main Lemma - version 2]\label{fe:lem-main2}\label{lem:6.6}
+\begin{lem}[Main Lemma - version 2]\label{fe:lem-main2}\label{lem:6.6}
Let $A,A'$ be types in a univalent universe. For each type $C$ and each
$f:A\ra A'$ define
$G_f := (-)\circ f:(A'\ra C)\ra (A\ra C)$;
i.e.
\[ G_f := \lambda_{g:A'\ra C}\; g\circ f\]
If $f$ is an equivalence then so is $G_f$.
-\en{lem}
+\end{lem}
%\pagebreak
-\proof Let $\bbU$ be a univalent universe and let $f:A\ra A'$ be an
+\begin{proof} Let $\bbU$ be a univalent universe and let $f:A\ra A'$ be an
equivalence, where $A,A':\bbU$. We want to show that $G_f$ is an equivalence.
-By Corollary~\ref{cor:3.5} $E_{A,A'}:(A=A')\ra (A\simeq A')$ is surjective.
+By \autoref{cor:3.5} $E_{A,A'}:(A=A')\ra (A\simeq A')$ is surjective.
As $f$ is an equivalence there is $q:EQ(f)$ so that $(f,q):A\simeq A'$. Let $u:=(f,q):(A\simeq A')$. So there is $Z_f:(A=A')$ such that $E_{A,A'}Z_f = u$.
-\be{description}
+\begin{description}
\item[Proof that $G_f$ is an equivalence:]
-By part 2 of Definition~\ref{def:6.4}
+By part 2 of \autoref{def:6.4}
\[ (Z_f)_B^*:(A'\ra B)\ra (A\ra B)\]
-is an equivalence. So, by Lemma~\ref{lem:6.5}
+is an equivalence. So, by \autoref{lem:6.5}
\[(Z_f)_B^*g = g\circ (Z_f)_0\mbox{ for } g:A'\ra B.\]
-By Proposition~\ref{prop:4.5}, as $EQ((Z_f)_B^*)$, $EQ(G_{(Z_f)_0})$; i.e. $G_{(Z_f)_0}$ is an equivalence.
+By \autoref{prop:4.5}, as $EQ((Z_f)_B^*)$, $EQ(G_{(Z_f)_0})$; i.e. $G_{(Z_f)_0}$ is an equivalence.
As $u:=(f,q):A\simeq A'$
\[(Z_f)_0 = \pi_{A,A'}(E_{A,A'}(Z_f)) =\pi_{A,A'}u = f.\]
So $G_f$ is an equivalence.
-\en{description}
-\qed
+\end{description}
+\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Proof of the Theorem - version 2: }
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
We use the following lemma.
-\be{lem}\label{fe:lem-projs}\label{lem:6.7}
+\begin{lem}\label{fe:lem-projs}\label{lem:6.7}
If $B$ is a type in a univalent universe then $\pi^1_B = \pi^2_B$.
-\en{lem}
-\proof By part 3 of Lemma~\ref{lem:4.9} $\delta_B$ is an isomorphism and hence an equivalence, by $2\ra 1$ of Proposition~\ref{prop:6.3}. So, by part 1 of the Main Lemma, if $B$ is in a univalent universe then
+\end{lem}
+\begin{proof} By part 3 of \autoref{lem:4.9} $\delta_B$ is an isomorphism and hence an equivalence, by $2\ra 1$ of \autoref{prop:6.3}. So, by part 1 of the Main Lemma, if $B$ is in a univalent universe then
\[ (-)\circ\delta_B : (Id(B)\ra B)\ra (B\ra B)\]
-is an equivalence and hence is injective. By part 1 of Lemma~\ref{lem:4.9},
+is an equivalence and hence is injective. By part 1 of \autoref{lem:4.9},
\[ \pi^1_B\circ\delta_B = \pi^2_B\circ\delta_B\]
so that $\pi^1_B=\pi^2_B$.
-\qed
+\end{proof}
Let $f_1,f_2:A\ra B$, where $B$ is a type in a univalent universe, and let
@@ -122,17 +122,17 @@ \subsection{Proof of the Theorem - version 2: }
\[ f := \lambda_{x:A}(f_1x,f_2x,qx).\]
For $i=1,2$, if $x:A$ then
- \[\be{array}{lll}
+ \[\begin{array}{lll}
\pi^i_B(fx)&=_{def}&\pi^i_B(f_1x,f_2x,qx)\\
&=_{def}& f_ix
- \en{array}\]
+ \end{array}\]
so that
- \[\be{array}{lll}
+ \[\begin{array}{lll}
\pi^i_B\circ f&=_{def}& \lambda_{x:A}\pi^i_B(fx)\\
&=_{def}& \lambda_{x:A}f_ix,\mbox{by } {\bf (FE_{def}})_{A\ra B}\\
&=& f_i,\mbox{by }{(\eta_{\sf prop})}.
- \en{array}\]
-By Lemma~\ref{lem:6.7} $\pi^1_B\circ f = \pi^2_B\circ f$ so that $f_1=f_2$.
+ \end{array}\]
+By \autoref{lem:6.7} $\pi^1_B\circ f = \pi^2_B\circ f$ so that $f_1=f_2$.
\qed
% Local Variables:
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