working on uniformizing notation and terminology

equivalences.tex

@@ -2,33 +2,50 @@ \chapter{Equivalences}
 \label{cha:equivalences}
 
 \section{Equivalences and Homotopies}
-\begin{defn}\label{defn:contractible} A type $A$ is {\em contractible} if there is $a:A$, called the {\em center of contraction} such that $a=x$ for all $x:A$.
+
+\begin{defn}\label{defn:contractible}
+  A type $A$ is \textbf{contractible}, or a \textbf{singleton}, if there is $a:A$, called the \textbf{center of contraction}, such that $a=x$ for all $x:A$.
 \end{defn}
-\begin{defn}\label{defn:equivalence} A map $f:A\ra B$ is an {\em equivalence}
-if, for $y:B$, its {\em fiber}, $\{x:A\mid fx = y\}$, is contractible.
-We write $A\simeq B$ if  there is an equivalence $A\ra B$.
+
+\begin{defn}\label{defn:equivalence}
+  A map $f:A\ra B$ is an {\em equivalence} if, for $y:B$, its {\em fiber}, $\setof{x:A | fx = y}$, is contractible.
+  We write $A\eqv B$ for the type of equivalences $A\ra B$.
 \end{defn}
 
 
-\begin{lem}\label{lem:id-map}  For each type $A$, the identity map 
-$1_A:= \lambda_{x:A}x :A\ra A$ is an equivalence.
+\begin{lem}\label{lem:id-map}
+  For each type $A$, the identity map $\idfunc[A]\defeq \lambda_{x:A}x :A\ra A$ is an equivalence.
 \end{lem}
-\begin{proof}  Let $a:A$ and let $\{ a\}_A:= \{ x:A\mid x=a\}$ be its fiber with respect to $1_A$.  We show that $\{ a\}_A$ is contractible.  Then, discharging $a:A$ we get that $\forall_{x:A}(\{ x\}_A$ is contractible); i.e. $1_A$ is an equivalence.
-
-Let $\oa =(a,\refl{a}):\{ a\}_A$.  As $(a,\refl{a}) = \oa$ we can use Id-induction \`{a} la Christine-Paulin to get
-  \[\forall_{x:A}\forall_{z:(x=a)}\; (x,z)=\oa.\]
-Hence, by $\Sigma$-folding, $\forall_{u:\{ a\}_A}\; u=\oa$ so that $\{ a\}_A$ is contractible, as desired.
+\begin{proof}
+  Let $a:A$ and let $\{a\}_A\defeq \setof{ x:A | x=a }$ be its fiber with respect to $\idfunc[A]$.
+  We show that $\{ a\}_A$ is contractible.
+  Then, discharging $a:A$, we get that $\fall{x:A} (\{ x\}_A$ is contractible); i.e.\ $\idfunc[A]$ is an equivalence.
+
+  Let $\oa \defeq (a,\refl{a}):\{ a\}_A$.  As $(a,\refl{a}) = \oa$ we can use Id-induction \`{a} la Christine-Paulin to get
+  \[\fall{x:A}{z:x=a} ((x,z)=\oa).\]
+  Hence, by $\Sigma$-folding, $\fall{u:\{ a\}_A} (u=\oa)$.
+  Thus $\{ a\}_A$ is contractible, as desired.
 \end{proof}
-\begin{defn}\label{defn:homotopy-between-functions} A {\em homotopy, $f\simeq f'$} between maps $f,f':A\ra B$ is a function $h:\prod_{x:A}\; f(x)=f'(x)$.  
+
+\begin{defn}\label{defn:homotopy-between-functions}
+  A \textbf{homotopy, $f\sim f'$} between maps $f,f':A\ra B$ is a function $h:\prd{x:A} (f(x)=f'(x))$.
 \end{defn}
-\begin{lem}\label{lem:homotopy-props} $\;$
+
+\begin{lem}\label{lem:homotopy-props}\ 
 \begin{enumerate}
-\item $\simeq$ is an equivalence relation on each function type $A\ra B$.
-\item If $f:A\ra B$ then $f\circ 1_A\simeq f\simeq 1_B\circ f$.
-\item If $f:A\ra B, g:B\ra C$ and $h:c\ra D$ then $h\circ (g\circ f)\;\simeq\; (h\circ g)\circ f$.
+\item $\sim$ is an equivalence relation on each function type $A\ra B$.
+  That is, we have terms of types
+  \begin{gather*}
+    \prd{f:A\to B} (f\sim f)\\
+    \prd{f,g:A\to B} (f\sim g) \to (g\sim f)\\
+    \prd{f,g,h:A\to B} (f\sim g) \to (g\sim h) \to (f\sim h).
+  \end{gather*}
+\item If $f:A\ra B$ then $f\circ \idfunc[A]\sim f\sim \idfunc[B]\circ f$.
+\item If $f:A\ra B, g:B\ra C$ and $h:C\ra D$ then $h\circ (g\circ f)\;\sim\; (h\circ g)\circ f$.
 \end{enumerate}
 \end{lem}
-\newpage
+
+%% TODO: This should probably be folded into wherever we discuss function extensionality, so that $(f\sim g)\eqv (f=g)$.
 
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
@@ -46,11 +63,11 @@ \section{Bijections and Isomorphisms}
 \item $f$ is {\em an isomorphism} if it has $g:B\ra A$ that is both a left and a right
 inverse where,
 \begin{itemize} \item $g:B\ra A$ is a {\em left/right inverse} to $f$ if 
-$g\circ f\simeq 1_A$/$f\circ g\simeq 1_B$.
+$g\circ f\sim \idfunc[A]$/$f\circ g\sim \idfunc[B]$.
 \end{itemize}
 \item $f$ is {\em (left/right)-invertible} if it has a (left/right) inverse.
 \item $f$ is a {\em weak isomorphism} if it is both left invertible and right invertible.
-\item We write $A\cong B/A\cong_w B$ if there is a(n) isomorphism/weak isomorphism $A\ra B$.
+\item We write $A\cong B$ if there is an) isomorphism $A\ra B$, and $A\cong_w B$ if there is a weak isomorphism $A\to B$.
 \end{enumerate}
 \end{defn}
 
@@ -59,7 +76,7 @@ \section{Bijections and Isomorphisms}
 \begin{enumerate}
 \item $f:A\ra B$ is a surjection iff it is right-invertible.
 \item If $f:A\ra B$ is left-invertible then it is injective.
-\item $1_A:A\ra A$ is an isomorphism.
+\item $\idfunc[A]:A\ra A$ is an isomorphism.
 \item If $f:A\ra B$ and $g:B\ra C$ are isomorphisms then so is $g\circ f:A\ra C$.
 \end{enumerate}
 \end{thm}
@@ -91,7 +108,7 @@ \section{Adjoint  Isomorphisms}
 }%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 
 \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 
-  \[\forall_{x:A}\;[(\map{f}{\eta x}\; \ct\;\epsilon(fx))=1_{fx}].\]
+  \[\forall_{x:A}\;[(\map{f}{\eta x}\; \ct\;\epsilon(fx))=\refl{fx}].\]
 We write $A\cong_a B$ if there is an adjoint isomorphism $A\ra B$.
 \end{defn}
 \begin{thm}\label{thm:equiv-iso-adj} The following are logically equivalent for $f:A\ra B$.
@@ -105,7 +122,7 @@ \section{Adjoint  Isomorphisms}
 
 \begin{description}
 \item[$(i)\ra (ii)$] 
-Let $f:A\ra B$ be an equivalence.  So each type $Cy$ is contractible, where for $y:B$, $Cy := \Sigma_{x:A}\; fx=y$.  This means that
+Let $f:A\ra B$ be an equivalence.  So each type $Cy$ is contractible, where for $y:B$, $Cy \defeq \Sigma_{x:A}\; fx=y$.  This means that
   \[ \forall_{y:B}\exists_{u:Cy}\forall_{u':Cy}\; u'=u.\]
 By $\Sigma$-unfolding we get
   \[ \forall_{y:B}\exists_{x:A}\exists_{z:(fx=y)}
@@ -122,7 +139,7 @@ \section{Adjoint  Isomorphisms}
 \begin{cor}\label{cor:equivs-equiv}
 For types $A,B$ we have the following sequence of two equivalences and a logical equivalence
 between our four equivalence relations.
-  \[ (A\simeq B) \simeq (A\cong_a B) \simeq (A\cong_w B) \leftrightarrow (A\cong B).\]
+  \[ (\eqv A B) \simeq (A\cong_a B) \simeq (A\cong_w B) \leftrightarrow (A\cong B).\]
 \end{cor}
 \begin{proof} ??
 
@@ -148,9 +165,9 @@ \section{Identity Systems on a Type Universe}
 \begin{enumerate}
 \item $R_{A,B}\ra \eqv{A}{B}$ for $A,B:\bbU$
 \item If $f_e:A\ra B$ and $g_e:B\ra A$ for $A,B:\bbU$ and $e:R_{A,B}$ such that 
-  \[ f_{\sfr{A}} = g_{\sfr{A}}=1_A\mbox{ for } A:\bbU\] 
+  \[ f_{\sfr{A}} = g_{\sfr{A}}=\idfunc[A]\mbox{ for } A:\bbU\] 
 then
-  \[ g_e\circ f_e =1_A\mbox{ and } f_e\circ g_e = 1_B
+  \[ g_e\circ f_e =\idfunc[A]\mbox{ and } f_e\circ g_e = \idfunc[B]
       \mbox{ for }  A,B:\bbU \mbox{ and } e:R_{A,B} \]
 \end{enumerate}
 \end{thm}
@@ -160,8 +177,8 @@ \section{Identity Systems on a Type Universe}
 \begin{thm}
 $\bbU$ is a univalent type universe iff $(Equiv^\bbU,\sfequiv{})$ is an identity system on $\bbU$, where
   \[ Equiv^\bbU_{A,B}\defeq (\eqv{A}{B})\mbox{ for } A,B:\bbU\]
-and, if $s_A: (1_A\mbox{ is an equivalence }A\ra A)$,
-  \[ \sfequiv{A}\defeq (1_A,s_A):\eqv{A}{A}\mbox{ for } A:\bbU.\]
+and, if $s_A: (\idfunc[A]\mbox{ is an equivalence }A\ra A)$,
+  \[ \sfequiv{A}\defeq (\idfunc[A],s_A):\eqv{A}{A}\mbox{ for } A:\bbU.\]
 \end{thm}
 
 

hlevels.tex

@@ -332,7 +332,7 @@ \section{Definition \& First Properties of H--Levels}
 \end{thm}
 
 \begin{defn}
- A type $X$ is an {\em h-set}, if it is of h-level $2$.
+ A type $X$ is a {\em set}, if it is of h-level $2$.
 \end{defn}
 
 \begin{defn}
@@ -346,7 +346,7 @@ \section{Definition \& First Properties of H--Levels}
 \begin{thm}\label{thm:h-set-uip-K}
  For a type $X$ the following are equivalent:
  \begin{enumerate}
-  \item $X$ is an h-set.
+  \item $X$ is a set.
   \item $X$ has UIP.
   \item $X$ satisfies Axiom K.
  \end{enumerate}
@@ -358,15 +358,15 @@ \section{Definition \& First Properties of H--Levels}
 \end{defn}
 
 \begin{thm}
- Let $X : \Type$. If $X$ has decidable equality, then $X$ is an h-set.
+ Let $X : \Type$. If $X$ has decidable equality, then $X$ is a set.
 \end{thm}
 
 \begin{proof}
  By Theorem \ref{thm:h-set-uip-K} it suffices to show that $X$ satisfies Axiom K. This is so trivial that I can't even write it!
 \end{proof}
 
 \begin{thm}\label{prop:nat-is-set}
- The type $\Nat$ of natural numbers is an h-set.
+ The type $\Nat$ of natural numbers is a set.
 \end{thm}
 
 \begin{proof}
@@ -414,4 +414,10 @@ \section{H--Level of types of h--level $n$}
  \[\eqv{X}{X'} \hookrightarrow X \rightarrow X'\]
  i.e. a map of h-level $1$. Thus it suffices to show that $X \rightarrow X'$ is of h-level $n$. 
 Since h-levels are preserved under $\prod$ and hence under the arrow type, this reduces to an assumption that $X'$ is of h-level $n$.
-\end{proof}
+\end{proof}
+
+
+% Local Variables:
+% mode: latex
+% TeX-master: "main"
+% End: