--- /dev/null
+\documentclass[10pt,letterpaper]{article}
+\DeclareSymbolFont{AMSb}{U}{msb}{m}{n}
+\DeclareMathAlphabet{\mathbbm}{U}{bbm}{m}{n}
+
+\usepackage{amsmath,amssymb,amsthm,latexsym}
+\usepackage{fancyhdr}
+\usepackage[tiny,center,compact,sc]{titlesec}
+\usepackage[cm]{fullpage}
+\usepackage{pstricks}
+\usepackage{graphicx}
+\usepackage{verbatim}
+\usepackage{bm}
+\usepackage{ifthen}
+\usepackage{epsfig}
+\usepackage[all]{xypic}
+\usepackage{textcomp}
+\usepackage{url}
+\usepackage{multirow}
+\usepackage{hyperref}
+\usepackage{breakurl}
+
+\renewcommand{\baselinestretch}{0.9}
+
+%\newtheorem{thm}{Thm}[section]
+%\newtheorem{dfn}{Def}[section]
+
+\setlength{\parindent}{0pt}
+\setlength{\parskip}{3pt}
+
+%Scalable bracket-like
+\newcommand{\paren}[1]{\left({#1}\right)}
+\newcommand{\brak}[1]{\left[{#1}\right]}
+\newcommand{\abs}[1]{\left\lvert{#1}\right\rvert}
+\newcommand{\ang}[1]{\left\langle{#1}\right\rangle}
+\newcommand{\set}[1]{\left\{#1\right\}}
+
+%Mathematics
+\newcommand{\condexp}[1]{\ifthenelse{\equal{#1}{false}}{}{^{#1}}}
+\newcommand{\dd}[3][false]{\frac{d\condexp{#1}{#2}}{d{#3}\condexp{#1}}}
+\newcommand{\pd}[3][false]{\frac{\partial\condexp{#1}{#2}}{\partial{#3}\condexp{#1}}}
+
+\newcommand{\ifrac}[2]{{#1}/{#2}}
+
+%Quantum Mechanics
+\newcommand{\ket}[1]{\left\lvert{#1}\right\rangle}
+\newcommand{\bra}[1]{\left\langle{#1}\right\rvert}
+\newcommand{\braket}[2]{\left\langle{#1}\middle\vert{#2}\right\rangle}
+\newcommand{\Braket}[3]{\left\langle{#1}\middle\vert{#2}\middle\vert{#3}\right\rangle}
+\newcommand{\dyad}[2]{\left\lvert{#1}\middle\rangle\middle\langle{#2}\right\rvert}
+
+\DeclareMathOperator{\mm}{\mid\mid}
+
+\newcommand{\defn}[1]{{\bf #1}}
+
+\begin{document}
+
+\section{Covariant Hom functors}
+
+These are defined in \S3.20.4, but a longer example never hurt anybody, right?
+
+Suppose $a^2 = id_A$, $b^2 = id_{B'}$, and $f' = g \circ f$ in this category $\mathbf{A}$.
+\[ \xymatrix@C=1in{
+ & B \ar[d]^g \ar@(dr,ur)_{id_B} \\
+ A \ar@(dr,d)^{id_A} \ar@(dl,l)^{a} \ar[ru]^f \ar[r]^{f'} & B' \ar@(dl,d)_{b} \ar@(dr,r)_{id_{B'}} \\
+ } \]
+
+If we actually draw out all the arrows, we get this diagram:
+\[ \xymatrix@C=1in{
+ & B \ar@<1pt>[d]_{b \circ g} \ar@<-1pt>[d]^g \ar@(dr,ur)_{id_B} \\
+ A \ar@(dr,d)^{id_A} \ar@(dl,l)^{a}
+ \ar@<1pt>[ru]^f \ar@<-1pt>[ru]_{f \circ a}
+ \ar@<3pt>[r] \ar@<1pt>[r] \ar@<-1pt>[r] \ar@<-3pt>[r]
+ & B' \ar@(dl,d)_{b} \ar@(dr,r)_{id_{B'}} \\
+ } \]
+where the four arrows from $A$ to $B'$ are $\set{f', f' \circ a, b \circ f', b \circ f' \circ a}$.
+
+$\mbox{hom}(A,-)$ is the following full subcategory of $\mathbf{Set}$; please
+note the similarities---the {\em representation} of the structure reachable from $A$:
+\[ \xymatrix@C=1in{
+ & \set{f, f \circ a} \ar@<1pt>[d] \ar@<-1pt>[d] \\
+ \set{id_A, a} \ar@<1pt>[ru] \ar@<-1pt>[ru]
+ \ar@<3pt>[r] \ar@<1pt>[r] \ar@<-1pt>[r] \ar@<-3pt>[r]
+ & \txt{$\{f', f' \circ a,$\\ $b \circ f', b \circ f' \circ a\}$} \\
+} \]
+The two arrows from $\mbox{hom}(A,A) = \set{id_A, a}$ to $\mbox{hom}(A,B)
+= \set{f, f \circ a}$ are obtained by pre-composition:
+\begin{align*}
+ \mbox{hom}(A,f) &= \set{ id_A \mapsto f, a \mapsto f \circ a } \\
+ \mbox{hom}(A,f \circ a) &= \set{ id_a \mapsto f \circ a, a \mapsto f }
+\end{align*}
+(the last entry holds because $(f \circ a) \circ a = f \circ (a \circ a) = f
+\circ id_A = f$). The four arrows from $\mbox{hom}(A,A)$ to
+$\mbox{hom}(A,B') = \set{f', f' \circ a, b \circ f', b \circ f' \circ a}$ are
+again obtained by pre-composition:
+\begin{align*}
+ \mbox{hom}(A,f') &= \set{id_A \mapsto f', a \mapsto f' \circ a} \\
+ \mbox{hom}(A,f' \circ a) &= \set{id_a \mapsto f' \circ a, a \mapsto f'} \\
+ \mbox{hom}(A,b \circ f') &= \set{id_a \mapsto b \circ f', a \mapsto b \circ f' \circ a} \\
+ \mbox{hom}(A,b \circ f' \circ a) &= \set{id_a \mapsto b \circ f' \circ a, a \mapsto b \circ f'}
+\end{align*}
+The two vertical arrows are (again by precomposition, and recall that $f' = g \circ f$):
+\begin{align*}
+ \mbox{hom}(A,g) &= \set{f \mapsto f', f \circ a \mapsto f' \circ a} \\
+ \mbox{hom}(A,b \circ g) &= \set{f \mapsto b \circ f', f \circ a \mapsto b \circ f' \circ a}
+\end{align*}
+It is easy to check that, indeed, composition still holds: the four horizontal
+arrows are each the result of composition of a choice of vertical and diagonal arrows,
+and we haven't missed any.
+
+\pagebreak
+\section{Proposition 6.18 and The Yoneda Lemma}
+
+Let's restrict our attention to this category $\mathbf{A}$
+(to truly appreciate the significance of this result, I encourage you to work out
+the details in full for a slightly larger category!):
+\[ \xymatrix{ A \ar[r]^f & B } \]
+
+The image of this in $\mathbf{Set}$ under $\mbox{hom}(A,-)$ is just
+\[ \xymatrix@C=.5in{ \set{id_A} \ar[r]^{\mbox{hom}(A,f)} & \set{id_B} } \]
+
+Now suppose we have some other functor $F : \mathbf{A} \to \mathbf{Set}$,
+whose image is
+\[ \xymatrix@C=.5in{ \set{a_0, \dots} \ar[r]^{Ff} & \set{b_0, \dots} } \]
+(where $FA = \set{a_0, \dots}$ and $FB = \set{b_0, \dots}$.)
+
+Now, the claim of Proposition 6.18 is that there exists a unique natural
+transformation $\tau : \mbox{hom}(A,-) \stackrel{\cdot}{\to} F$ if we additionally
+constrain $\tau_A(id_A) = a_0$. OK, so, first off: what does that mean? $\tau$ being
+natural means $\forall B,C,g : B \to C$, this commutes:
+\[ \xymatrix{
+ \mbox{hom}(A,B) \ar[r]^{\tau_B} \ar[d]_{\mbox{hom}(A,g)} & FB \ar[d]^{Fg} \\
+ \mbox{hom}(A,C) \ar[r]^{\tau_C} & FC
+} \]
+or more specifically, at $A,B,f$ (first generically, then expanding some computations):
+\[ \xymatrix{
+ \mbox{hom}(A,A) \ar[r]^{\tau_A} \ar[d]_{\mbox{hom}(A,f)} & FA \ar[d]^{Ff} \\
+ \mbox{hom}(A,B) \ar[r]^{\tau_B} & FB
+} \qquad \xymatrix{
+ \set{id_A} \ar[r]^{\tau_A} \ar[d]_{\set{id_A \mapsto f}} & \set{a_0,\dots} \ar[d]^{Ff} \\
+ \set{f} \ar[r]^{\tau_B} & \set{b_0,\dots}
+} \]
+and so requiring $\tau_A(id_A) = a_0$ makes sense. If this is to be natural, it
+must be the case (for all $B$ and $f : A \to B$; note that this works even to
+define $\tau_A$ at inputs other than $id_A$ just as well!) that
+\begin{align*}
+ \tau_B(f) &= \tau_B(f \circ id_A) \\
+ &= \tau_B(\mbox{hom}(A,f)(id_A)) & \forall_x . f \circ x = \mbox{hom}(A,f)(x) \\
+ &= F(f)(\tau_A(id_A)) & \text{naturality of $\tau$} \\
+ &= F(f)(a_0) & \text{requirement}
+\end{align*}
+So $\tau$ is fully determined by naturality and the requirement given,
+precisely because $\mbox{hom}(A,-)$ on arrows captures pre-composition.
+So: given a choice of $a_0 \in FA$, we can fully specify a natural transformation $\tau$.
+
+
+Conversely, given a $\tau'$, it must pick out some $\tau_A(id_A) \in FA$. Therefore,
+the Yoneda lemma:
+
+\begin{quote}{\em
+ Given a functor $F : \mathbf{A} \to \mathbf{Set}$, the set
+ $\set{\tau \middle\vert \tau : \mbox{hom}(A,-) \stackrel{\cdot}{\to} F}$
+ is isomorphic (in $\mathbf{Set}$) to $FA$.
+ The isomorphism is witnessed by the function $Y(\tau) = \tau_A(id_A)$.
+}\end{quote}
+
+\end{document}
projects / ctcheat / commitdiff