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yoneda.tex
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\documentclass[10pt,letterpaper]{article}
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\usepackage{amsmath,amssymb,amsthm,latexsym}
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\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 post-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 post-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 post-composition, 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
\[ \xymatrix@C=.5in{ \set{id_A} \ar[r]^{\mbox{hom}(A,f)} & \set{f} } \]
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 post-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}