\documentclass[10pt,letterpaper]{article}
\input homework.tex
\usepackage{amssymb}
\usepackage{color}
\def\sc{{\bf Sc}}
\def\A{{\goth A}}
\def\B{{\goth B}}
\begin{document}
\title{Notes on Split Coequalizers}
\author{Jason Reed}
\maketitle

Let $\sc$ be the category whose objects are $\A, \B, 0$ and whose
arrows are
$$\delta_0,\delta_1 : \A\to \B$$
$$\Delta_0,\Delta_1,\id_\A : \A\to \A$$
$$\sigma : \B \to \A$$
$$\id_\B : \B\to \B$$
$$\{0_{M,N} : M \to N \st M,N \in \{\A, \B, 0\}\}$$
The composition law is given as follows:
$$\delta_0 \sigma = \id_\B \qquad \delta_1 \sigma = 0_{\B\B}$$
$$\delta_0 \Delta_i = \delta_i \qquad \delta_1 \Delta_i = 0_{\A\B}$$
$$ \sigma \delta_i = \Delta_i \qquad $$
$$\Delta_0 \Delta_i = \Delta_i\qquad \Delta_1 \Delta_i = 0_{\A\A}$$
$$f0_{M,N} = 0_{M,\cod f} \qquad 0_{M,N}f = 0_{\dom f, N}$$
$\sc$ sort of looks like this, without all the zeroes:
\begin{diagram}
\A&\pile{\rTo^{\delta_0} \\ \rTo_{\delta_1}}&\B\\
\uTo<{\Delta_0} \uTo>{\Delta_1}&&\dCong\\
\A&\lTo_\sigma&\B\\
\end{diagram}
A {\em fork} is a diagram
\begin{diagram}
A&\pile{\rTo^{d_0} \\ \rTo_{d_1}}&B&\rTo^e&C
\end{diagram}
such that $ed_0 = ed_1$. A {\em split fork} is a fork together with a {\em splitting}
\begin{diagram}
A&\lTo^{s}&B&\lTo^t&C
\end{diagram}
such that
\begin{diagram}
B&\rTo^s&A&\rTo^{d_0}&B\\
\dTo<e&&\dTo<{d_1}&&\dTo>e\\
C&\rTo_t&B&\rTo_e&C\\
\end{diagram}
commutes, and the top two rows compose to identities.
\begin{proposition}
$\sc$ is isomorphic to the category generated by the graph
\begin{diagram}
G &:= & A&\pile{\rTo~{d_0} \\ \rTo~{d_1} \\ \lTo~s}&B&\pile{\rTo~e \\ \lTo~t}&C
\end{diagram}
with the identities $ed_0 = ed_1$, $d_1s = te$, $d_0s = \id_B$, and $et = \id_C$.
\end{proposition}
\begin{proof}
\def\2{{\bf 2}}
Let $F \colon \Grph \to \Cat$ be the left adjoint of the forgetful functor
$U\colon \Cat \to \Grph$. Let $J = \2 + \2 + \2 + \2$, and define $f_0, f_1 : J \to FG$
by $f_0 := [ed_0, d_1s, d_0s, et]$ and $f_1 := [ed_1, te, \id_B, \id_C]$. (That is, each arrow
is defined by a copairing of arrows from $\2$ into $FG$, i.e. particular paths in $G$) Then $Q$, the category
generated by $G$ subject to the identities is by definition the coequalizer of $f_0,f_1$.
\begin{diagram}
J&\pile{\rTo^{f_0} \\ \rTo_{f_1}}&F G&\rTo^q&Q
\end{diagram}
To show $Q \cong \sc$, it sufficies that $\sc$ has the universal property of the coequalizer.
We need a $r \colon FG \to \sc$, and for every $h \colon FG \to H$ such that $hf_0 = hf_1$, we want to show that
there is a unique $k$ such that
\begin{diagram}
&&H\\
&\ruTo^h&\uDashto>k\\       
F G&\rTo_r&\sc
\end{diagram}
commutes. Since $FG$ is a free category, we only need to specify $r$'s action on $G$.
The action on objects is obvious: we take $A$ to $\A$, $B$ to $\B$, $C$ to $0$.
The action of arrows is equally apparent: we take $d_i$ to $\delta_i$, $s$ to $\sigma$,
$e$ to $0_{\B,0}$ and $t$ to $0_{0,\B}$.

Now if we are to have $kr = h$, then we must have $k(r(x)) = h(x)$ for all $x\in \{d_0,d_1,s,e,t\}$.
That is,
$$k(\delta_i) = h(d_i)$$
$$k(\sigma) = h(s)$$
$$k(0_{\B,0}) = h(e)$$
$$k(0_{0,\B}) = h(t)$$
If $k$ is to be a functor, then
$$k(\Delta_i) = k(\sigma \delta_i) = k(\sigma) k(\delta_i) = h(s) h(d_i)$$
$$k(0_{0,\A}) = k(\sigma 0_{0,\B}) = k(\sigma) k(0_{0,\B}) = h(s) h(t)$$
$$k(0_{\A,0}) = k(0_{\B,0}\delta_0) = k(0_{\B,0})k(\delta_0) = h(e) h(d_0)$$
$$k(0_{\A,\B}) = k(0_{0,\B})k(0_{\A,0}) = h(t)h(e) h(d_0)$$
$$k(0_{\B,\A}) = k(0_{0,\A})k(0_{\B,0}) = h(s)h(t)h(e)$$
$$k(0_{0,0}) = k(0_{\B,0})k(0_{0,\B}) = h(e)h(t)$$
$$k(0_{\A,\A}) = k(0_{0,\A})k(0_{\A,0}) = h(s)h(t)h(e)h(d_0)$$
$$k(0_{\B,\B}) = k(0_{0,\B})k(0_{\B,0}) = h(t)h(e)$$
Using the fact that $h$ coequalizes $f_0, f_1$, we can show that this indeed defines a functor. \cqed
\end{proof}

So of course any split coequalizer is an arrow out of $FG$ that coequalizes $f_0,f_1$, hence
factors uniquely through $r$. So the split coequalizers in $\Sets$, say, are the presheaves
$\Sets^{\sc} = \Sets^{(\sc^\op)^\op}$. The structure of the full subcategory of representable presheaves
looks like the following: 
(4 stands for $\{0_{AA},\Delta_1,\Delta_0,\id_A\}$ and 3 stands for $\{0_{AB},\delta_1,\delta_0\}$)

\begin{diagram}
4&\pile{\rTo^{0122}\\\rTo_{0001}}&3&\rTo^!&1\\
\uTo<{02}\uTo>{00}&&\uTo<{02}\uTo>{00}&&\uCong\\
2&\pile{\rCong\\\rTo_{00}}&2&\rTo^!&1\\
\uTo<0&&\uTo<0&&\uCong\\
1&\pile{\rCong\\\rCong}&1&\lCong&1\\
\end{diagram}

\begin{diagram}
4&\lInto&3&\lInto&1\\
\dTo<{0011}&&\dTo~{001}&&\dCong\\
2&\lCong&2&\lInto&1\\
\dTo<!&&\dTo<!&&\dTo>!\\
1&\rCong&1&\rCong&1\\
\end{diagram}
\end{document}
