\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\\ \dToe\\ 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!\\ 1&\rCong&1&\rCong&1\\ \end{diagram} \end{document}