Item 1 (00BS) — The Clowder Project

Let $f,g\colon (X,x_{0})\rightrightarrows (Y,y_{0})$ be morphisms of pointed sets.

The coequaliser of $(f,g)$ is the pointed set $(\operatorname {\mathrm{CoEq}}(f,g),[y_{0}])$.

The coequaliser of $(f,g)$ is the pair $\smash {((\operatorname {\mathrm{CoEq}}(f,g),[y_{0}]),\operatorname {\mathrm{coeq}}(f,g))}$ consisting of:

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The Colimit. The pointed set $(\operatorname {\mathrm{CoEq}}(f,g),[y_{0}])$, where $\operatorname {\mathrm{CoEq}}(f,g)$ is the coequaliser of $f$ and $g$ as in Chapter 4: Constructions With Sets, Definition 4.2.5.1.1.
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The Cocone. The map

\[ \operatorname {\mathrm{coeq}}(f,g)\colon Y\twoheadrightarrow (\operatorname {\mathrm{CoEq}}(f,g),[y_{0}]) \]

given by the quotient map, as in Chapter 4: Constructions With Sets, Item 2 of Construction 4.2.5.1.2.

We claim that $(\operatorname {\mathrm{CoEq}}(f,g),[y_{0}])$ is the categorical coequaliser of $f$ and $g$ in $\mathsf{Sets}_{*}$. First we need to check that the relevant coequaliser diagram commutes, i.e. that we have

\[ \operatorname {\mathrm{coeq}}(f,g)\circ f=\operatorname {\mathrm{coeq}}(f,g)\circ g. \]

Indeed, we have

\begin{align*} [\operatorname {\mathrm{coeq}}(f,g)\circ f](x) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[\operatorname {\mathrm{coeq}}(f,g)](f(x))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[f(x)]\\ & = [g(x)]\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[\operatorname {\mathrm{coeq}}(f,g)](g(x))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[\operatorname {\mathrm{coeq}}(f,g)\circ g](x)\end{align*}

for each $x\in X$. Next, we prove that $\operatorname {\mathrm{CoEq}}(f,g)$ satisfies the universal property of the coequaliser. Suppose we have a diagram of the form

in $\mathsf{Sets}$. Then, since $c(f(a))=c(g(a))$ for each $a\in A$, it follows from Chapter 10: Conditions on Relations, Item 4 and Item 5 of Proposition 10.6.2.1.3 that there exists a unique map $\phi \colon \operatorname {\mathrm{CoEq}}(f,g)\overset {\exists !}{\to }C$ making the diagram

commute, where we note that $\phi $ is indeed a morphism of pointed sets since

\begin{align*} \phi ([y_{0}]) & = [\phi \circ \operatorname {\mathrm{coeq}}(f,g)]([y_{0}])\\ & = c([y_{0}])\\ & = *, \end{align*}

where we have used that $c$ is a morphism of pointed sets.

Let $(X,x_{0})$ and $(Y,y_{0})$ be pointed sets and let $f,g,h\colon (X,x_{0})\to (Y,y_{0})$ be morphisms of pointed sets.

6.3.5 Coequalisers