Definition 6.3.5.1.1 (00BQ): Coequalisers of Pointed Sets

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

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

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

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

We claim that $\webleft (\operatorname {\mathrm{CoEq}}\webleft (f,g\webright ),\webleft [y_{0}\webright ]\webright )$ 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

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

Indeed, we have

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

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

in $\mathsf{Sets}$. Then, since $c\webleft (f\webleft (a\webright )\webright )=c\webleft (g\webleft (a\webright )\webright )$ 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 $\phi \colon \operatorname {\mathrm{CoEq}}\webleft (f,g\webright )\overset {\exists !}{\to }C$ making the diagram

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

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

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

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

1.
Associativity. We have isomorphisms of pointed sets

\[ \underbrace{\operatorname {\mathrm{CoEq}}\webleft (\operatorname {\mathrm{coeq}}\webleft (f,g\webright )\circ f,\operatorname {\mathrm{coeq}}\webleft (f,g\webright )\circ h\webright )}_{{}=\operatorname {\mathrm{CoEq}}\webleft (\operatorname {\mathrm{coeq}}\webleft (f,g\webright )\circ g,\operatorname {\mathrm{coeq}}\webleft (f,g\webright )\circ h\webright )}\cong \operatorname {\mathrm{CoEq}}\webleft (f,g,h\webright ) \cong \underbrace{\operatorname {\mathrm{CoEq}}\webleft (\operatorname {\mathrm{coeq}}\webleft (g,h\webright )\circ f,\operatorname {\mathrm{coeq}}\webleft (g,h\webright )\circ g\webright )}_{{}=\operatorname {\mathrm{CoEq}}\webleft (\operatorname {\mathrm{coeq}}\webleft (g,h\webright )\circ f,\operatorname {\mathrm{coeq}}\webleft (g,h\webright )\circ h\webright )}, \]
where $\operatorname {\mathrm{CoEq}}\webleft (f,g,h\webright )$ is the colimit of the diagram
in $\mathsf{Sets}_{*}$.
2.
Unitality. We have an isomorphism of pointed sets

\[ \operatorname {\mathrm{CoEq}}\webleft (f,f\webright )\cong B. \]
3.
Commutativity. We have an isomorphism of pointed sets

\[ \operatorname {\mathrm{CoEq}}\webleft (f,g\webright ) \cong \operatorname {\mathrm{CoEq}}\webleft (g,f\webright ). \]

The Clowder Project

6.3.5 Coequalisers