The coequaliser of $\webleft (f,g\webright )$ is the pointed set $\webleft (\text{CoEq}\webleft (f,g\webright ),\webleft [y_{0}\webright ]\webright )$.
Proof of Definition 3.3.5.1.1.
We claim that $\webleft (\text{CoEq}\webleft (f,g\webright ),\webleft [y_{0}\webright ]\webright )$ 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
\[ \text{coeq}\webleft (f,g\webright )\circ f=\text{coeq}\webleft (f,g\webright )\circ g. \]
Indeed, we have
\begin{align*} \webleft [\text{coeq}\webleft (f,g\webright )\circ f\webright ]\webleft (x\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [\text{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 [\text{coeq}\webleft (f,g\webright )\webright ]\webleft (g\webleft (x\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [\text{coeq}\webleft (f,g\webright )\circ g\webright ]\webleft (x\webright )\end{align*}
for each $x\in X$. Next, we prove that $\text{CoEq}\webleft (f,g\webright )$ 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 7: Equivalence Relations and Apartness Relations, Item 4 and Item 5 of Proposition 7.5.2.1.3 that there exists a unique map $\phi \colon \text{CoEq}\webleft (f,g\webright )\overset {\exists !}{\to }C$ 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 \text{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.
