The coequaliser of $(f,g)$ is the pointed set $(\operatorname {\mathrm{CoEq}}(f,g),[y_{0}])$.
6.3.5 Coequalisers
Let $f,g\colon (X,x_{0})\rightrightarrows (Y,y_{0})$ be morphisms of pointed sets.
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.
Proof of Construction 6.3.5.1.2.
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
\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.
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Associativity. We have isomorphisms of pointed sets
where $\operatorname {\mathrm{CoEq}}(f,g,h)$ is the colimit of the diagram\[ \underbrace{\operatorname {\mathrm{CoEq}}(\operatorname {\mathrm{coeq}}(f,g)\circ f,\operatorname {\mathrm{coeq}}(f,g)\circ h)}_{{}=\operatorname {\mathrm{CoEq}}(\operatorname {\mathrm{coeq}}(f,g)\circ g,\operatorname {\mathrm{coeq}}(f,g)\circ h)}\cong \operatorname {\mathrm{CoEq}}(f,g,h) \cong \underbrace{\operatorname {\mathrm{CoEq}}(\operatorname {\mathrm{coeq}}(g,h)\circ f,\operatorname {\mathrm{coeq}}(g,h)\circ g)}_{{}=\operatorname {\mathrm{CoEq}}(\operatorname {\mathrm{coeq}}(g,h)\circ f,\operatorname {\mathrm{coeq}}(g,h)\circ h)}, \]in $\mathsf{Sets}_{*}$.
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Unitality. We have an isomorphism of pointed sets
\[ \operatorname {\mathrm{CoEq}}(f,f)\cong B. \] -
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Commutativity. We have an isomorphism of pointed sets
\[ \operatorname {\mathrm{CoEq}}(f,g) \cong \operatorname {\mathrm{CoEq}}(g,f). \]
Proof of Proposition 6.3.5.1.3.
Item 1: Associativity
This follows from Chapter 4: Constructions With Sets, Item 1 of Proposition 4.2.5.1.5.
Item 2: Unitality
This follows from Chapter 4: Constructions With Sets, Item 2 of Proposition 4.2.5.1.5.
Item 3: Commutativity
This follows from Chapter 4: Constructions With Sets, Item 3 of Proposition 4.2.5.1.5.
