The equaliser of $\webleft (f,g\webright )$ is the equaliser of $f$ and $g$ in $\mathsf{Sets}_{*}$ as in ,
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6.2.5 Equalisers
Let $f,g\colon \webleft (X,x_{0}\webright )\rightrightarrows \webleft (Y,y_{0}\webright )$ be morphisms of pointed sets.
Concretely, the equaliser of $\webleft (f,g\webright )$ is the pair consisting of:
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The Limit. The pointed set $\webleft (\operatorname {\mathrm{Eq}}\webleft (f,g\webright ),x_{0}\webright )$.
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The Cone. The morphism of pointed sets
\[ \operatorname {\mathrm{eq}}\webleft (f,g\webright )\colon \webleft (\operatorname {\mathrm{Eq}}\webleft (f,g\webright ),x_{0}\webright )\hookrightarrow \webleft (X,x_{0}\webright ) \]given by the canonical inclusion $\operatorname {\mathrm{eq}}\webleft (f,g\webright )\hookrightarrow \operatorname {\mathrm{Eq}}\webleft (f,g\webright )\hookrightarrow X$.
We claim that $\webleft (\operatorname {\mathrm{Eq}}\webleft (f,g\webright ),x_{0}\webright )$ is the categorical equaliser of $f$ and $g$ in $\mathsf{Sets}_{*}$. First we need to check that the relevant equaliser diagram commutes, i.e. that we have
which indeed holds by the definition of the set $\operatorname {\mathrm{Eq}}\webleft (f,g\webright )$. Next, we prove that $\operatorname {\mathrm{Eq}}\webleft (f,g\webright )$ satisfies the universal property of the equaliser. Suppose we have a diagram of the form
making the diagram
via
for each $x\in E$, where we note that $e\webleft (x\webright )\in A$ indeed lies in $\operatorname {\mathrm{Eq}}\webleft (f,g\webright )$ by the condition
which gives
for each $x\in E$, so that $e\webleft (x\webright )\in \operatorname {\mathrm{Eq}}\webleft (f,g\webright )$. Lastly, we note that $\phi $ is indeed a morphism of pointed sets, as we have
where we have used that $e$ 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.
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Associativity. We have isomorphisms of pointed sets
\[ \underbrace{\operatorname {\mathrm{Eq}}\webleft (f\circ \operatorname {\mathrm{eq}}\webleft (g,h\webright ),g\circ \operatorname {\mathrm{eq}}\webleft (g,h\webright )\webright )}_{{}=\operatorname {\mathrm{Eq}}\webleft (f\circ \operatorname {\mathrm{eq}}\webleft (g,h\webright ),h\circ \operatorname {\mathrm{eq}}\webleft (g,h\webright )\webright )}\cong \operatorname {\mathrm{Eq}}\webleft (f,g,h\webright ) \cong \underbrace{\operatorname {\mathrm{Eq}}\webleft (f\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright ),h\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright )\webright )}_{{}=\operatorname {\mathrm{Eq}}\webleft (g\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright ),h\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright )\webright )}, \]in $\mathsf{Sets}_{*}$, being explicitly given by\[ \operatorname {\mathrm{Eq}}\webleft (f,g,h\webright )\cong \left\{ a\in A\ \middle |\ f\webleft (a\webright )=g\webleft (a\webright )=h\webleft (a\webright )\right\} . \] -
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Unitality. We have an isomorphism of pointed sets
\[ \operatorname {\mathrm{Eq}}\webleft (f,f\webright )\cong X. \] -
3.
Commutativity. We have an isomorphism of pointed sets
\[ \operatorname {\mathrm{Eq}}\webleft (f,g\webright ) \cong \operatorname {\mathrm{Eq}}\webleft (g,f\webright ). \]