The equaliser of $(f,g)$ is the equaliser of $f$ and $g$ in $\mathsf{Sets}_{*}$ as in 
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The Limit. The pointed set $(\operatorname {\mathrm{Eq}}(f,g),x_{0})$.
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The Cone. The morphism of pointed sets
\[ \operatorname {\mathrm{eq}}(f,g)\colon (\operatorname {\mathrm{Eq}}(f,g),x_{0})\hookrightarrow (X,x_{0}) \]given by the canonical inclusion $\operatorname {\mathrm{eq}}(f,g)\hookrightarrow \operatorname {\mathrm{Eq}}(f,g)\hookrightarrow X$.
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1.
Associativity. We have isomorphisms of pointed sets
where $\operatorname {\mathrm{Eq}}(f,g,h)$ is the limit of the diagram\[ \underbrace{\operatorname {\mathrm{Eq}}(f\circ \operatorname {\mathrm{eq}}(g,h),g\circ \operatorname {\mathrm{eq}}(g,h))}_{{}=\operatorname {\mathrm{Eq}}(f\circ \operatorname {\mathrm{eq}}(g,h),h\circ \operatorname {\mathrm{eq}}(g,h))}\cong \operatorname {\mathrm{Eq}}(f,g,h) \cong \underbrace{\operatorname {\mathrm{Eq}}(f\circ \operatorname {\mathrm{eq}}(f,g),h\circ \operatorname {\mathrm{eq}}(f,g))}_{{}=\operatorname {\mathrm{Eq}}(g\circ \operatorname {\mathrm{eq}}(f,g),h\circ \operatorname {\mathrm{eq}}(f,g))}, \]in $\mathsf{Sets}_{*}$, being explicitly given by
\[ \operatorname {\mathrm{Eq}}(f,g,h)\cong \left\{ a\in A\ \middle |\ f(a)=g(a)=h(a)\right\} . \] -
2.
Unitality. We have an isomorphism of pointed sets
\[ \operatorname {\mathrm{Eq}}(f,f)\cong X. \] -
3.
Commutativity. We have an isomorphism of pointed sets
\[ \operatorname {\mathrm{Eq}}(f,g) \cong \operatorname {\mathrm{Eq}}(g,f). \]
6.2.5 Equalisers
Let $f,g\colon (X,x_{0})\rightrightarrows (Y,y_{0})$ be morphisms of pointed sets.
Concretely, the equaliser of $(f,g)$ is the pair consisting of:
Proof of Construction 6.2.5.1.2.
We claim that $(\operatorname {\mathrm{Eq}}(f,g),x_{0})$ 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
\[ f\circ \operatorname {\mathrm{eq}}(f,g)=g\circ \operatorname {\mathrm{eq}}(f,g), \]
which indeed holds by the definition of the set $\operatorname {\mathrm{Eq}}(f,g)$. Next, we prove that $\operatorname {\mathrm{Eq}}(f,g)$ satisfies the universal property of the equaliser. Suppose we have a diagram of the form
\[ \phi \colon (E,*)\to (\operatorname {\mathrm{Eq}}(f,g),x_{0}) \]
making the diagram
\[ \operatorname {\mathrm{eq}}(f,g)\circ \phi =e \]
via
\[ \phi (x)=e(x) \]
for each $x\in E$, where we note that $e(x)\in A$ indeed lies in $\operatorname {\mathrm{Eq}}(f,g)$ by the condition
\[ f\circ e=g\circ e, \]
which gives
\[ f(e(x))=g(e(x)) \]
for each $x\in E$, so that $e(x)\in \operatorname {\mathrm{Eq}}(f,g)$. Lastly, we note that $\phi $ is indeed a morphism of pointed sets, as we have
\begin{align*} \phi (*) & = e(*)\\ & = x_{0},\end{align*}
where we have used that $e$ 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.
Proof of Proposition 6.2.5.1.3.
Item 1: Associativity
This follows from Chapter 4: Constructions With Sets, Item 1 of Proposition 4.1.5.1.3.
Item 2: Unitality
This follows from Chapter 4: Constructions With Sets, Item 2 of Proposition 4.1.5.1.3.
Item 3: Commutativity
This follows from Chapter 4: Constructions With Sets, Item 3 of Proposition 4.1.5.1.3.
