The equaliser of $f$ and $g$ is the equaliser of $f$ and $g$ in $\mathsf{Sets}$ as in 
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1.
The Limit. The set $\operatorname {\mathrm{Eq}}\webleft (f,g\webright )$ defined by
\[ \operatorname {\mathrm{Eq}}\webleft (f,g\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ a\in A\ \middle |\ f\webleft (a\webright )=g\webleft (a\webright )\right\} . \] -
2.
The Cone. The inclusion map
\[ \operatorname {\mathrm{eq}}\webleft (f,g\webright )\colon \operatorname {\mathrm{Eq}}\webleft (f,g\webright )\hookrightarrow A. \] -
1.
Associativity. We have isomorphisms of sets1
where $\operatorname {\mathrm{Eq}}\webleft (f,g,h\webright )$ is the limit of the diagram\[ \underbrace{\operatorname {\mathrm{Eq}}\webleft (f\circ \operatorname {\mathrm{eq}}\webleft (g,h\webright ),g\circ \operatorname {\mathrm{eq}}\webleft (g,h\webright )\webright )}_{{}=\operatorname {\mathrm{Eq}}\webleft (f\circ \operatorname {\mathrm{eq}}\webleft (g,h\webright ),h\circ \operatorname {\mathrm{eq}}\webleft (g,h\webright )\webright )}\cong \operatorname {\mathrm{Eq}}\webleft (f,g,h\webright ) \cong \underbrace{\operatorname {\mathrm{Eq}}\webleft (f\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright ),h\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright )\webright )}_{{}=\operatorname {\mathrm{Eq}}\webleft (g\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright ),h\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright )\webright )}, \]in $\mathsf{Sets}$, being explicitly given by
\[ \operatorname {\mathrm{Eq}}\webleft (f,g,h\webright )\cong \left\{ a\in A\ \middle |\ f\webleft (a\webright )=g\webleft (a\webright )=h\webleft (a\webright )\right\} . \] -
2.
Unitality. We have an isomorphism of sets
\[ \operatorname {\mathrm{Eq}}\webleft (f,f\webright )\cong A. \] -
3.
Commutativity. We have an isomorphism of sets
\[ \operatorname {\mathrm{Eq}}\webleft (f,g\webright ) \cong \operatorname {\mathrm{Eq}}\webleft (g,f\webright ). \] -
4.
Interaction With Composition. Let
\[ A \underset {g}{\overset {f}{\rightrightarrows }} B \underset {k}{\overset {h}{\rightrightarrows }} C \]be functions. We have an inclusion of sets
\[ \operatorname {\mathrm{Eq}}\webleft (h\circ f\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright ),k\circ g\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright )\webright ) \subset \operatorname {\mathrm{Eq}}\webleft (h\circ f,k\circ g\webright ), \]where $\operatorname {\mathrm{Eq}}\webleft (h\circ f\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright ),k\circ g\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright )\webright )$ is the equaliser of the composition
\[ \operatorname {\mathrm{Eq}}\webleft (f,g\webright )\overset {\operatorname {\mathrm{eq}}\webleft (f,g\webright )}{\hookrightarrow }A\underset {g}{\overset {f}{\rightrightarrows }}B\underset {k}{\overset {h}{\rightrightarrows }}C. \] -
1That is, the following three ways of forming “the” equaliser of $\webleft (f,g,h\webright )$ agree:
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(a)
Take the equaliser of $\webleft (f,g,h\webright )$, i.e. the limit of the diagram
in $\mathsf{Sets}$.
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(b)
First take the equaliser of $f$ and $g$, forming a diagram
\[ \operatorname {\mathrm{Eq}}\webleft (f,g\webright )\overset {\operatorname {\mathrm{eq}}\webleft (f,g\webright )}{\hookrightarrow }A\underset {g}{\overset {f}{\rightrightarrows }}B \]and then take the equaliser of the composition
\[ \operatorname {\mathrm{Eq}}\webleft (f,g\webright )\overset {\operatorname {\mathrm{eq}}\webleft (f,g\webright )}{\hookrightarrow }A\underset {h}{\overset {f}{\rightrightarrows }}B, \]obtaining a subset
\[ \operatorname {\mathrm{Eq}}\webleft (f\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright ),h\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright )\webright )=\operatorname {\mathrm{Eq}}\webleft (g\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright ),h\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright )\webright ) \]of $\operatorname {\mathrm{Eq}}\webleft (f,g\webright )$.
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(c)
First take the equaliser of $g$ and $h$, forming a diagram
\[ \operatorname {\mathrm{Eq}}\webleft (g,h\webright )\overset {\operatorname {\mathrm{eq}}\webleft (g,h\webright )}{\hookrightarrow }A\underset {h}{\overset {g}{\rightrightarrows }}B \]and then take the equaliser of the composition
\[ \operatorname {\mathrm{Eq}}\webleft (g,h\webright )\overset {\operatorname {\mathrm{eq}}\webleft (g,h\webright )}{\hookrightarrow }A\underset {g}{\overset {f}{\rightrightarrows }}B, \]obtaining a subset
\[ \operatorname {\mathrm{Eq}}\webleft (f\circ \operatorname {\mathrm{eq}}\webleft (g,h\webright ),g\circ \operatorname {\mathrm{eq}}\webleft (g,h\webright )\webright )=\operatorname {\mathrm{Eq}}\webleft (f\circ \operatorname {\mathrm{eq}}\webleft (g,h\webright ),h\circ \operatorname {\mathrm{eq}}\webleft (g,h\webright )\webright ) \]of $\operatorname {\mathrm{Eq}}\webleft (g,h\webright )$.
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(a)
4.1.5 Equalisers
Let $A$ and $B$ be sets and let $f,g\colon A\rightrightarrows B$ be functions.
Concretely, the equaliser of $f$ and $g$ is the pair $\webleft (\operatorname {\mathrm{Eq}}\webleft (f,g\webright ),\operatorname {\mathrm{eq}}\webleft (f,g\webright )\webright )$ consisting of:
Proof of Construction 4.1.5.1.2.
We claim that $\operatorname {\mathrm{Eq}}\webleft (f,g\webright )$ 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
\[ f\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright )=g\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright ), \]
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
\[ \operatorname {\mathrm{eq}}\webleft (f,g\webright )\circ \phi =e \]
via
\[ \phi \webleft (x\webright )=e\webleft (x\webright ) \]
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
\[ f\circ e=g\circ e, \]
which gives
\[ f\webleft (e\webleft (x\webright )\webright )=g\webleft (e\webleft (x\webright )\webright ) \]
for each $x\in E$, so that $e\webleft (x\webright )\in \operatorname {\mathrm{Eq}}\webleft (f,g\webright )$.
Let $A$, $B$, and $C$ be sets.
Proof of Proposition 4.1.5.1.3.
Item 1: Associativity
We first prove that $\operatorname {\mathrm{Eq}}\webleft (f,g,h\webright )$ is indeed given by
\[ \operatorname {\mathrm{Eq}}\webleft (f,g,h\webright )\cong \left\{ a\in A\ \middle |\ f\webleft (a\webright )=g\webleft (a\webright )=h\webleft (a\webright )\right\} . \]
Indeed, suppose we have a diagram of the form
\[ \operatorname {\mathrm{eq}}\webleft (f,g\webright )\circ \phi =e \]
being necessarily given by
\[ \phi \webleft (x\webright )=e\webleft (x\webright ) \]
for each $x\in E$, where we note that $e\webleft (x\webright )\in A$ indeed lies in $\operatorname {\mathrm{Eq}}\webleft (f,g,h\webright )$ by the condition
\[ f\circ e=g\circ e=h\circ e, \]
which gives
\[ f\webleft (e\webleft (x\webright )\webright )=g\webleft (e\webleft (x\webright )\webright )=h\webleft (e\webleft (x\webright )\webright ) \]
for each $x\in E$, so that $e\webleft (x\webright )\in \operatorname {\mathrm{Eq}}\webleft (f,g,h\webright )$.
We now check the equalities
\[ \operatorname {\mathrm{Eq}}\webleft (f\circ \operatorname {\mathrm{eq}}\webleft (g,h\webright ),g\circ \operatorname {\mathrm{eq}}\webleft (g,h\webright )\webright )\cong \operatorname {\mathrm{Eq}}\webleft (f,g,h\webright ) \cong \operatorname {\mathrm{Eq}}\webleft (f\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright ),h\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright )\webright ). \]
Indeed, we have
\begin{align*} \operatorname {\mathrm{Eq}}\webleft (f\circ \operatorname {\mathrm{eq}}\webleft (g,h\webright ),g\circ \operatorname {\mathrm{eq}}\webleft (g,h\webright )\webright ) & \cong \left\{ x\in \operatorname {\mathrm{Eq}}\webleft (g,h\webright )\ \middle |\ \webleft [f\circ \operatorname {\mathrm{eq}}\webleft (g,h\webright )\webright ]\webleft (a\webright )=\webleft [g\circ \operatorname {\mathrm{eq}}\webleft (g,h\webright )\webright ]\webleft (a\webright )\right\} \\ & \cong \left\{ x\in \operatorname {\mathrm{Eq}}\webleft (g,h\webright )\ \middle |\ f\webleft (a\webright )=g\webleft (a\webright )\right\} \\ & \cong \left\{ x\in A\ \middle |\ \text{$f\webleft (a\webright )=g\webleft (a\webright )$ and $g\webleft (a\webright )=h\webleft (a\webright )$}\right\} \\ & \cong \left\{ x\in A\ \middle |\ \text{$f\webleft (a\webright )=g\webleft (a\webright )=h\webleft (a\webright )$}\right\} \\ & \cong \operatorname {\mathrm{Eq}}\webleft (f,g,h\webright ).\end{align*}
\begin{align*} \operatorname {\mathrm{Eq}}\webleft (f\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright ),h\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright )\webright ) & \cong \left\{ x\in \operatorname {\mathrm{Eq}}\webleft (f,g\webright )\ \middle |\ \webleft [f\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright )\webright ]\webleft (a\webright )=\webleft [h\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright )\webright ]\webleft (a\webright )\right\} \\ & \cong \left\{ x\in \operatorname {\mathrm{Eq}}\webleft (f,g\webright )\ \middle |\ f\webleft (a\webright )=h\webleft (a\webright )\right\} \\ & \cong \left\{ x\in A\ \middle |\ \text{$f\webleft (a\webright )=h\webleft (a\webright )$ and $f\webleft (a\webright )=g\webleft (a\webright )$}\right\} \\ & \cong \left\{ x\in A\ \middle |\ \text{$f\webleft (a\webright )=g\webleft (a\webright )=h\webleft (a\webright )$}\right\} \\ & \cong \operatorname {\mathrm{Eq}}\webleft (f,g,h\webright ).\end{align*}
Item 2: Unitality
Indeed, we have
\begin{align*} \operatorname {\mathrm{Eq}}\webleft (f,f\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ a\in A\ \middle |\ f\webleft (a\webright )=f\webleft (a\webright )\right\} \\ & = A. \end{align*}
Item 3: Commutativity
Indeed, we have
\begin{align*} \operatorname {\mathrm{Eq}}\webleft (f,g\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ a\in A\ \middle |\ f\webleft (a\webright )=g\webleft (a\webright )\right\} \\ & = \left\{ a\in A\ \middle |\ g\webleft (a\webright )=f\webleft (a\webright )\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\operatorname {\mathrm{Eq}}\webleft (g,f\webright ). \end{align*}
Item 4: Interaction With Composition
Indeed, we have
\begin{align*} \operatorname {\mathrm{Eq}}\webleft (h\circ f\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright ),k\circ g\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright )\webright ) & \cong \left\{ a\in \operatorname {\mathrm{Eq}}\webleft (f,g\webright )\ \middle |\ h\webleft (f\webleft (a\webright )\webright )=k\webleft (g\webleft (a\webright )\webright )\right\} \\ & \cong \left\{ a\in A\ \middle |\ \text{$f\webleft (a\webright )=g\webleft (a\webright )$ and $h\webleft (f\webleft (a\webright )\webright )=k\webleft (g\webleft (a\webright )\webright )$}\right\} .\end{align*}
\[ \operatorname {\mathrm{Eq}}\webleft (h\circ f,k\circ g\webright )\cong \left\{ a\in A\ \middle |\ h\webleft (f\webleft (a\webright )\webright )=k\webleft (g\webleft (a\webright )\webright )\right\} , \]
and thus there’s an inclusion from $\operatorname {\mathrm{Eq}}\webleft (h\circ f\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright ),k\circ g\circ \operatorname {\mathrm{eq}}\webleft (f,g\webright )\webright )$ to $\operatorname {\mathrm{Eq}}\webleft (h\circ f,k\circ g\webright )$.
