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