Subsection 2.2.5 (002M): Coequalisers

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2.2.5 Coequalisers

Let $A$ and $B$ be sets and let $f,g\colon A\rightrightarrows B$ be functions.

Definition 2.2.5.1.1 ▶ Coequalisers of Sets

The coequaliser of $f$ and $g$ is the coequaliser of $f$ and $g$ in $\mathsf{Sets}$ as in , .

Construction 2.2.5.1.2 ▶ Construction of Coequalisers of Sets

The coequaliser of $f$ and $g$ is the pair $\webleft (\text{CoEq}\webleft (f,g\webright ),\text{coeq}\webleft (f,g\webright )\webright )$ consisting of:

The Colimit. The set $\text{CoEq}\webleft (f,g\webright )$ defined by
\[ \text{CoEq}\webleft (f,g\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}B/\mathord {\sim }, \]

where $\mathord {\sim }$ is the equivalence relation on $B$ generated by $f\webleft (a\webright )\sim g\webleft (a\webright )$.
The Cocone. The map
\[ \text{coeq}\webleft (f,g\webright )\colon B\to \text{CoEq}\webleft (f,g\webright ) \]

given by the quotient map $\pi \colon B\twoheadrightarrow B/\mathord {\sim }$ with respect to the equivalence relation generated by $f\webleft (a\webright )\sim g\webleft (a\webright )$.

Proof of Construction 2.2.5.1.2.

We claim that $\text{CoEq}\webleft (f,g\webright )$ 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

\[ \text{coeq}\webleft (f,g\webright )\circ f=\text{coeq}\webleft (f,g\webright )\circ g. \]

Indeed, we have

\begin{align*} \webleft [\text{coeq}\webleft (f,g\webright )\circ f\webright ]\webleft (a\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [\text{coeq}\webleft (f,g\webright )\webright ]\webleft (f\webleft (a\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [f\webleft (a\webright )\webright ]\\ & = \webleft [g\webleft (a\webright )\webright ]\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [\text{coeq}\webleft (f,g\webright )\webright ]\webleft (g\webleft (a\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [\text{coeq}\webleft (f,g\webright )\circ g\webright ]\webleft (a\webright )\end{align*}

for each $a\in A$. Next, we prove that $\text{CoEq}\webleft (f,g\webright )$ satisfies the universal property of the coequaliser. Suppose we have a diagram of the form

in $\mathsf{Sets}$. Then, since $c\webleft (f\webleft (a\webright )\webright )=c\webleft (g\webleft (a\webright )\webright )$ for each $a\in A$, it follows from , and of that there exists a unique map $\text{CoEq}\webleft (f,g\webright )\overset {\exists !}{\to }C$ making the diagram

commute.

Remark 2.2.5.1.3 ▶ Unwinding Definition 2.2.5.1.1

In detail, by , , the relation $\mathord {\sim }$ of Definition 2.2.5.1.1 is given by declaring $a\sim b$ iff one of the following conditions is satisfied:

We have $a=b$;
There exist $x_{1},\ldots ,x_{n}\in B$ such that $a\sim 'x_{1}\sim '\cdots \sim 'x_{n}\sim 'b$, where we declare $x\sim 'y$ if one of the following conditions is satisfied:
1. There exists $z\in A$ such that $x=f\webleft (z\webright )$ and $y=g\webleft (z\webright )$.
2. There exists $z\in A$ such that $x=g\webleft (z\webright )$ and $y=f\webleft (z\webright )$.
That is: we require the following condition to be satisfied:
- There exist $x_{1},\ldots ,x_{n}\in B$ satisfying the following conditions:
  1. There exists $z_{0}\in A$ satisfying one of the following conditions:
    1. We have $a=f\webleft (z_{0}\webright )$ and $x_{1}=g\webleft (z_{0}\webright )$.
    2. We have $a=g\webleft (z_{0}\webright )$ and $x_{1}=f\webleft (z_{0}\webright )$.
  2. For each $1\leq i\leq n-1$, there exists $z_{i}\in A$ satisfying one of the following conditions:
    1. We have $x_{i}=f\webleft (z_{i}\webright )$ and $x_{i+1}=g\webleft (z_{i}\webright )$.
    2. We have $x_{i}=g\webleft (z_{i}\webright )$ and $x_{i+1}=f\webleft (z_{i}\webright )$.
  3. There exists $z_{n}\in A$ satisfying one of the following conditions:
    1. We have $x_{n}=f\webleft (z_{n}\webright )$ and $b=g\webleft (z_{n}\webright )$.
    2. We have $x_{n}=g\webleft (z_{n}\webright )$ and $b=f\webleft (z_{n}\webright )$.

Example 2.2.5.1.4 ▶ Examples of Coequalisers of Sets

Here are some examples of coequalisers of sets.

Quotients by Equivalence Relations. Let $R$ be an equivalence relation on a set $X$. We have a bijection of sets
\[ X/\mathord {\sim }_{R}\cong \text{CoEq}\webleft (R\hookrightarrow X\times X\underset {\text{pr}_{2}}{\overset {\text{pr}_{1}}{\rightrightarrows }}X\webright ). \]

Proof of Example 2.2.5.1.4.

Item 1: Quotients by Equivalence Relations

See [Proof Wiki, Quotient Mapping is Coequalizer].

Proposition 2.2.5.1.5 ▶ Properties of Coequalisers of Sets

Let $A$, $B$, and $C$ be sets.

Associativity. We have isomorphisms of sets¹

\[ \underbrace{\text{CoEq}\webleft (\text{coeq}\webleft (f,g\webright )\circ f,\text{coeq}\webleft (f,g\webright )\circ h\webright )}_{{}=\text{CoEq}\webleft (\text{coeq}\webleft (f,g\webright )\circ g,\text{coeq}\webleft (f,g\webright )\circ h\webright )}\cong \text{CoEq}\webleft (f,g,h\webright ) \cong \underbrace{\text{CoEq}\webleft (\text{coeq}\webleft (g,h\webright )\circ f,\text{coeq}\webleft (g,h\webright )\circ g\webright )}_{{}=\text{CoEq}\webleft (\text{coeq}\webleft (g,h\webright )\circ f,\text{coeq}\webleft (g,h\webright )\circ h\webright )}, \]

where $\text{CoEq}\webleft (f,g,h\webright )$ is the colimit of the diagram

in $\mathsf{Sets}$.
Unitality. We have an isomorphism of sets
\[ \text{CoEq}\webleft (f,f\webright )\cong B. \]
Commutativity. We have an isomorphism of sets
\[ \text{CoEq}\webleft (f,g\webright ) \cong \text{CoEq}\webleft (g,f\webright ). \]
Interaction With Composition. Let
\[ A \underset {g}{\overset {f}{\rightrightarrows }} B \underset {k}{\overset {h}{\rightrightarrows }} C \]

be functions. We have a surjection

\[ \text{CoEq}\webleft (h\circ f,k\circ g\webright )\twoheadrightarrow \text{CoEq}\webleft (\text{coeq}\webleft (h,k\webright )\circ h\circ f,\text{coeq}\webleft (h,k\webright )\circ k\circ g\webright ) \]

exhibiting $\text{CoEq}\webleft (\text{coeq}\webleft (h,k\webright )\circ h\circ f,\text{coeq}\webleft (h,k\webright )\circ k\circ g\webright )$ as a quotient of $\text{CoEq}\webleft (h\circ f,k\circ g\webright )$ by the relation generated by declaring $h\webleft (y\webright )\sim k\webleft (y\webright )$ for each $y\in B$.

¹That is, the following three ways of forming “the” coequaliser of $\webleft (f,g,h\webright )$ agree:

Take the coequaliser of $\webleft (f,g,h\webright )$, i.e. the colimit of the diagram

in $\mathsf{Sets}$.
First take the coequaliser of $f$ and $g$, forming a diagram

\[ A\underset {g}{\overset {f}{\rightrightarrows }}B\overset {\text{coeq}\webleft (f,g\webright )}{\twoheadrightarrow }\text{CoEq}\webleft (f,g\webright ) \]

and then take the coequaliser of the composition

\[ A\underset {h}{\overset {f}{\rightrightarrows }}B\overset {\text{coeq}\webleft (f,g\webright )}{\twoheadrightarrow }\text{CoEq}\webleft (f,g\webright ), \]

obtaining a quotient

\[ \text{CoEq}\webleft (\text{coeq}\webleft (f,g\webright )\circ f,\text{coeq}\webleft (f,g\webright )\circ h\webright )=\text{CoEq}\webleft (\text{coeq}\webleft (f,g\webright )\circ g,\text{coeq}\webleft (f,g\webright )\circ h\webright ) \]

of $\text{CoEq}\webleft (f,g\webright )$
First take the coequaliser of $g$ and $h$, forming a diagram

\[ A\underset {h}{\overset {g}{\rightrightarrows }}B\overset {\text{coeq}\webleft (g,h\webright )}{\twoheadrightarrow }\text{CoEq}\webleft (g,h\webright ) \]

and then take the coequaliser of the composition

\[ A\underset {g}{\overset {f}{\rightrightarrows }}B\overset {\text{coeq}\webleft (g,h\webright )}{\twoheadrightarrow }\text{CoEq}\webleft (g,h\webright ), \]

obtaining a quotient

\[ \text{CoEq}\webleft (\text{coeq}\webleft (g,h\webright )\circ f,\text{coeq}\webleft (g,h\webright )\circ g\webright )=\text{CoEq}\webleft (\text{coeq}\webleft (g,h\webright )\circ f,\text{coeq}\webleft (g,h\webright )\circ h\webright ) \]

of $\text{CoEq}\webleft (g,h\webright )$.