nLab
coequalizer

Idea

The concept of coequalizer in a general category is the generalization of the construction where for two functions $f,g$ between sets $X$ and $Y$

$$X \stackrel{\overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}} Y$$

one forms the set $Y/_\sim$ of equivalence classes induced by the equivalence relation generated by the relation

$$f(x)\sim g(x)$$

for all $x \in X$. This means that the projection function $p \colon Y \longrightarrow Y/_\sim$ satisfies

$$p \circ f = p \circ g$$

and in fact $p$ is universal with this property, hence it “co-equalizes” $f$ and $g$.

In this form this may be phrased generally in any category,

Definition

Definition

In some category $\mathcal{C}$, the coequalizer $coeq(f,g)$ of two parallel morphisms $f$ and $g$ between two objects $X$ and $Y$ is (if it exists), the colimit under the diagram formed by these two morphisms

$$\array{ X && \stackrel{\overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}} && Y \\ & \searrow && \swarrow_{\mathrlap{p}} \\ && coeq(f,g) } \,.$$

Equivalently:

Definition

In a category $\mathcal{C}$ a diagram

$$X \underoverset{g}{f}{\rightrightarrows} Y \overset{p}{\rightarrow} Z$$

is called a coequalizer diagram if

1. $p \circ f = p \circ g$;

2. $p$ is universal for this property: i.e. if $q \colon Y \to W$ is a morphism of $\mathcal{C}$ such that $q \circ f = q \circ g$, then there is a unique morphism $q' \colon Z \to W$ such that $q' \circ p = q$

$$\array{ X &\stackrel{\overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}}& Y &\overset{p}{\longrightarrow}& Z \\ && {}^{\mathllap{q}}\downarrow & \swarrow_{\mathrlap{\exists ! \, q'}} \\ && W }$$

Remark

By formal duality, a coequalizer in $\mathcal{C}$ is equivalently an equalizer in the opposite category $\mathcal{C}^{op}$.

Properties

Relation to pushouts

Coequalizers are closely related to pushouts:

Proposition

A diagram

$$X \stackrel{\overset{f}{\longrightarrow}}{\underset{g}{\longrightarrow}} Y \overset{p}{\longrightarrow} Z$$

is a coequalizer diagram, def. 2, precisely if

$$\array{ X \sqcup X &\overset{(f,g)}{\longrightarrow}& Y \\ \downarrow && \downarrow^{\mathrlap{p}} \\ X &\underset{}{\longrightarrow}& Z }$$

is a pushout diagram.

Conversely:

Proposition

A diagram

$$\array{ A &\overset{f_1}{\longrightarrow}& B \\ {}^{\mathllap{f_2}}\downarrow && \downarrow^{\mathrlap{p_1}} \\ C &\underset{p_2}{\longrightarrow}& D }$$

is a pushout square, precisely if

$$A \stackrel{\overset{\iota_1 \circ f_1}{\longrightarrow}}{\underset{\iota_2 \circ f_2}{\longrightarrow}} B \sqcup C \overset{(p_1,p_2)}{\longrightarrow} D$$

is a coequalizer diagram.

Examples

Example

For $\mathcal{C} =$ Set, the coequalizer of two functions $f$, $g$ is the quotient set by the equivalence relation generated by the relation $f(x) \sim g(x)$ for all $x \in X$.

Example

For $\mathcal{C} =$ Top, the coequalizer of two continuous functions $f$, $g$ is the topological space whose underlying set is the quotient set from example 1, and whose topology is the corresponding quotient topology.