---
tag: 956E1C2VE6
kind: definition
name: Equalizer
msc: [18A30]
defines: CTJ5AD8ST5
uses: [A2DE66937D, 859F0TFVA3, CHS5AFJGWE, 0W88CEGQP1]
sources:
  - url: https://en.wikipedia.org/wiki/Equaliser_(mathematics)
---

## Definition

Let $\mathcal{C}$ be a [[category|A2DE66937D]] and let $f, g \colon X \to Y$ be two [[morphisms|859F0TFVA3]] of $\mathcal{C}$.
An *equalizer* of $f$ and $g$ is an object $E$ with a morphism $e \colon E \to X$ such that:

1. $f \circ e = g \circ e$;
2. for every object $Z$ and every morphism $k \colon Z \to X$ with $f \circ k = g \circ k$, there is exactly one morphism $u \colon Z \to E$ with $e \circ u = k$.

## Remarks

An equalizer is the [[limit|CHS5AFJGWE]] of the diagram $X \rightrightarrows Y$ formed by $f$ and $g$, so it is unique up to unique isomorphism.
It is a monomorphism: if $e \circ a = e \circ b$, then $a$ and $b$ both solve clause 2 for $k = e \circ a$, so $a = b$.

Example: in the category of sets, let $E = \set{x \in X : f(x) = g(x)}$ and let $e$ be the inclusion.
Clause 1 holds by the choice of $E$.
If $f \circ k = g \circ k$, every $k(z)$ lies in $E$, so $u(z) = k(z)$ works, and it is the only choice since $e$ is injective.
In the category of abelian groups the equalizer of $f$ and $g$ is the [[kernel|0W88CEGQP1]] of $f - g$ with its inclusion.

Non-example: for $f, g \colon \R \to \R$ with $f(x) = x^2$ and $g(x) = x$, the inclusion $\{0\} \to \R$ satisfies clause 1 but not clause 2.
The map $k \colon \{*\} \to \R$ with $k(*) = 1$ has $f \circ k = g \circ k$ but does not factor through $\{0\}$.