Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've seen in several places without further comment that if an equalizer is epic, it's an isomorphism. I've only proved one half of this:

Suppose $e:X \rightarrow A$ is an epimorphism and an equalizer for $f$ and $g$. Then $f \circ e = g \circ e \implies f = g$. Then any function $e': X' \rightarrow A$ trivially equalizes $f$ and $g$, so take $id_A: A \rightarrow A$. $e$ is an equalizer, so there exists a unique $k: A \rightarrow X$ such that $e \circ k = id_A$.

That gets me one side of the inverse, but how do I prove that $k \circ e = id_X$?

share|cite|improve this question
up vote 5 down vote accepted

$id_A$ is an equalizer of $f$ and $g$.

share|cite|improve this answer
I fail to proceed from there. – IARI Jan 9 '14 at 23:07
@IARI Equalizers being defined by a universal property, they are unique up to isomorphism. – Najib Idrissi Sep 6 '15 at 11:56

Suppose $B$ is the target of $f$ and $g$.

Since $e : X \to A$ is an equalizer of $f$ and $g$ and $X\xrightarrow{e} A \xrightarrow{f} B = X \xrightarrow{e} A \xrightarrow{g} B$ , hence there exists a unique $\ell : X \to X$ such that $X\xrightarrow{\ell} X \xrightarrow{e} A = X \xrightarrow{e} A$.

On one hand $X\xrightarrow{\mathrm{id}_X} X \xrightarrow{e} A = X \xrightarrow{e} A$. Hence, by the uniqueness, $$ \ell \text{ is } X\xrightarrow{\mathrm{id}_X} X $$

On the other hand, since you have already proved that $A \xrightarrow{k} X \xrightarrow{ e } A = id_A$, hence $X \xrightarrow{e} A \xrightarrow{k} X \xrightarrow{e} A = X \xrightarrow{e} A$. Hence, by the uniqueness, $$ \ell \text{ is } X \xrightarrow{e} A \xrightarrow{k} X $$

Therefore, comparing the two displayed equations, the result that $ X \xrightarrow{e} A \xrightarrow{k} X = X\xrightarrow{\mathrm{id}_X} X $ follows.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.