# Why is an epic equalizer an isomorphism?

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 $k \circ e = id_X$?

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

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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.

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