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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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up vote 5 down vote accepted

$id_A$ 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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