Epic Equalizer is Isomorphism

Theorem

Let $\mathbf C$ be a metacategory.

Let $e: E \to C$ be the equalizer of two morphisms $f, g: C \to D$.

Let $e$ be an epimorphism.

Then $e$ is an isomorphism.

As $e$ equalises $f$ and $g$, $f \circ e = g \circ e$.

Since $e$ is epic, it follows that $f = g$.

Then in the equaliser diagram:

$\begin{xy}\xymatrix{

E 
 \ar[r]^*{e}

&

A
 \ar[r]<2pt>^*{f}
 \ar[r]<-2pt>_*{g}

&

\\

A
 \ar@{.>}[u]^*{k}
 \ar[ur]_*{\operatorname{id}_A}

}\end{xy}$

We have that:

$f \circ 1_A = g \circ 1_A$

so there is a unique $k$ with $e \circ k = \operatorname{id}_A$.

Then:

$\ds e \circ k \circ e$

$=$

$\ds {\operatorname {id}_A} \circ e$

$\ds $

$=$

$\ds e$

$\ds $

$=$

$\ds e \circ \operatorname{id}_E$

By Equalizer is Monomorphism, it follows that $k \circ e = \operatorname{id}_E$.

This gives $k$ as an inverse to $e$.

Thus $e$ is an isomorphism.

$\blacksquare$