# 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.

## Proof

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} & B \\ A \[email protected]{.>}[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:

 $\displaystyle e \circ k \circ e$ $=$ $\displaystyle \operatorname {id}_A \circ e$ $\displaystyle$ $=$ $\displaystyle e$ $\displaystyle$ $=$ $\displaystyle e$ 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.

$\blacksquare$