# Epic Equalizer is Isomorphism

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## 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}$

$f \circ 1_A = g \circ 1_A$, so there is a unique $k$ with $e \circ k = \operatorname{id}_A$.

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

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

This gives $k$ as an inverse to $e$, so $e$ is an isomorphism.

$\blacksquare$

## Sources

- 1984: Robert Goldblatt:
*Topoi: The Categorical Analysis of Logic*: $\S 3.10$: Theorem $2$