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:
- $\begin{xy}\xymatrix{
E \ar[r]^*{e}
&
A \ar[r]<2pt>^*{f} \ar[r]<-2pt>_*{g}
&
B
\\
A \ar@{.>}[u]^*{k} \ar[ur]_*{\operatorname{id}_A}
}\end{xy}$
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.
$\blacksquare$
Sources
- 1984: Robert Goldblatt: Topoi: The Categorical Analysis of Logic: $\S 3.10$: Theorem $2$