Equalizer is Monomorphism

Theorem
Let $\mathbf C$ be a metacategory.

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

Then $e$ is a monomorphism.

Proof
Suppose that for morphisms $x,y: Z \to E$, it holds that:


 * $e \circ y = e \circ x$

Putting $z = e \circ x$, the following commutative diagram applies:


 * $\begin{xy}\xymatrix{

E \ar[r]^*{e} & C \ar[r] ^*{f} \ar[r]<-2pt>_*{g} & D

\\ Z \ar[u] ^*{x} \ar[u]<-2pt>_*{y} \ar[ur]_*{z} }\end{xy}$

It follows that $f \circ z = g \circ z$.

Since $e$ is an equalizer, there thus exists a unique $u: Z \to E$ with:


 * $z = e \circ u$

Hence $x = u = y$, and it follows that $e$ is a monomorphism.