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:

$\quad\quad \begin{xy}\xymatrix{ E \ar[r]^*{e} & C \ar[r]<2pt>^*{f} \ar[r]<-2pt>_*{g} & D \\ Z \ar[u]<2pt>^*{x} \ar[u]<-2pt>_*{y} \ar[ur]_*{z} }\end{xy}$

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.

$\blacksquare$

Sources

• 2010: Steve Awodey: Category Theory (2nd ed.) ... (previous) ... (next): $\S 3.3$: Proposition $3.16$