Definition 4.13.1. Let $\mathcal{C}$ be a category and let $f : X \to Y$ be a morphism of $\mathcal{C}$.
We say that $f$ is a monomorphism if for every object $W$ and every pair of morphisms $a, b : W \to X$ such that $f \circ a = f \circ b$ we have $a = b$.
We say that $f$ is an epimorphism if for every object $W$ and every pair of morphisms $a, b : Y \to W$ such that $a \circ f = b \circ f$ we have $a = b$.
Example 4.13.2. In the category of sets the monomorphisms correspond to injective maps and the epimorphisms correspond to surjective maps.
Lemma 4.13.3. Let $\mathcal{C}$ be a category, and let $f : X \to Y$ be a morphism of $\mathcal{C}$. Then
$f$ is a monomorphism if and only if $X$ is the fibre product $X \times _ Y X$, and
$f$ is an epimorphism if and only if $Y$ is the pushout $Y \amalg _ X Y$.
Proof. Let suppose that $f$ is a monomorphism. Let $W$ be an object of $\mathcal C$ and $\alpha , \beta \in \mathop{\mathrm{Mor}}\nolimits _\mathcal C(W,X)$ such that $f\circ \alpha = f\circ \beta $. Therefore $\alpha = \beta $ as $f$ is monic. In addition, we have the commutative diagram
which verify the universal property with $\gamma := \alpha = \beta $. Thus $X$ is indeed the fibre product $X\times _ Y X$.
Suppose that $X \times _ Y X \cong X $. The diagram
commutes and if $W \in \mathop{\mathrm{Ob}}\nolimits (\mathcal C)$ and $\alpha , \beta : W \to X$ such that $f \circ \alpha = f \circ \beta $, we have a unique $\gamma $ verifying
which proves that $\alpha = \beta $.
The proof is exactly the same for the second point, but with the pushout $Y\amalg _ X Y = Y$. $\square$
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Comment #8860 by Francisco Gallardo on
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