Variant 10.2.2.9. Let $\operatorname{\mathcal{C}}$ be a category and let $f: X \rightarrow Y$ be a morphism of $\operatorname{\mathcal{C}}$. If $f$ is a quotient morphism (in the sense of Definition 10.2.2.1), then it is an epimorphism.

$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$

**Proof.**
Suppose that we are given a pair of morphisms $e_0, e_1: Y \rightarrow Z$ in $\operatorname{\mathcal{C}}$ satisfying $e_0 \circ f = e_1 \circ f$; we wish to show that $e_0 = e_1$. By virtue of our assumption that $f$ is a quotient morphism, it will suffice to show that $e_0 \circ h = e_1 \circ h$ for every morphism $h: C \rightarrow Y$ which belongs to the sieve $\operatorname{\mathcal{C}}^{0}_{/Y} \subseteq \operatorname{\mathcal{C}}_{/Y}$ generated by $f$. In this case, we can write $h = f \circ g$ for some morphism $g: C \rightarrow X$; the desired result then follows from the calculation

\[ e_0 \circ h = e_0 \circ (f \circ g) = (e_0 \circ f) \circ g = (e_1 \circ f) \circ g = e_1 \circ (f \circ g) = e_1 \circ h. \]

$\square$