The Stacks project

Proof. There is a canonical morphism $\mathop{\mathrm{Coim}}(f) \to y$ because $\mathop{\mathrm{Ker}}(f) \to x \to y$ is zero. The composition $\mathop{\mathrm{Coim}}(f) \to y \to \mathop{\mathrm{Coker}}(f)$ is zero, because it is the unique morphism which gives rise to the morphism $x \to y \to \mathop{\mathrm{Coker}}(f)$ which is zero (the uniqueness follows from Lemma 12.3.11 (3)). Hence $\mathop{\mathrm{Coim}}(f) \to y$ factors uniquely through $\mathop{\mathrm{Im}}(f) \to y$, which gives us the desired map. $\square$