Lemma 12.3.12. Let $f : x \to y$ be a morphism in a preadditive category such that the kernel, cokernel, image and coimage all exist. Then $f$ can be factored uniquely as $x \to \mathop{\mathrm{Coim}}(f) \to \mathop{\mathrm{Im}}(f) \to y$.

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. $\square$

Comment #2614 by on

I don't understand where the word "unique" in "it is the unique morphism which gives rise to the morphism $x \to y \to \Coker(f)$ which is zero". I can prove this (see https://github.com/stacks/stacks-project/pull/36 , more precisely this commit: https://github.com/stacks/stacks-project/pull/36/commits/2862030f02057c657b447d5707023c747e9b8662 ), but it takes me a lemma (which I guess is useful anyway -- it doesn't hurt to know that kernels are monos, etc.).

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