# The Stacks Project

## Tag 0107

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. $\square$

The code snippet corresponding to this tag is a part of the file homology.tex and is located in lines 287–293 (see updates for more information).

\begin{lemma}
\label{lemma-coim-im-map}
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
$x \to \Coim(f) \to \Im(f) \to y$.
\end{lemma}

\begin{proof}
There is a canonical morphism $\Coim(f) \to y$
because $\Ker(f) \to x \to y$ is zero.
The composition $\Coim(f) \to y \to \Coker(f)$
is zero, because it is the unique morphism which gives
rise to the morphism $x \to y \to \Coker(f)$ which
is zero
(the uniqueness follows from
Lemma \ref{lemma-kernel-mono} (3)).
Hence $\Coim(f) \to y$ factors uniquely through
$\Im(f) \to y$, which gives us the desired map.
\end{proof}

Comment #2614 by Darij Grinberg (site) on July 2, 2017 a 12:53 pm UTC

I don't understand where the word "unique" in "it is the unique morphism which gives rise to the morphism $x \to y \to \mathop{\mathrm{Coker}}(f)$ 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.).

Comment #2635 by Johan (site) on July 7, 2017 a 12:56 pm UTC

OK, I think this got fixed now.

There are also 4 comments on Section 12.3: Homological Algebra.

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