## Tag `0107`

Chapter 12: Homological Algebra > Section 12.3: Preadditive and additive categories

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}
```

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