The Stacks project

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. $\square$


Comments (2)

Comment #2614 by on

I don't understand where the word "unique" in "it is the unique morphism which gives rise to the morphism 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.).

There are also:

  • 5 comment(s) on Section 12.3: Preadditive and additive categories

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0107. Beware of the difference between the letter 'O' and the digit '0'.