The Stacks project

Lemma 12.19.2. Let $\mathcal{A}$ be an abelian category. The category of filtered objects $\text{Fil}(\mathcal{A})$ has the following properties:

  1. It is an additive category.

  2. It has a zero object.

  3. It has kernels and cokernels, images and coimages.

  4. In general it is not an abelian category.

Proof. It is clear that $\text{Fil}(\mathcal{A})$ is additive with direct sum given by $(A, F) \oplus (B, F) = (A \oplus B, F)$ where $F^ p(A \oplus B) = F^ pA \oplus F^ pB$. The kernel of a morphism $f : (A, F) \to (B, F)$ of filtered objects is the injection $\mathop{\mathrm{Ker}}(f) \subset A$ where $\mathop{\mathrm{Ker}}(f)$ is endowed with the induced filtration. The cokernel of a morphism $f : A \to B$ of filtered objects is the surjection $B \to \mathop{\mathrm{Coker}}(f)$ where $\mathop{\mathrm{Coker}}(f)$ is endowed with the quotient filtration. Since all kernels and cokernels exist, so do all coimages and images. See Example 12.3.13 for the last statement. $\square$


Comments (0)

There are also:

  • 6 comment(s) on Section 12.19: Filtrations

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Your email address will not be published.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 0122. Beware of the difference between the letter 'O' and the digit '0'.