The Stacks project

Lemma 13.6.9. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{B}$ be a full triangulated subcategory. The kernel of the quotient functor $Q : \mathcal{D} \to \mathcal{D}/\mathcal{B}$ is the strictly full subcategory of $\mathcal{D}$ whose objects are

\[ \mathop{\mathrm{Ob}}\nolimits (\mathop{\mathrm{Ker}}(Q)) = \left\{ \begin{matrix} Z \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}) \text{ such that there exists a }Z' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}) \\ \text{ such that }Z \oplus Z'\text{ is isomorphic to an object of }\mathcal{B} \end{matrix} \right\} \]

In other words it is the smallest strictly full saturated triangulated subcategory of $\mathcal{D}$ containing $\mathcal{B}$.

Proof. First note that the kernel is automatically a strictly full triangulated subcategory containing summands of any of its objects, see Lemma 13.6.2. The description of its objects follows from the definitions and Lemma 13.5.7 part (4). $\square$


Comments (1)

Comment #331 by arp on

Typo: Delete the word stable in the proof.

There are also:

  • 2 comment(s) on Section 13.6: Quotients of triangulated 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Your email address will not be published.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 05RK. Beware of the difference between the letter 'O' and the digit '0'.