The Stacks project

Lemma 13.10.5. Let $\mathcal{A}$ be an additive category. The categories $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, and $K^ b(\mathcal{A})$ are full triangulated subcategories of $K(\mathcal{A})$.

Proof. Each of the categories mentioned is a full additive subcategory. We use the criterion of Lemma 13.4.16 to show that they are triangulated subcategories. It is clear that each of the categories $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, and $K^ b(\mathcal{A})$ is preserved under the shift functors $[1], [-1]$. Finally, suppose that $f : A^\bullet \to B^\bullet $ is a morphism in $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^ b(\mathcal{A})$. Then $(A^\bullet , B^\bullet , C(f)^\bullet , f, i, -p)$ is a distinguished triangle of $K(\mathcal{A})$ with $C(f)^\bullet \in K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^ b(\mathcal{A})$ as is clear from the construction of the cone. Thus the lemma is proved. (Alternatively, $K^\bullet \to L^\bullet $ is isomorphic to an termwise split injection of complexes in $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^ b(\mathcal{A})$, see Lemma 13.9.6 and then one can directly take the associated distinguished triangle.) $\square$

Comments (2)

Comment #1706 by Keenan Kidwell on

In the statement of the lemma, the first instance of "subcategory" should be "category."

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Your email address will not be published.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 05RQ. Beware of the difference between the letter 'O' and the digit '0'.