The Stacks project

Lemma 13.10.7. Let $\mathcal{A}$ be an additive category. Let $(A^\bullet , B^\bullet , C^\bullet , a, b, c)$ be a distinguished triangle in $K(\mathcal{A})$. Then there exists an isomorphic distinguished triangle $(A^\bullet , (B')^\bullet , C^\bullet , a', b', c)$ such that $0 \to A^ n \to (B')^ n \to C^ n \to 0$ is a split short exact sequence for all $n$.

Proof. We will use that $K(\mathcal{A})$ is a triangulated category by Proposition 13.10.3. Let $W^\bullet $ be the cone on $c : C^\bullet \to A^\bullet [1]$ with its maps $i : A^\bullet [1] \to W^\bullet $ and $p : W^\bullet \to C^\bullet [1]$. Then $(C^\bullet , A^\bullet [1], W^\bullet , c, i, -p)$ is a distinguished triangle by Lemma 13.9.14. Rotating backwards twice we see that $(A^\bullet , W^\bullet [-1], C^\bullet , -i[-1], p[-1], c)$ is a distinguished triangle. By TR3 there is a morphism of distinguished triangles $(\text{id}, \beta , \text{id}) : (A^\bullet , B^\bullet , C^\bullet , a, b, c) \to (A^\bullet , W^\bullet [-1], C^\bullet , -i[-1], p[-1], c)$ which must be an isomorphism by Lemma 13.4.3. This finishes the proof because $0 \to A^\bullet \to W^\bullet [-1] \to C^\bullet \to 0$ is a termwise split short exact sequence of complexes by the very construction of cones in Section 13.9. $\square$


Comments (0)


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Your email address will not be published.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 0G6C. Beware of the difference between the letter 'O' and the digit '0'.