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. $\square$

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).