Proposition 13.10.3. Let $\mathcal{A}$ be an additive category. The category $K(\mathcal{A})$ of complexes up to homotopy with its natural translation functors and distinguished triangles as defined above is a triangulated category.

Proof. Proof of TR1. By definition every triangle isomorphic to a distinguished one is distinguished. Also, any triangle $(A^\bullet , A^\bullet , 0, 1, 0, 0)$ is distinguished since $0 \to A^\bullet \to A^\bullet \to 0 \to 0$ is a termwise split sequence of complexes. Finally, given any morphism of complexes $f : K^\bullet \to L^\bullet$ the triangle $(K, L, C(f), f, i, -p)$ is distinguished by Lemma 13.9.14.

Proof of TR2. Let $(X, Y, Z, f, g, h)$ be a triangle. Assume $(Y, Z, X[1], g, h, -f[1])$ is distinguished. Then there exists a termwise split sequence of complexes $A^\bullet \to B^\bullet \to C^\bullet$ such that the associated triangle $(A^\bullet , B^\bullet , C^\bullet , \alpha , \beta , \delta )$ is isomorphic to $(Y, Z, X[1], g, h, -f[1])$. Rotating back we see that $(X, Y, Z, f, g, h)$ is isomorphic to $(C^\bullet [-1], A^\bullet , B^\bullet , -\delta [-1], \alpha , \beta )$. It follows from Lemma 13.9.16 that the triangle $(C^\bullet [-1], A^\bullet , B^\bullet , \delta [-1], \alpha , \beta )$ is isomorphic to $(C^\bullet [-1], A^\bullet , C(\delta [-1])^\bullet , \delta [-1], i, p)$. Precomposing the previous isomorphism of triangles with $-1$ on $Y$ it follows that $(X, Y, Z, f, g, h)$ is isomorphic to $(C^\bullet [-1], A^\bullet , C(\delta [-1])^\bullet , \delta [-1], i, -p)$. Hence it is distinguished by Lemma 13.9.14. On the other hand, suppose that $(X, Y, Z, f, g, h)$ is distinguished. By Lemma 13.9.14 this means that it is isomorphic to a triangle of the form $(K^\bullet , L^\bullet , C(f), f, i, -p)$ for some morphism of complexes $f$. Then the rotated triangle $(Y, Z, X[1], g, h, -f[1])$ is isomorphic to $(L^\bullet , C(f), K^\bullet [1], i, -p, -f[1])$ which is isomorphic to the triangle $(L^\bullet , C(f), K^\bullet [1], i, p, f[1])$. By Lemma 13.9.17 this triangle is distinguished. Hence $(Y, Z, X[1], g, h, -f[1])$ is distinguished as desired.

Proof of TR3. Let $(X, Y, Z, f, g, h)$ and $(X', Y', Z', f', g', h')$ be distinguished triangles of $K(\mathcal{A})$ and let $a : X \to X'$ and $b : Y \to Y'$ be morphisms such that $f' \circ a = b \circ f$. By Lemma 13.9.14 we may assume that $(X, Y, Z, f, g, h) = (X, Y, C(f), f, i, -p)$ and $(X', Y', Z', f', g', h') = (X', Y', C(f'), f', i', -p')$. At this point we simply apply Lemma 13.9.2 to the commutative diagram given by $f, f', a, b$.

Proof of TR4. At this point we know that $K(\mathcal{A})$ is a pre-triangulated category. Hence we can use Lemma 13.4.15. Let $A^\bullet \to B^\bullet$ and $B^\bullet \to C^\bullet$ be composable morphisms of $K(\mathcal{A})$. By Lemma 13.9.15 we may assume that $A^\bullet \to B^\bullet$ and $B^\bullet \to C^\bullet$ are split injective morphisms. In this case the result follows from Lemma 13.10.2. $\square$

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

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 014S. Beware of the difference between the letter 'O' and the digit '0'.