The Stacks project

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. $\square$