Lemma 22.10.1. Let $(A, \text{d})$ be a differential graded algebra. The homotopy category $K(\text{Mod}_{(A, \text{d})})$ with its natural translation functors and distinguished triangles is a pre-triangulated category.

Proof. Proof of TR1. By definition every triangle isomorphic to a distinguished one is distinguished. Also, any triangle $(K, K, 0, 1, 0, 0)$ is distinguished since $0 \to K \to K \to 0 \to 0$ is an admissible short exact sequence. Finally, given any homomorphism $f : K \to L$ of differential graded $A$-modules the triangle $(K, L, C(f), f, i, -p)$ is distinguished by Lemma 22.9.4.

Proof of TR2. Let $(X, Y, Z, f, g, h)$ be a triangle. Assume $(Y, Z, X, g, h, -f)$ is distinguished. Then there exists an admissible short exact sequence $0 \to K \to L \to M \to 0$ such that the associated triangle $(K, L, M, \alpha , \beta , \delta )$ is isomorphic to $(Y, Z, X, g, h, -f)$. Rotating back we see that $(X, Y, Z, f, g, h)$ is isomorphic to $(M[-1], K, L, -\delta [-1], \alpha , \beta )$. It follows from Lemma 22.9.2 that the triangle $(M[-1], K, L, \delta [-1], \alpha , \beta )$ is isomorphic to $(M[-1], K, C(\delta [-1]), \delta [-1], i, p)$. Precomposing the previous isomorphism of triangles with $-1$ on $Y$ it follows that $(X, Y, Z, f, g, h)$ is isomorphic to $(M[-1], K, C(\delta [-1]), \delta [-1], i, -p)$. Hence it is distinguished by Lemma 22.9.4. On the other hand, suppose that $(X, Y, Z, f, g, h)$ is distinguished. By Lemma 22.9.4 this means that it is isomorphic to a triangle of the form $(K, L, C(f), f, i, -p)$ for some morphism $f$ of $\text{Mod}_{(A, \text{d})}$. Then the rotated triangle $(Y, Z, X, g, h, -f)$ is isomorphic to $(L, C(f), K, i, -p, -f)$ which is isomorphic to the triangle $(L, C(f), K, i, p, f)$. By Lemma 22.9.1 this triangle is distinguished. Hence $(Y, Z, X, g, h, -f)$ 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 22.9.4 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 22.6.2 to the commutative diagram given by $f, f', a, b$. $\square$

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