Lemma 22.27.14. In Situation 22.27.2 the homotopy category K(\mathcal{A}) with its natural translation functors and distinguished triangles is a pre-triangulated category.
Proof. We will verify each of TR1, TR2, and TR3.
Proof of TR1. By definition every triangle isomorphic to a distinguished one is distinguished. Since
\xymatrix{x\ar[r]^{1_ x} & x\ar[r] & 0}
is an admissible short exact sequence, (x, x, 0, 1_ x, 0, 0) is a distinguished triangle. Moreover, given a morphism \alpha : x \to y in \text{Comp}(\mathcal{A}), the triangle given by (x, y, c(\alpha ), \alpha , i, -p) is distinguished by Lemma 22.27.13.
Proof of TR2. Let (x,y,z,\alpha ,\beta ,\gamma ) be a triangle and suppose (y,z,x[1],\beta ,\gamma ,-\alpha [1]) is distinguished. Then there exists an admissible short exact sequence 0 \to x' \to y' \to z' \to 0 such that the associated triangle (x',y',z',\alpha ',\beta ',\gamma ') is isomorphic to (y,z,x[1],\beta ,\gamma ,-\alpha [1]). After rotating, we conclude that (x,y,z,\alpha ,\beta ,\gamma ) is isomorphic to (z'[-1],x',y', \gamma '[-1], \alpha ',\beta '). By Lemma 22.27.11, we deduce that (z'[-1],x',y', \gamma '[-1], \alpha ',\beta ') is isomorphic to (z'[-1],x',c(\gamma '[-1]), \gamma '[-1], i, p). Composing the two isomorphisms with sign changes as indicated in the following diagram:
\xymatrix@C=3pc{ x\ar[r]^{\alpha }\ar[d] & y\ar[r]^{\beta }\ar[d] & z\ar[r]^{\gamma }\ar[d] & x[1]\ar[d] \\ z'[-1]\ar[r]^{-\gamma '[-1]}\ar[d]_{-1_{z'[-1]}} & x \ar[r]^{\alpha '}\ar@{=}[d] & y' \ar[r]^{\beta '} \ar[d] & z'\ar[d]^{-1_{z'}} \\ z'[-1]\ar[r]^{\gamma '[-1]} & x \ar[r]^{\alpha '} & c(\gamma '[-1]) \ar[r]^{-p} & z' }
We conclude that (x,y,z,\alpha ,\beta ,\gamma ) is distinguished by Lemma 22.27.13 (2). Conversely, suppose that (x,y,z,\alpha ,\beta ,\gamma ) is distinguished, so that by Lemma 22.27.13 (1), it is isomorphic to a triangle of the form (x',y', c(\alpha '), \alpha ', i, -p) for some morphism \alpha ': x' \to y' in \text{Comp}(\mathcal{A}). The rotated triangle (y,z,x[1],\beta ,\gamma , -\alpha [1]) is isomorphic to the triangle (y',c(\alpha '), x'[1], i, -p, -\alpha [1]) which is isomorphic to (y',c(\alpha '), x'[1], i, p, \alpha [1]). By Lemma 22.27.10, this triangle is distinguished, from which it follows that (y,z,x[1], \beta ,\gamma , -\alpha [1]) is distinguished.
Proof of TR3: Suppose (x,y,z, \alpha ,\beta ,\gamma ) and (x',y',z',\alpha ',\beta ',\gamma ') are distinguished triangles of \text{Comp}(\mathcal{A}) and let f: x \to x' and g: y \to y' be morphisms such that \alpha ' \circ f = g \circ \alpha . By Lemma 22.27.13, we may assume that (x,y,z,\alpha ,\beta ,\gamma )= (x,y,c(\alpha ),\alpha , i, -p) and (x',y',z', \alpha ',\beta ',\gamma ')= (x',y',c(\alpha '), \alpha ',i',-p'). Now apply Lemma 22.27.3 and we are done. \square
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