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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