Lemma 22.20.11. In Situation 22.20.2 let $\alpha : x \to y$ and $\beta : y \to z$ define an admissible short exact sequence

$\xymatrix{ x \ar[r] & y\ar[r] & z }$

in $\text{Comp}(\mathcal{A})$. Let $(x, y, z, \alpha , \beta , \delta )$ be the associated triangle in $K(\mathcal{A})$. Then, the triangles

$(z[-1], x, y, \delta [-1], \alpha , \beta ) \quad \text{and}\quad (z[-1], x, c(\delta [-1]), \delta [-1], i, p)$

are isomorphic.

Proof. We have a diagram of the form

$\xymatrix{ z[-1]\ar[r]^{\delta [-1]}\ar[d]^1 & x\ar@<0.5ex>[r]^{\alpha }\ar[d]^1 & y\ar@<0.5ex>[r]^{\beta }\ar@{.>}[d]\ar@<0.5ex>[l]^{\tilde{\alpha }} & z\ar[d]^1\ar@<0.5ex>[l]^{\tilde\beta } \\ z[-1] \ar[r]^{\delta [-1]} & x\ar@<0.5ex>[r]^ i & c(\delta [-1]) \ar@<0.5ex>[r]^ p\ar@<0.5ex>[l]^{\tilde i} & z\ar@<0.5ex>[l]^{\tilde p} }$

with splittings to $\alpha , \beta , i$, and $p$ given by $\tilde{\alpha }, \tilde{\beta }, \tilde{i},$ and $\tilde{p}$ respectively. Define a morphism $y \to c(\delta [-1])$ by $i\tilde{\alpha } + \tilde{p}\beta$ and a morphism $c(\delta [-1]) \to y$ by $\alpha \tilde{i} + \tilde{\beta } p$. Let us first check that these define morphisms in $\text{Comp}(\mathcal{A})$. We remark that by identities from Lemma 22.20.1, we have the relation $\delta [-1] = \tilde{\alpha }d(\tilde{\beta }) = -d(\tilde{\alpha })\tilde{\beta }$ and the relation $\delta [-1] = \tilde{i}d(\tilde{p})$. Then

\begin{align*} d(\tilde{\alpha }) & = d(\tilde{\alpha })\tilde{\beta }\beta \\ & = -\delta [-1]\beta \end{align*}

where we have used equation (6) of Lemma 22.20.1 for the first equality and the preceeding remark for the second. Similarly, we obtain $d(\tilde{p}) = i\delta [-1]$. Hence

\begin{align*} d(i\tilde{\alpha } + \tilde{p}\beta ) & = d(i)\tilde{\alpha } + id(\tilde{\alpha }) + d(\tilde{p})\beta + \tilde{p}d(\beta ) \\ & = id(\tilde{\alpha }) + d(\tilde{p})\beta \\ & = -i\delta [-1]\beta + i\delta [-1]\beta \\ & = 0 \end{align*}

so $i\tilde{\alpha } + \tilde{p}\beta$ is indeed a morphism of $\text{Comp}(\mathcal{A})$. By a similar calculation, $\alpha \tilde{i} + \tilde{\beta } p$ is also a morphism of $\text{Comp}(\mathcal{A})$. It is immediate that these morphisms fit in the commutative diagram. We compute:

\begin{align*} (i\tilde{\alpha } + \tilde{p}\beta )(\alpha \tilde{i} + \tilde{\beta } p) & = i\tilde{\alpha }\alpha \tilde{i} + i\tilde{\alpha }\tilde{\beta }p + \tilde{p}\beta \alpha \tilde{i} + \tilde{p}\beta \tilde{\beta }p \\ & = i\tilde{i} + \tilde{p}p \\ & = 1_{c(\delta [-1])} \end{align*}

where we have freely used the identities of Lemma 22.20.1. Similarly, we compute $(\alpha \tilde{i} + \tilde{\beta } p)(i\tilde{\alpha } + \tilde{p}\beta ) = 1_ y$, so we conclude $y \cong c(\delta [-1])$. Hence, the two triangles in question are isomorphic. $\square$

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).