Lemma 22.27.9. In Situation 22.27.2 let $0 \to x \to y \to z \to 0$ be an admissible short exact sequence in $\text{Comp}(\mathcal{A})$. The triangle
\[ \xymatrix{x\ar[r] & y\ar[r] & z\ar[r]^{\delta } & x[1]} \]
with $\delta : z \to x[1]$ as defined in Lemma 22.27.1 is up to canonical isomorphism in $K(\mathcal{A})$, independent of the choices made in Lemma 22.27.1.
Proof.
Suppose $\delta $ is defined by the splitting
\[ \xymatrix{ x \ar@<0.5ex>[r]^{a} & y \ar@<0.5ex>[r]^ b\ar@<0.5ex>[l]^{\pi } & z \ar@<0.5ex>[l]^ s } \]
and $\delta '$ is defined by the splitting with $\pi ',s'$ in place of $\pi ,s$. Then
\[ s'-s = (a\pi + sb)(s'-s) = a\pi s' \]
since $bs' = bs = 1_ z$ and $\pi s = 0$. Similarly,
\[ \pi ' - \pi = (\pi ' - \pi )(a\pi + sb) = \pi 'sb \]
Since $\delta = \pi d(s)$ and $\delta ' = \pi 'd(s')$ as constructed in Lemma 22.27.1, we may compute
\[ \delta ' = \pi 'd(s') = (\pi + \pi 'sb)d(s + a\pi s') = \delta + d(\pi s') \]
using $\pi a = 1_ x$, $ba = 0$, and $\pi 'sbd(s') = \pi 'sba\pi d(s') = 0$ by formula (5) in Lemma 22.27.1.
$\square$
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