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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