
Remark 13.18.5. Let $\mathcal{A}$ be an abelian category. Using the fact that $K(\mathcal{A})$ is a triangulated category we may use Lemma 13.18.4 to obtain proofs of some of the lemmas below which are usually proved by chasing through diagrams. Namely, suppose that $\alpha : K^\bullet \to L^\bullet$ is a quasi-isomorphism of complexes. Then

$(K^\bullet , L^\bullet , C(\alpha )^\bullet , \alpha , i, -p)$

is a distinguished triangle in $K(\mathcal{A})$ (Lemma 13.9.14) and $C(\alpha )^\bullet$ is an acyclic complex (Lemma 13.11.2). Next, let $I^\bullet$ be a bounded below complex of injective objects. Then

$\xymatrix{ \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(C(\alpha )^\bullet , I^\bullet ) \ar[r] & \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(L^\bullet , I^\bullet ) \ar[r] & \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I^\bullet ) \ar[lld] \\ \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(C(\alpha )^\bullet [-1], I^\bullet ) }$

is an exact sequence of abelian groups, see Lemma 13.4.2. At this point Lemma 13.18.4 guarantees that the outer two groups are zero and hence $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(L^\bullet , I^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I^\bullet )$.

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