The Stacks project

Proof. Proof of (1). Note that

is a distinguished triangle in $K(\text{Fil}^ f(\mathcal{A}))$ (Lemma 13.9.14) and $C(\alpha )^\bullet $ is a filtered acyclic complex (Lemma 13.13.4). Then

is an exact sequence of abelian groups, see Lemma 13.4.2. At this point Lemma 13.26.10 guarantees that the outer two groups are zero and hence $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(L^\bullet , I^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(K^\bullet , I^\bullet )$.

Proof of (2). Let $a$ be an element of the right hand side. We may represent $a = \gamma \alpha ^{-1}$ where $\alpha : K^\bullet \to L^\bullet $ is a filtered quasi-isomorphism and $\gamma : K^\bullet \to I^\bullet $ is a map of complexes. By part (1) we can find a morphism $\beta : L^\bullet \to I^\bullet $ such that $\beta \circ \alpha $ is homotopic to $\gamma $. This proves that the map is surjective. Let $b$ be an element of the left hand side which maps to zero in the right hand side. Then $b$ is the homotopy class of a morphism $\beta : L^\bullet \to I^\bullet $ such that there exists a filtered quasi-isomorphism $\alpha : K^\bullet \to L^\bullet $ with $\beta \circ \alpha $ homotopic to zero. Then part (1) shows that $\beta $ is homotopic to zero also, i.e., $b = 0$. $\square$