**Proof.**
Proof of (1). Note that

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

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

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

is an exact sequence of abelian groups, see Lemma 13.4.2. At this point Lemma 13.26.10 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., $b = 0$.
$\square$

## Comments (2)

Comment #3769 by Owen B on

Comment #3899 by Johan on