Lemma 13.26.11. Let $\mathcal{A}$ be an abelian category. Let $I^\bullet \in K(\text{Fil}^ f(\mathcal{A}))$ be a bounded below complex consisting of filtered injective objects.

1. Let $\alpha : K^\bullet \to L^\bullet$ in $K(\text{Fil}^ f(\mathcal{A}))$ be a filtered quasi-isomorphism. Then the map

$\mathop{\mathrm{Hom}}\nolimits _{K(\text{Fil}^ f(\mathcal{A}))}(L^\bullet , I^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits _{K(\text{Fil}^ f(\mathcal{A}))}(K^\bullet , I^\bullet )$

is bijective.

2. Let $L^\bullet \in K(\text{Fil}^ f(\mathcal{A}))$. Then

$\mathop{\mathrm{Hom}}\nolimits _{K(\text{Fil}^ f(\mathcal{A}))}(L^\bullet , I^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{DF(\mathcal{A})}(L^\bullet , I^\bullet ).$

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., $b = 0$. $\square$

Comment #3769 by Owen B on

typo: $C(f)$ should be $C(\alpha)$. additionally, in the statement of the lemma, I believe we can take $L^\bullet\in K(\operatorname{Fil}^f(A))$.

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