Lemma 13.26.10. Let \mathcal{A} be an abelian category. Let K^\bullet , I^\bullet \in K(\text{Fil}^ f(\mathcal{A})). Assume K^\bullet is filtered acyclic and I^\bullet bounded below and consisting of filtered injective objects. Any morphism K^\bullet \to I^\bullet is homotopic to zero: \mathop{\mathrm{Hom}}\nolimits _{K(\text{Fil}^ f(\mathcal{A}))}(K^\bullet , I^\bullet ) = 0.
Proof. Let \alpha : K^\bullet \to I^\bullet be a morphism of complexes. Assume that \alpha ^ j = 0 for j < n. We will show that there exists a morphism h : K^{n + 1} \to I^ n such that \alpha ^ n = h \circ d. Thus \alpha will be homotopic to the morphism of complexes \beta defined by
\beta ^ j = \left\{ \begin{matrix} 0
& \text{if}
& j \leq n
\\ \alpha ^{n + 1} - d \circ h
& \text{if}
& j = n + 1
\\ \alpha ^ j
& \text{if}
& j > n + 1
\end{matrix} \right.
This will clearly prove the lemma (by induction). To prove the existence of h note that \alpha ^ n \circ d_ K^{n - 1} = 0 since \alpha ^{n - 1} = 0. Since K^\bullet is filtered acyclic we see that d_ K^{n - 1} and d_ K^ n are strict and that
0 \to \mathop{\mathrm{Im}}(d_ K^{n - 1}) \to K^ n \to \mathop{\mathrm{Im}}(d_ K^ n) \to 0
is an exact sequence of the exact category \text{Fil}^ f(\mathcal{A}), see Homology, Lemma 12.19.15. Hence we can think of \alpha ^ n as a map into I^ n defined on \mathop{\mathrm{Im}}(d_ K^ n). Using that \mathop{\mathrm{Im}}(d_ K^ n) \to K^{n + 1} is a strict monomorphism and that I^ n is filtered injective we may lift this map to a map h : K^{n + 1} \to I^ n as desired, see Lemma 13.26.4. \square
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