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

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

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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