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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