Lemma 13.26.9. Let $\mathcal{A}$ be an abelian category with enough injectives. For every $K^\bullet \in K^{+}(\text{Fil}^ f(\mathcal{A}))$ there exists a filtered quasi-isomorphism $K^\bullet \to I^\bullet $ with $I^\bullet $ bounded below, each $I^ n$ a filtered injective object, and each $K^ n \to I^ n$ a strict monomorphism.

**Proof.**
After replacing $K^\bullet $ by a shift (which is harmless for the proof) we may assume that $K^ n = 0$ for $n < 0$. Consider the short exact sequences

of the exact category $\text{Fil}^ f(\mathcal{A})$ and the maps $u_ i : \mathop{\mathrm{Coim}}(d_ K^ i) \to \mathop{\mathrm{Ker}}(d_ K^{i + 1})$. For each $i \geq 0$ we may choose filtered quasi-isomorphisms

with $I_{ker, i}^ n, I_{coim, i}^ n$ filtered injective and zero for $n < 0$, see Lemma 13.26.6. By Lemma 13.26.7 we may lift $u_ i$ to a morphism of complexes $u_ i^\bullet : I_{coim, i}^\bullet \to I_{ker, i + 1}^\bullet $. Finally, for each $i \geq 0$ we may complete the diagrams

with the lower sequence a termwise split exact sequence, see Lemma 13.26.8. For $i \geq 0$ set $d_ i : I_ i^\bullet \to I_{i + 1}^\bullet $ equal to $d_ i = \alpha _{i + 1} \circ u_ i^\bullet \circ \beta _ i$. Note that $d_ i \circ d_{i - 1} = 0$ because $\beta _ i \circ \alpha _ i = 0$. Hence we have constructed a commutative diagram

Here the vertical arrows are filtered quasi-isomorphisms. The upper row is a complex of complexes and each complex consists of filtered injective objects with no nonzero objects in degree $< 0$. Thus we obtain a double complex by setting $I^{a, b} = I_ a^ b$ and using

the map $d_ a^ b$ and using for

the map $d_{I_ a}^ b$. Denote $\text{Tot}(I^{\bullet , \bullet })$ the total complex associated to this double complex, see Homology, Definition 12.18.3. Observe that the maps $K^ n[0] \to I_ n^\bullet $ come from maps $K^ n \to I^{n, 0}$ which give rise to a map of complexes

We claim this is a filtered quasi-isomorphism. As $\text{gr}(-)$ is an additive functor, we see that $\text{gr}(\text{Tot}(I^{\bullet , \bullet })) = \text{Tot}(\text{gr}(I^{\bullet , \bullet }))$. Thus we can use Homology, Lemma 12.25.4 to conclude that $\text{gr}(K^\bullet ) \to \text{gr}(\text{Tot}(I^{\bullet , \bullet }))$ is a quasi-isomorphism as desired. $\square$

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