Lemma 13.26.7. Let $\mathcal{A}$ be an abelian category with enough injectives. Let $f : A \to B$ be a morphism of $\text{Fil}^ f(\mathcal{A})$. Given filtered quasi-isomorphisms $A[0] \to I^\bullet $ and $B[0] \to J^\bullet $ where $I^\bullet , J^\bullet $ are complexes of filtered injective objects with $I^ n = J^ n = 0$ for $n < 0$, then there exists a commutative diagram

**Proof.**
As $A[0] \to I^\bullet $ and $C[0] \to J^\bullet $ are filtered quasi-isomorphisms we conclude that $a : A \to I^0$, $b : B \to J^0$ and all the morphisms $d_ I^ n$, $d_ J^ n$ are strict, see Homology, Lemma 12.19.15. We will inductively construct the maps $f^ n$ in the following commutative diagram

Because $A \to I^0$ is a strict monomorphism and because $J^0$ is filtered injective, we can find a morphism $f^0 : I^0 \to J^0$ such that $f^0 \circ a = b \circ f$, see Lemma 13.26.4. The composition $d_ J^0 \circ b \circ f$ is zero, hence $d_ J^0 \circ f^0 \circ a = 0$, hence $d_ J^0 \circ f^0$ factors through a unique morphism

As $\mathop{\mathrm{Im}}(d_ I^0) \to I^1$ is a strict monomorphism we can extend the displayed arrow to a morphism $f^1 : I^1 \to J^1$ by Lemma 13.26.4 again. And so on. $\square$

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