Lemma 13.26.8. Let $\mathcal{A}$ be an abelian category with enough injectives. Let $0 \to A \to B \to C \to 0$ be a short exact sequence in $\text{Fil}^ f(\mathcal{A})$. Given filtered quasi-isomorphisms $A \to I^\bullet$ and $C \to J^\bullet$ 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

$\xymatrix{ 0 \ar[r] & A \ar[r] \ar[d] & B \ar[r] \ar[d] & C \ar[r] \ar[d] & 0 \\ 0 \ar[r] & I^\bullet \ar[r] & M^\bullet \ar[r] & J^\bullet \ar[r] & 0 }$

where the lower row is a termwise split sequence of complexes.

Proof. As $A \to I^\bullet$ and $C \to J^\bullet$ are filtered quasi-isomorphisms we conclude that $a : A \to I^0$, $c : C \to J^0$ and all the morphisms $d_ I^ n$, $d_ J^ n$ are strict, see Homology, Lemma 13.13.4. We are going to step by step construct the south-east and the south arrows in the following commutative diagram

$\xymatrix{ B \ar[r]_\beta \ar[rd]^ b & C \ar[r]_ c \ar[rd]^{\overline{b}} & J^0 \ar[d]^{\delta ^0} \ar[r] & J^1 \ar[d]^{\delta ^1} \ar[r] & \ldots \\ A \ar[u]^\alpha \ar[r]^ a & I^0 \ar[r] & I^1 \ar[r] & I^2 \ar[r] & \ldots }$

As $A \to B$ is a strict monomorphism, we can find a morphism $b : B \to I^0$ such that $b \circ \alpha = a$, see Lemma 13.26.4. As $A$ is the kernel of the strict morphism $I^0 \to I^1$ and $\beta = \mathop{\mathrm{Coker}}(\alpha )$ we obtain a unique morphism $\overline{b} : C \to I^1$ fitting into the diagram. As $c$ is a strict monomorphism and $I^1$ is filtered injective we can find $\delta ^0 : J^0 \to I^1$, see Lemma 13.26.4. Because $B \to C$ is a strict epimorphism and because $B \to I^0 \to I^1 \to I^2$ is zero, we see that $C \to I^1 \to I^2$ is zero. Hence $d_ I^1 \circ \delta ^0$ is zero on $C \cong \mathop{\mathrm{Im}}(c)$. Hence $d_ I^1 \circ \delta ^0$ factors through a unique morphism

$\mathop{\mathrm{Coker}}(c) = \mathop{\mathrm{Coim}}(d_ J^0) = \mathop{\mathrm{Im}}(d_ J^0) \longrightarrow I^2.$

As $I^2$ is filtered injective and $\mathop{\mathrm{Im}}(d_ J^0) \to J^1$ is a strict monomorphism we can extend the displayed morphism to a morphism $\delta ^1 : J^1 \to I^2$, see Lemma 13.26.4. And so on. We set $M^\bullet = I^\bullet \oplus J^\bullet$ with differential

$d_ M^ n = \left( \begin{matrix} d_ I^ n & (-1)^{n + 1}\delta ^ n \\ 0 & d_ J^ n \end{matrix} \right)$

Finally, the map $B \to M^\bullet$ is given by $b \oplus c \circ \beta : M \to I^0 \oplus J^0$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).