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The Stacks project

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[0] \to I^\bullet and C[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

\xymatrix{ 0 \ar[r] & A[0] \ar[r] \ar[d] & B[0] \ar[r] \ar[d] & C[0] \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[0] \to I^\bullet and C[0] \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[0] \to M^\bullet is given by b \oplus c \circ \beta : M \to I^0 \oplus J^0. \square


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