Lemma 13.18.9 (013T)—The Stacks project

Lemma 13.18.9. Let $\mathcal{A}$ be an abelian category. Assume $\mathcal{A}$ has enough injectives. For any short exact sequence $0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$ of $\text{Comp}^{+}(\mathcal{A})$ there exists a commutative diagram in $\text{Comp}^{+}(\mathcal{A})$

\[ \xymatrix{ 0 \ar[r] & A^\bullet \ar[r] \ar[d] & B^\bullet \ar[r] \ar[d] & C^\bullet \ar[r] \ar[d] & 0 \\ 0 \ar[r] & I_1^\bullet \ar[r] & I_2^\bullet \ar[r] & I_3^\bullet \ar[r] & 0 } \]

where the vertical arrows are injective resolutions and the rows are short exact sequences of complexes. In fact, given any injective resolution $A^\bullet \to I^\bullet $ we may assume $I_1^\bullet = I^\bullet $.

Proof. Step 1. Choose an injective resolution $A^\bullet \to I^\bullet $ (see Lemma 13.18.3) or use the given one. Recall that $\text{Comp}^{+}(\mathcal{A})$ is an abelian category, see Homology, Lemma 12.13.9. Hence we may form the pushout along the map $A^\bullet \to I^\bullet $ to get

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

Because of the $5$-lemma and the last assertion of Homology, Lemma 12.13.12 the map $B^\bullet \to E^\bullet $ is a quasi-isomorphism. Note that the lower short exact sequence is termwise split, see Homology, Lemma 12.27.2. Hence it suffices to prove the lemma when $0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$ is termwise split.

Step 2. Choose splittings. In other words, write $B^ n = A^ n \oplus C^ n$. Denote $\delta : C^\bullet \to A^\bullet [1]$ the morphism as in Homology, Lemma 12.14.10. Choose injective resolutions $f_1 : A^\bullet \to I_1^\bullet $ and $f_3 : C^\bullet \to I_3^\bullet $. (If $A^\bullet $ is a complex of injectives, then use $I_1^\bullet = A^\bullet $.) We may assume $f_3$ is injective in every degree. By Lemma 13.18.6 we may find a morphism $\delta ' : I_3^\bullet \to I_1^\bullet [1]$ such that $\delta ' \circ f_3 = f_1[1] \circ \delta $ (equality of morphisms of complexes). Set $I_2^ n = I_1^ n \oplus I_3^ n$. Define

\[ d_{I_2}^ n = \left( \begin{matrix} d_{I_1}^ n & (\delta ')^ n \\ 0 & d_{I_3}^ n \end{matrix} \right) \]

and define the maps $B^ n \to I_2^ n$ to be given as the sum of the maps $A^ n \to I_1^ n$ and $C^ n \to I_3^ n$. Everything is clear. $\square$

Comments (4)

Comment #7872 by Anonymous on November 13, 2022 at 13:07

In Step 1, it says "Hence we may form the pushout along the injective map $A^{\bullet} \rightarrow I^{\bullet}$ ''. But is this map really injective? Maybe what was meant is "Hence we may form the pushout along $A^{\bullet} \to I^{\bullet}$ to get...'' And then "Since $A^{\bullet} \to B^{\bullet}$ is an injection, the pushout square implies that $B^{\bullet} \to E^{\bullet}$ is a quasi-isomorphism." (Assuming that was the intention...also was the latter fact proved in the Stacks project?)

Comment #8143 by Aise Johan de Jong on February 27, 2023 at 13:41

Thanks and slightly differently fixed here.

Comment #8256 by Nicolas Weiss on March 13, 2023 at 13:48

One of the typos pointed out above hasn't been fixed in the correction, namely that $B\rightarrow E$ is a q-iso (and not something about $B\rightarrow A$ ).

Comment #8898 by Stacks project on April 09, 2024 at 16:47

Thanks and fixed here.

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3 comment(s) on Section 13.18: Injective resolutions

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