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

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

Comment #8256 by Nicolas Weiss on

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

There are also:

  • 3 comment(s) on Section 13.18: Injective resolutions

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