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 (8)

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.

Comment #9944 by Elías Guisado on January 23, 2025 at 06:14

One could add rewrite the last claim of the statement to be:

Moreover: 1. If $A^{n}$ (resp., $B^{n}$ , $C^n$ ) vanishes for all values of $n$ less than $n_1$ (resp., $n_2$ , $n_3$ ), then $I_1^\bullet$ , $I_2^\bullet$ , $I_3^\bullet$ might be chosen so that $I_k^n=0$ for all $n<n_k$ . 2. Given any injective resolution $A^\bullet\to I^\bullet$ we may assume $I_1^\bullet=I^\bullet$ .

And we can assume 1+2 provided that $I^n=0$ for all $n<n_1$ .

Comment #9948 by Elías Guisado on January 24, 2025 at 09:24

One could also add to the statement that it is possible to pick the resolutions $A^\bullet\to I_1^\bullet$ , $B^\bullet\to I_2^\bullet$ , $C^\bullet\to I_3^\bullet$ to be term-wise injective (i.e., $A^n\to I_1^n$ , $B^n\to I_2^n$ , $C^n\to I_3^n$ are monomorphisms, for every $n$ ). Indeed, at the first step of the proof, one chooses $A^\bullet\to I^\bullet$ to be an injective resolution which term-wise injective by Lemma 13.18.3 (or uses the given one assuming it is term-wise injective). When forming the pushout $E^\bullet$ , one uses Homology, Lemma 12.5.13 to guarantee term-wise injectivity of $B^\bullet\to E^\bullet$ . The rest is clear.

Comment #9950 by Elías Guisado on January 24, 2025 at 09:51

Proposal of additional result (either as a second part of this lemma or in a new lemma):

Lemma. Let $\mathcal{A}$ be an abelian category. Let this be a morphism of s.e.s. in $\operatorname{Comp}^+(\mathcal{A})$ . Using Lemma 13.18.9, pick injective resolutions of these s.e.s as in the solid diagram here, and assume that the injective resolutions of $A^\bullet,B^\bullet,C^\bullet,X^\bullet,Y^\bullet,Z^\bullet$ are term-wise injective (the diagonal maps in last diagram). Then there exist dashed morphisms $f_k:I_k\to J_k$ as in the diagram making it fully commutative (in $\operatorname{Comp}^+(\mathcal{A})$ ). Moreover, if $g_k:I_k\to J_k$ are other such morphisms, then $g_k$ is homotopic to $f_k$ .

Proof. Ignoring for a moment $f_2$ , there are unique up to homotopy morphisms $f_k:I_k\to J_k$ , $k=1,3$ making the diagram commute by Lemma 13.18.6. By same result, there is a unique up to homotopy morphism $f_2:I_2\to J_2$ such that this square commutes. It is left to check that $f_2$ is compatible with $f_1,f_3$ in the sense that it renders the diagram fully commutative, that is, that the front squares commute. This might be checked after precomposing by the monomorphism $A^\bullet\to I_1^\bullet$ for the left front square and by the mono $B^\bullet\to I_2^\bullet$ for the right front face. $\square$

Comment #9951 by Elías Guisado on January 24, 2025 at 09:54

(I forgot adding the hypothesis of $\mathcal{A}$ has enough injectives in the statement of #9950.)

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