Lemma 13.31.8 (070L)—The Stacks project

Lemma 13.31.8. Let $\mathcal{A}$ be an abelian category. Let

$\ldots \to I_3^\bullet \to I_2^\bullet \to I_1^\bullet$

be an inverse system of complexes. Assume

each $I_ n^\bullet$ is $K$ -injective,
each map $I_{n + 1}^ m \to I_ n^ m$ is a split surjection,
the limits $I^ m = \mathop{\mathrm{lim}}\nolimits I_ n^ m$ exist.

Then the complex $I^\bullet$ is K-injective.

Proof. We urge the reader to skip the proof of this lemma. Let $M^\bullet$ be an acyclic complex. Let us abbreviate $H_ n(a, b) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(M^ a, I_ n^ b)$ . With this notation $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(M^\bullet , I^\bullet )$ is the cohomology of the complex

$\prod _ m \mathop{\mathrm{lim}}\nolimits \limits _ n H_ n(m, m - 2) \to \prod _ m \mathop{\mathrm{lim}}\nolimits \limits _ n H_ n(m, m - 1) \to \prod _ m \mathop{\mathrm{lim}}\nolimits \limits _ n H_ n(m, m) \to \prod _ m \mathop{\mathrm{lim}}\nolimits \limits _ n H_ n(m, m + 1)$

in the third spot from the left. We may exchange the order of $\prod$ and $\mathop{\mathrm{lim}}\nolimits$ and each of the complexes

$\prod _ m H_ n(m, m - 2) \to \prod _ m H_ n(m, m - 1) \to \prod _ m H_ n(m, m) \to \prod _ m H_ n(m, m + 1)$

is exact by assumption (1). By assumption (2) the maps in the systems

$\ldots \to \prod _ m H_3(m, m - 2) \to \prod _ m H_2(m, m - 2) \to \prod _ m H_1(m, m - 2)$

are surjective. Thus the lemma follows from Homology, Lemma 12.31.4. $\square$

Comments (3)

Comment #853 by Bhargav Bhatt on July 26, 2014 at 03:27

Suggested slogan: The limit of a "split" tower of K-injective complexes is K-injective.

Comment #4337 by Manuel Hoff on August 09, 2019 at 12:15

In the formulation of the Lemma, it is mentioned two times that the complexes $I_k^\bullet$ are K-injective (one time in the second sentence and a second time in (1)).

Comment #4487 by Johan on September 02, 2019 at 12:17

Thanks and fixed here.

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