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
Comment #4337 by Manuel Hoff on
Comment #4487 by Johan on
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