Lemma 13.31.6 (070Y)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 13: Derived Categories
Section 13.31: K-injective complexes
Lemma 13.31.6 (cite)

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Lemma 13.31.6. Let $\mathcal{A}$ be an abelian category. Let $F : K(\mathcal{A}) \to \mathcal{D}'$ be an exact functor of triangulated categories. Then $RF$ is defined at every complex in $K(\mathcal{A})$ which is quasi-isomorphic to a K-injective complex. In fact, every K-injective complex computes $RF$ .

Proof. By Lemma 13.14.4 it suffices to show that $RF$ is defined at a K-injective complex, i.e., it suffices to show a K-injective complex $I^\bullet$ computes $RF$ . Any quasi-isomorphism $I^\bullet \to N^\bullet$ is a homotopy equivalence as it has an inverse by Lemma 13.31.2. Thus $I^\bullet \to I^\bullet$ is a final object of $I^\bullet /\text{Qis}(\mathcal{A})$ and we win. $\square$

Comments (1)

Comment #9464 by Elías Guisado on June 23, 2024 at 12:01

For anyone that might care: from Lemma 13.31.2 we get that $\label{1}\tag{1} \operatorname{Hom}_{K(\mathcal{A})}(N^\bullet,I^\bullet)\to\operatorname{Hom}_{K(\mathcal{A})}(I^\bullet,I^\bullet)$ is bijective. In particular, $I^\bullet\to N^\bullet$ has a left inverse. This alone already implies that $I^\bullet\to I^\bullet$ is a final object of $I^\bullet/\operatorname{Qis}(\mathcal{A})$ .

Even though one does not need it in the proof, actually $I^\bullet\to N^\bullet$ is invertible in $K(\mathcal{A})$ . Bijectivity of $\eqref{1}$ means that $I^\bullet\to N^\bullet$ has a unique left inverse, and this implies existence of an inverse in a preadditive category. The dual result is proven here.

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