The Stacks project

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 $RF$. Consider a quasi-isomorphism $I^\bullet \to N^\bullet $. By Lemma 13.31.2 it has a left inverse. Thus $I^\bullet \to I^\bullet $ is a final object of $I^\bullet /\text{Qis}(\mathcal{A})$ and we win. $\square$


Comments (4)

Comment #9464 by on

For anyone that might care: from Lemma 13.31.2 we get that is bijective. In particular, has a left inverse. This alone already implies that is a final object of .

Even though one does not need it in the proof, actually is invertible in . Bijectivity of \eqref{1} means that has a unique left inverse, and this implies existence of an inverse in a preadditive category. The dual result is proven here.

Comment #10816 by Jonas Frank on

I don't think the statement in the second sentence of the proof is correct: Let and be acyclic. Then the map is a quasi-isomorphism, but, in general, not a homotopy equivalence. (However, as Elías Guisado points out, the existence of a left inverse is sufficient to prove the lemma anyway.)

Comment #10870 by on

Sorry, yes, I forgot to answer Elias comment earlier this summer. Thanks for insisting! Now fixed here.

Comment #10882 by on

Correcting myself in #9464:

  1. “In a preadditive category, a morphism with a unique left inverse is invertible.” This is false; for an example, consider the inclusion in the category of chain complexes over a preadditive category , where is not zero. Then has a unique left inverse, but is not invertible. What is true is: “in a preadditive category, an endomorphism with a unique left inverse is invertible.”

  2. It is also not the case that a quasi-isomorphism , where is K-injective, needs to be a homotopy equivalence. For an example, take and take to be an acyclic but not contractible cochain complex. For instance, could be any non-split short exact sequence of locally finite-free sheaves over a ringed space (here's one). What is true is a quasi-isomorphism with K-injective domain has a unique left homotopy inverse.

There are also:

  • 5 comment(s) on Section 13.31: K-injective complexes

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