7
$\begingroup$

This is a crosspost of this MSE question according to the recommendation in the comments. I know this question is elementary, but I'm hoping the author of these notes could provide a more detailed explanation.

Here is an excerpt from some notes I stumbled upon online: enter image description here

From what I understand, the "elementary proof" is just the fundamental lemma of homological algebra which says the homotopy type of chain maps out of projective resolutions is determined by maps between the objects being resolved.

It seems that the author says that the image of every $F$-acyclic resolution is homotopic to some injective/projective resolution, but I don't think this follows from the fundamental lemma.

Why is this true? How does it prove one may compute derived functors using $F$-acyclics? Does this approach circumvent dimension shifting?

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ Perhaps I am misunderstanding something or misremembering some definition but: suppose we take $F$ to be the identity functor, so every object is $F$-acyclic. The $0$ complex is a perfectly good injective resolution of the zero object, and any acyclic complex living in degrees $\geq 0$ is an $F$-acyclic resolution. But we can, in general, find an acyclic complex which is not contractible, i.e. not homotopic to the $0$ complex... $\endgroup$
    – Greg Stevenson
    Feb 17, 2017 at 22:13
  • $\begingroup$ Hmm, it would seem to me that you're right (though I should probably check in more detail). If we cannot show that the map of chain complexes is a homotopy equivalence, then, we must still be able to show that it is a quasi-isomorphism. How do we do that? $\endgroup$
    – Dominic Wynter
    Feb 19, 2017 at 4:48

1 Answer 1

Reset to default
2
+50
$\begingroup$

This is explained on the n-lab here. In particular in Theorem 3.15.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.