# Acyclic resolutions

Hallo,

I have to worry you one more time with these acyclicity problems, but as I am currently working on derived functors in a.g., I really need to understand derived functors in a very general form.

So my question is just: I have an exact functor $F: K^{+}(A)\rightarrow K(B)$ of the homotopy categories of abelian categories A and B and I know that the derived $RF: D^{+}(A)\rightarrow D(B)$ exists (in the sense of: it is exact and has universal property). Then this does not a priori imply that for each complex in $K^{+}(A)$ I can find a quasiiso to a complex of F-acyclics? I myself would guess that one has to have a triangulated subcategory L of $K^{+}(A)$ which is adapted to F, in the sense of Hartshorne, Residues and Duality. A short hint to if I am right is totally enough, thanks.

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Why do you want to do this? – Mariano Suárez-Alvarez Aug 7 '11 at 22:51
To follow up on Mariano's question: Can you give an example where you find yourself this situation? Moreover, I don't think Hartshorne's R&D uses the word adapted at all (I think this appears in Gelfand-Manin only). What is your main reference and what do you actually want to do? Why are the references you were given so far not sufficient? – t.b. Aug 8 '11 at 8:03
That's right, Hartsorne just speaks of a triangulated subcategory L "as in Theorem I, 5.1.". So as an abbreviation I would like to call such L simply adapted, as in Gelfand-Manin. Well, I want to know this because it seems to me of principal interest: in the simple situation when you have a left exact functor $F:A\rightarrow B$ and an F-adapted class, then you know that the derived exists. Furthermore one knows that then the class of F-acyclic objects also is an F-adapted class, i.e. one can find acyclic resolutions. I am asking whether the existence of the derived implies already enough acycs – Descartes Aug 8 '11 at 10:20
Okay, you clarified the terminology and the academic motivation. Still, you haven't addressed my main question: can you give me an example in which you know that the derived functor exists and you don't know that there exist enough acyclics? – t.b. Aug 8 '11 at 10:40
Well, I don't know such an example, but that's exactly my question: can there be such examples or does the existence of the derived automatically imply enough acyclics. – Descartes Aug 8 '11 at 10:47

If your functor $F$ is exact, meaning that it sends quasi-isomorphisms (quis) to quasi-isomorphisms, then you don't need anything to derive it: by definition of the derived categories, it induces a functor $F':D^+(A) \longrightarrow D(B)$ defined on objects as $F'( \gamma (a)) = \gamma (F(a))$ and on morphisms as $F'(\gamma (f )\circ \gamma (s^{-1})) = \gamma ( F(f) )\circ \gamma (F(s)^{-1})$.

Here, $\gamma$ denotes both the localising functors $K(A) \longrightarrow D(A)$ and $K(B) \longrightarrow D(B)$

This definition makes sense, since you can represent morphisms on the derived category as compositions $f\circ s^{-1}$, where $f, s$ are morphisms of $K(A)$ and $s$ is a quis, going in the opposite direction, but you're inverting them when you construct the derived category, so $s^{-1}$ is a real morphism in the derived category. Now, since $F$ sends quis to quis, $F(s)$ is also a quis, so you can also invert them.

Exercise 1. Verify that $F'$ is well-defined.

Exercise 2. Once convinced, it is handy to delete all the $\gamma$'s in the previous formulas: everybody understands what do you mean.

It is obvious that this $F'$ verifies $\gamma F = F' \gamma$. So it is the right and left derived functor for $F$ simultaneously.

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I don't think that this is the intention. Many people (confusingly) call a triangle functor (i.e. one sending distinguished triangles to distinguished triangles) an exact functor. The idea being "distinguished triangles correspond to short exact sequences". The precise definition of course involves a pair $(F,\alpha)$ of a functor and a natural transformation $\alpha : F\Sigma \to \Sigma F$ "commuting $F$ with the shift $\Sigma$". – t.b. Aug 7 '11 at 18:06
Yes, sorry, I thought the notion "exact" in this context should be clear. Yes, what I mean is also called a triangulated or delta functor, i.e. commuting with shift and sending distinguished triangles into distinguished ones. – Descartes Aug 8 '11 at 6:18
Thanks Theo and Descartes. So my answer doesn't apply to your problem. – a.r. Aug 8 '11 at 12:34