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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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    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.2011-08-08

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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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    Thanks Theo and Descartes. So my answer doesn't apply to your problem.2011-08-08