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.
 A: 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.
