Let ${\mathcal D}$ be a triangulated category, ${\mathcal C}$ a triangulated subcategory and $Q: {\mathcal D}\to {\mathcal D}/{\mathcal C}$ the corresponding Verdier-localization. Now suppose we have a triangulated functor ${\mathbb F}: {\mathcal D}\to {\mathcal T}$ to some other triangulated category ${\mathcal T}$.


My question is the following: Under which circumstances do we have some kind of "right derived" functor of ${\mathbb F}$ with respect to ${\mathcal C}$? By that I mean a triangulated functor $\textbf{R}{\mathbb F}: {\mathcal D}/{\mathcal C}\to {\mathcal T}$ together with a natural transformation ${\mathbb F}\Rightarrow \textbf{R}{\mathbb F}\circ Q$ which is initial with this property.


Does there exist such a treatment of derived functors in arbitrary triangulated categories? 


Thank you.



Derived Functors in arbitrary triangulated categories
Yes, there exists such a treatment by Deligne, see "Cohomologie a supports propres", SGA4, Tome 3, Lect. Notes Math. 305, subsections 1.2.1-1.2.2. Basically, what one needs is that for any object X in D there exists a morphism X→Y in D with a cone in C such that for any morphism Y→Z in D with a cone in C there exists a morphism Z→W in D with a cone in C such that F(Y)→F(W) is an isomorphism. Then one defines RF(X) as F(Y).


