MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 7 down vote accepted

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).

share|cite|improve this answer
Thank you, Leonid! – Hanno Becker Jan 10 '10 at 8:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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