Derived Functors in arbitrary triangulated categories - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T07:55:37Z http://mathoverflow.net/feeds/question/11268 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/11268/derived-functors-in-arbitrary-triangulated-categories Derived Functors in arbitrary triangulated categories Hanno Becker 2010-01-09T23:40:54Z 2010-01-10T00:41:39Z <p>Let <code>${\mathcal D}$</code> be a triangulated category, <code>${\mathcal C}$</code> a triangulated subcategory and <code>$Q: {\mathcal D}\to {\mathcal D}/{\mathcal C}$</code> the corresponding Verdier-localization. Now suppose we have a triangulated functor <code>${\mathbb F}: {\mathcal D}\to {\mathcal T}$</code> to some other triangulated category ${\mathcal T}$.</p> <p>My question is the following: Under which circumstances do we have some kind of "right derived" functor of <code>${\mathbb F}$</code> with respect to <code>${\mathcal C}$</code>? By that I mean a triangulated functor <code>$\textbf{R}{\mathbb F}: {\mathcal D}/{\mathcal C}\to {\mathcal T}$</code> together with a natural transformation <code>${\mathbb F}\Rightarrow \textbf{R}{\mathbb F}\circ Q$</code> which is initial with this property.</p> <p>Does there exist such a treatment of derived functors in arbitrary triangulated categories? </p> <p>Thank you.</p> http://mathoverflow.net/questions/11268/derived-functors-in-arbitrary-triangulated-categories/11272#11272 Answer by Leonid Positselski for Derived Functors in arbitrary triangulated categories Leonid Positselski 2010-01-10T00:33:59Z 2010-01-10T00:41:39Z <p>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&rarr;Y in D with a cone in C such that for any morphism Y&rarr;Z in D with a cone in C there exists a morphism Z&rarr;W in D with a cone in C such that F(Y)&rarr;F(W) is an isomorphism. Then one defines RF(X) as F(Y).</p>