Derived functors vs universal delta functors - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-27T02:56:13Z http://mathoverflow.net/feeds/question/2985 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/2985/derived-functors-vs-universal-delta-functors Derived functors vs universal delta functors Andrew Critch 2009-10-28T02:29:06Z 2009-10-28T15:04:41Z <p>I would like to understand the relationship between the derived category definition of a right derived functor Rf (which involves an initial natural transformation n: Qf &rarr; (Rf)Q, where Q is the map to the derived category) and the "universal delta functor" definition given in Hartshorne III.1.</p> <p>I already know that R^if(A) = H^i(Rf(A)). What I want to know most is:</p> <blockquote> <p>What is the role of the natural transformation n in this comparison?</p> </blockquote> <p>I guess it can be thought of as a natural map from a injective resolution of f(A) to f(an injective resolution of A), but I'm not sure what is the significance of this... Does anyone know a good reference explaining such things?</p> http://mathoverflow.net/questions/2985/derived-functors-vs-universal-delta-functors/2993#2993 Answer by Reid Barton for Derived functors vs universal delta functors Reid Barton 2009-10-28T04:22:15Z 2009-10-28T04:22:15Z <p>I don't have a complete answer, but maybe this is helpful: Unpacking the definition of "universal", a universal delta functor whose 0th functor is f is the same thing as an initial object in the category {delta functors T together with a natural transformation f &rarr; T^0} (provided, I guess, that the former object exists). Giving your n : Qf &rarr; (Rf)Q is the same as giving f &rarr; H^0 ∘ Rf ∘ Q, which looks rather similar.</p> http://mathoverflow.net/questions/2985/derived-functors-vs-universal-delta-functors/3065#3065 Answer by Agusti Roig for Derived functors vs universal delta functors Agusti Roig 2009-10-28T15:04:41Z 2009-10-28T15:04:41Z <p>I haven't checked all the details, but I think the story could go like this. (I have to apologize: it's a bit long.)</p> <p>(1) Let F: <strong>A</strong> ---> <strong>B</strong> be an additive left exact functor between two abelian categories. Take an injective resolution of an object A in <strong>A</strong> :</p> <p>0---> A ---> I^0 ---> I^1 ---> ...</p> <p>Let us call i: A ---> I^0 the first morphism. Apply F to this exact sequence:</p> <p>0---> FA ---> FI^0 ---> FI^1 ---> ...</p> <p>Now, the <em>total</em> right derived functor of F applied to A (thought as a complex concentrated in degree zero) is the complex</p> <p><strong>R</strong>F(A) = { FI^0 ---> FI^1 ---> FI^2 ---> ... }</p> <p>and the <em>classical</em> right derived functors of F are its cohomology:</p> <p>R^nF(A) = H^n(<strong>R</strong>F(A)) = H^n(FI^*) .</p> <p>These {R^nF}_n are a universal cohomological delta-functor and we have a natural transformation of functors</p> <p>qF ===> (<strong>R</strong>F)q</p> <p>which is essentially</p> <p>Fi: FA ---> <strong>R</strong>F(A) </p> <p>(here we have extended F degree-wise to the category of complexes, and this is the degree zero of the natural transformation, because <strong>R</strong>F(A)^0 = FI^0 ).</p> <p>(2) Now, let { T^n : <strong>A</strong> ---> <strong>B</strong> } be a cohomological delta-functor and f^0 : F ---> T^0 a natural transformation. We have to extend this f^0 to a unique morphism of delta-functors { f^n : R^nF ---> T^n } .</p> <p>To do this, observe that, in general, given two right-derivable functors between two, say, model categories </p> <p>F, G: <strong>C</strong> ---> <strong>D</strong> , </p> <p>and a natural transformation between them</p> <p>t: F ===> G </p> <p>we have a natural transformation between the total right derived functors</p> <p><strong>R</strong>t : <strong>R</strong>F ===> <strong>R</strong>G</p> <p>because of the universal property of the derived functors: indeed, if</p> <p>f : qF ===> (<strong>R</strong>F)q and g : qG ===> (<strong>R</strong>G)q</p> <p>are the universal morphisms of the derived functors, then we have a natural transformation</p> <p>gt : F ===> (<strong>R</strong>G)q</p> <p>and, so, because of the universal property of derived functors, a unique natural transformation</p> <p><strong>R</strong>t : <strong>R</strong>F ===> <strong>R</strong>G</p> <p>such that (<strong>R</strong>t)qf = g .</p> <p>(3) So, take our f^0 : F ---> T^0 , extend it to a natural transformation between the degree-wise induced functors between complexes. Passing to the derived functors, we obtain</p> <p><strong>R</strong>f^0 : <strong>R</strong>F ===> <strong>R</strong>T^0 .</p> <p>Taking cohomology, for each n , we get</p> <p>H^n(<strong>R</strong>f^0) : H^n (<strong>R</strong>F) ===> H^n (<strong>R</strong>T^0) .</p> <p>But these are the classical right derived functors, so we have natural transformations</p> <p>R^nf : R^n F ===> R^nT^0</p> <p>and because the classical right derived functors are universal delta-functors, we have unique natural transformations</p> <p>i^n : R^nT^0 ===> T^n</p> <p>which extend the identity </p> <p>i^0 : R^0T^0 = T^0 .</p> <p>The composition</p> <p>i^n R^f : R^F ===> T^n</p> <p>is, I think, the required morphisms of delta-functors that we need.</p>