Derived functor - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T07:08:35Z http://mathoverflow.net/feeds/question/47771 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47771/derived-functor Derived functor Messi 2010-11-30T09:16:24Z 2010-11-30T10:29:50Z <p>Let $F:A\longrightarrow B$ be a left exact functor of Abelian categories. My question is about the derived functor $RF: D(A)\longrightarrow D(B)$.</p> <p>Let $X$ be an object of $A$. If $0\longrightarrow X\longrightarrow I^0 \longrightarrow I^1\longrightarrow \cdots$ is an injective resolution of $X$, we know $RF(X)=F(I^n)$. When $X$ is a complex in $A$, can we write $RF(X)$ explicitly? The abstract definition of $RF$ is not easy to understand.</p> http://mathoverflow.net/questions/47771/derived-functor/47775#47775 Answer by Theo Buehler for Derived functor Theo Buehler 2010-11-30T09:31:09Z 2010-11-30T09:31:09Z <p>If $X$ is a left bounded complex, you may take a quasi-isomorphism $X \to I$ where $I$ is a left bounded complex of injectives. Such a complex exists and can be found explicitly in many books on homological algebra (see Weibel or Gelfand-Manin) - the most common approach seems to be using a "Cartan-Eilenberg resolution". An easier and more general construction is given in Lemma 4.1 of <a href="http://www.math.jussieu.fr/~keller/publ/csc.pdf" rel="nofollow">this article</a>. Then we simply have $RF(X) = F(I)$. For $X$ unbounded the question is much more subtle and depends on a host of assumption on the category $A$ (this was first treated systematically by Spaltenstein, as far as I know).</p> http://mathoverflow.net/questions/47771/derived-functor/47777#47777 Answer by Harry Gindi for Derived functor Harry Gindi 2010-11-30T09:52:17Z 2010-11-30T10:29:50Z <p>The "correct" abstract definition of a derived functor involves the theory of "homotopical categories". An abelian category $\cal A$ does not have enough structure alone to consider homotopical properties. To this end, we usually look at $Ch^+(\cal A)$ or $s\cal A$ of chain complexes in $\cal A$ or simplicial objects in $\cal A$ respectively. We equip these larger categories with the structure of a homotopical category by fixing the lluf subcategories consisting of quasi-isomorphisms or weak homotopy equivalences respectively (by the Dold-Kan correspondence, these are essentially identical). In the terminology of Dwyer-Hirschhorn-Kan-Smith, left-exact functors are deformable in the sense that they preserve weak equivalences (that is to say, they are homotopical) on a deformation retract of our original category. So in truth, a derived functor is simply the evaluation of a functor on a suitable approximation of our original object by a homotopically equivalent one on which the functor is well-behaved. </p> <p>When $\cal A$ has enough structure, we can upgrade the homotopical structure on our homotopical version of $\cal A$ (that is, chain complexes or simplicial objects) to a model structure, which makes it substantially easier to compute derived functors, since the existence of approximations is guaranteed, and further, such approximations can usually be taken to be functorial. For Quillen functors between model categories (these are functors that become homotopical on the deformation retracts consisting of cofibrant or fibrant objects), we can, in the presence of functorial factorization, define the derived functors to be the composition of the functorial approximation and the original functor. These approximations have the property that they descend to the usual Kan extensions at the level of the derived categories (for a proof, see Dwyer-Hirschhorn-Kan-Smith <em>Homotopy Limit Functors on Model Categories and Homotopical Categories</em>). </p> <p>That is, in your case, we approximate the chain complex $X$ (concentrated in degree zero, so it is constant) by an injective approximation (fibrant for the injective model structure) and evaluate the functor on this new complex. In fact, if injective approximations exist for all chain complexes, we see that when we compose with an appropriate functorial injective approximation, our functor becomes homotopical (it preserves quasi-isomorphisms) and therefore descends uniquely by the universal property of localization to a total derived functor between derived categories of chain complexes.</p> http://mathoverflow.net/questions/47771/derived-functor/47780#47780 Answer by Leo Alonso for Derived functor Leo Alonso 2010-11-30T10:23:28Z 2010-11-30T10:23:28Z <p>It's my opinion that derived functors are better understood through a universal property. On derived categories it is possible to express the condition of derivability in a concise way due to the additivity of everything in sight. My impression is that you do not need the full machinery of homotopical algebra unless you are in a non-additive situation (which is often the case, I should add).</p> <p>Let me spell it out. Let $F\colon A\to B$ be a left exact functor between Abelian categories. Being additive, this induces a (Delta) functor between their <em>homotopy</em> categories that by abuse of notation we keep calling it $F\colon K(A)\to K(B)$. Recall that $K(A)$ is the category of (possibly unbounded) complexes of objects of $A$ with maps homotopy classes of chain maps. Now if we localize a homotopy category making invertible the quasi-isomorphisms (i.e. those maps that induce isomorphisms in homology) we obtain the derived category $D(A)$. If we call the canonical localization functor $Q \colon K(A) \to D(A)$ and similarly for $B$, the problem of deriving $F$ is to extend "in an optimal way" $Q \circ F$ to $D(A)$. The solution, if it exists, it is expressed by the following universal property The (Delta) functor $RF \colon D(A)\to D(B)$ is the derived functor of $F$ if there is a natural transformation $Q \circ F \to RF \circ Q$ such that for any other (Delta) functor $G \colon D(A)\to D(B)$ the natural transformation induces a bijection $$Hom(RF, G) \longrightarrow Hom(Q \circ F, G \circ Q)$$ I suggest to draw a diagram (It is difficult to draw a digram here, but very instructive if you do). </p> <p>It turns out that if there are unbounded injective resolutions on $K(A)$ they provide a right adjoint to $Q \colon K(A) \to D(A)$ that automatically gives a way to express $RF$. This is related to the Bousfield localization philosophy in the context of derived categories. </p> <p>Yet another question is the description of the adjoint functor i.e. the unbounded injective resolution. If the complex is bounded below, then the resolution can be constructed step by step. In the case that $A$ is the category of modules over a ring, one can use the exactness of products and use countable homotopy limits as in Bökstedt-Neeman. In more general cases such as Grothendieck categories, one needs more sophisticated tools.</p>