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Let $F:A\longrightarrow B$ be a left exact functor of Abelian categories. My question is about the derived functor $RF: D(A)\longrightarrow D(B)$.

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.

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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 this article. 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).

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If the category $A$ has enough injectives, as well as infinite products, then every object has a K-injective resolution, so the right derived functor exists. –  Liran Shaul Nov 30 '10 at 10:42
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@Liran: one probably has to require a little more. If the category $A$ has enough injectives and exact infinite (or at least countable) products, then every complex has a K-injective resolution. The same conclusion also holds if $A$ is a Grothendieck abelian category, i.e., has exact colimits and a set of generators (even though countable products in such a category may not be exact). –  Leonid Positselski Nov 30 '10 at 12:13
    
Can you give a simple proof of the existence of $I$? –  Hao Sun Dec 1 '10 at 11:15
    
@Hao Sun: Sure, follow the link in my post (part b of Lemma 4.1). See also Theorem 12.8 on page 33 in esi.ac.at/preprints/esi2063.pdf for a more detailed proof (although it's stated there for projective resolutions). –  Theo Buehler Dec 1 '10 at 11:55
    
@Theo Buehler: Thank you so much! –  Hao Sun Dec 2 '10 at 4:18
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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).

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 homotopy 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).

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.

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.

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Is there really any reason to consider unbounded chain complexes (outside of homotopy theory)? Toën-Vezzosi restrict to bdd chain complexes and are still able to develop a pretty comprehensive theory of derived algebraic geometry. Bdd-below (or above) chain complexes have a nice homotopy theory by the Dold-Kan correspondence, so I'm curious about whether or not we really need the added complexity. Also, $K(A)$ should probably be defined to be be "Chain complexes modulo homotopy" or "triangulated", since at least in homotopy theory, "derived category" and "homotopy category" are synonymous. –  Harry Gindi Nov 30 '10 at 10:38
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If you are seriously interested in cohomology, Grothendieck duality, and these kind of things, the the closed monoidal structure is extremely useful. Compare the statement of the internal tensor-hom adjunction in classical references like Hartshorne's with the modern approach that you can find in Lipman's book. In topology, is as considering non-connective spectra. You gain the existence of arbitrary coproducts, and as a consequence the possibility of a sereies of constructions like, e.g. Brown representability. –  Leo Alonso Nov 30 '10 at 11:06
    
Dear Leo, is it not true that the monoidal structure on simplicial abelian groups and simplicial modules is actually closed (and also that this category is complete-cocomplete as well as homotopy-complete-cocomplete)? I could have sworn that it was. Is there an issue when considering simplicial abelian sheaves (or simplicial $\cal O_X$-modules) that I'm overlooking? Thanks. –  Harry Gindi Nov 30 '10 at 11:20
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You are right, and me too, I guess. In a few words, $D^{\leq 0}$ is closed monoidal but not triangulated (just suspended in the sense of Keller-Vossieck), while $D^{-}$ is triangulated but not closed monoidal. In either case, boundedness causes some kind of trouble. Simplicial (sheaves of) modules are OK, but not for duality, or these kind of results. –  Leo Alonso Nov 30 '10 at 11:40
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I don't think so. The treatments I know of unbounded complexes (Spaltenstein, Bökstedt-Neeman, Franke, myself with Jeremías and Souto, Serpé... to name a few) work mostly over Grothendieck abelian categories or sheaf categories regardless of any base field. The trouble might arise if you want to look at DG modules over a DG algebra but even in this case you may use Brown representability due to the existence of compact generators. So, perhaps, the short answer is "no". –  Leo Alonso Nov 30 '10 at 12:30
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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.

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 Homotopy Limit Functors on Model Categories and Homotopical Categories).

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.

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I have not read this and already I don't see a good reason for the downvote... –  Sean Tilson Feb 16 '11 at 19:20
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