Resolutions by Adapted Class of Objects and Model Categories - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T06:18:41Z http://mathoverflow.net/feeds/question/68979 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68979/resolutions-by-adapted-class-of-objects-and-model-categories Resolutions by Adapted Class of Objects and Model Categories Mikhail Gudim 2011-06-27T22:39:24Z 2011-12-23T17:30:39Z <p>My question is about the construction of derived functor in the language of model categories. (As it is done for example the paper by Dwyer and Spalinski "Homotopy Theories and Model Categories".) I've just started learning these things, so my question may be trivial.</p> <p>When one looks at the model category of (non-negatively graded) chain complexes, the procedure of taking cofibrant replacement is taking projective resolution of a given complex (Cartan-Eilenberg resolution). This gives a standard recipe for computing derived functor via projective resolution.</p> <p>However, it is a fact that in order to compute derived functors of an additive left exact functor $F: A \to B$ one can take resolution by any adapted to $F$ class of objects $R$. Actually, there is an equivalence (induced by inclusion) between the derived categories of (non-negatively graded) complexes of objects from $R$ and the derived category of $A$. Class of injective (projective) objects is adapted to any functor. (See Gelfand and Manin III.6 for example.)</p> <p>How can one prove these facts in the language of model categories?</p> <p>Thanks for your help,</p> <p>Mikhail</p> http://mathoverflow.net/questions/68979/resolutions-by-adapted-class-of-objects-and-model-categories/84172#84172 Answer by Emily Riehl for Resolutions by Adapted Class of Objects and Model Categories Emily Riehl 2011-12-23T17:30:39Z 2011-12-23T17:30:39Z <p>(Disclaimer: I don't have a copy of Gelfand and Manin to hand, so I'm only conjecturing that what I'm about to say is relevant to your question.)</p> <p>I think what you might be looking for is the construction of derived functors via deformations. This is a generalization of the construction that's used in the context of model categories to so-called homotopical categories, due originally to Dwyer, Kan, Hirschhorn, and Smith. A good summary can be found in the first few sections of <a href="http://arxiv.org/abs/math/0610194" rel="nofollow">this paper</a>.</p> <p>The main idea is the following: if $F \colon C \to D$ is a functor between categories equipped with some notion of weak equivalences satisfying the 2-of-3 property, to construct a left derived functor of $F$, you don't need a full model structure on $C$. Instead it suffices to have an endofunctor $Q$ equipped with a natural weak equivalence $Q \Rightarrow 1_C$ such that $F$ preserves all weak equivalences between objects in the image of $Q$.</p> <p>The theorem is that in this case $LF = FQ$ together with the natural transformation $FQ \Rightarrow F$ is a (point-set) left derived functor of $F$ (meaning, if you compose with the localization functor $D \to Ho(D)$, this becomes a left derived functor in the usual sense). It's really easy to prove once you know it's true. I suggest it as an exercise.</p> <p>The notation is meant to suggest cofibrant replacement. Assuming we have a functorial cofibration - trivial fibration factorization, we can factor maps $\emptyset \to X$ to obtain a cofibrant object $QX$ and a natural weak equivalence $QX \to X$. If $F$ is left Quillen (preserves (trivial) cofibrations and the initial object), then $F$ preserves all trivial cofibrations between cofibrant objects. By Ken Brown's lemma, $F$ then preserves all weak equivalences between cofibrant objects. So the left derived functor of $F$ can be constructed simply by precomposing with some cofibrant replacement.</p> <p>But the point is it doesn't matter what sort of cofibrant replacement we use, or even that it is a cofibrant replacement for some model structure. Maybe this is what's going on with the class of objects adapted to $F$.''</p>