Intuition for the satellite of a functor - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T15:50:49Z http://mathoverflow.net/feeds/question/29552 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29552/intuition-for-the-satellite-of-a-functor Intuition for the satellite of a functor Eric A. Bunch 2010-06-25T22:44:56Z 2010-12-09T11:01:33Z <p>Occasionally in math I come across constructions or tools that are a bit convoluted. I can look at these constructions and see that they indeed perform the task they were made to do, but sometimes I can't see why they $\textit{should}$ perform this task; why is that the logical thing to choose for the job? Right now, I'm having this issue with the satellite of a functor.</p> <p>Just to recall, given an additive functor between two abelian categories $F:\mathcal{C} \rightarrow \mathcal{D}$, the satellite is another functor $S_-^1(F): \mathcal{C} \rightarrow \mathcal{D}$ defined by </p> <p>$S_-^1(F)(M) = lim(ker(F(M) \rightarrow F(P)))$</p> <p>where $0 \rightarrow M \rightarrow P \rightarrow N \rightarrow 0$ is an exact sequence with $P$ a projective object. Then a derived functor is formed by taking iterations of the satellite: $S_-^n(F) = S_-(S_-^{n-1}(F))$. More information can be found on <a href="http://ncatlab.org/nlab/show/satellite" rel="nofollow">nlab</a>.</p> <p>I am learning about derived functors in a slightly different setting, namely with nonadditive categories where there are not necessarily enough projectives in the category $\mathcal{C}$, and so the definition is modified slightly; perhaps the definition is more transparent in the standard setting. </p> <p>So my question is</p> <blockquote> <p>Is there any intuition for why the satellite is the correct tool to use for obtaining derived functors? If I needed to create a derived functor out of a given functor, is there a logical progression that would lead me to define the satellite?</p> </blockquote> http://mathoverflow.net/questions/29552/intuition-for-the-satellite-of-a-functor/29830#29830 Answer by Mattia Talpo for Intuition for the satellite of a functor Mattia Talpo 2010-06-28T20:27:52Z 2010-06-28T20:27:52Z <p>I don't know if this can be considered as "intuition", anyway</p> <p>another way to think about derived functors is the following: given an abelian category $\mathcal{A}$, you can define its derived category $\mathcal{D}(\mathcal{A})$ (or its variants of bounded complexes in one or both directions). You get $\mathcal{D}(\mathcal{A})$ by "localizing" the homotopy category of complexes $\mathcal{K}(\mathcal{A})$, where the objects are complexes in $\mathcal{A}$ and maps are maps of complexes up to homotopy, at the system of quasi-isomorphisms.</p> <p>There is a natural localization functor $\pi_A:\mathcal{K}(\mathcal{A})\to \mathcal{D}(\mathcal{A})$. If $F:\mathcal{A}\to \mathcal{B}$ is your additive functor between abelian categories, and $\mathcal{K}(F):\mathcal{K}(\mathcal{A})\to \mathcal{K}(\mathcal{B})$ is the induced functor, it is natural to ask for an "extension" of $\mathcal{K}(F)$ to the derived category $\mathcal{D}(\mathcal{A})$, with values in $\mathcal{D}(\mathcal{B})$. In other words this would be a functor $RF:\mathcal{D}(\mathcal{A})\to \mathcal{D}(\mathcal{B})$, such that $\pi_B \circ \mathcal{K}(F) = RF \circ \pi_A$.</p> <p>This is not possible to find in general, and the problem is that $\mathcal{K}(F)$ may not send quasi-isomorphisms into quasi-isomorphisms. The best you can ask for is a functor $RF:\mathcal{D}(\mathcal{A})\to \mathcal{D}(\mathcal{B})$, with a natural transformation $\eta:\pi_B \circ \mathcal{K}(F) \to RF\circ \pi_A$ having a universal property among such functors and natural transformations (this is a particular case of a Kan extension).</p> <p>Such an $RF$ (unique up to isomorphism) is called the (total) right derived functor of $F$. One way to think about it is as "the" functor between the derived categories which approximates $\mathcal{K}(F)$ in the best possible way. You can recover the single derived functors $R^iF$ by taking the cohomology objects of $RF$. In most cases the derived functor is constructed by using resolutions, by injective (or projective, if you're defining left derived functors) objects, or more generally by suitable subcategories of $\mathcal{A}$.</p> <p>In your case, if you also assume that $F$ is right exact, then the satellite functors coincide with the (left) derived ones, and so I guess that it follows that they can be calculated by the formulas you wrote.</p> <p>The idea of the "best approximation" of $\mathcal{K}(F)$ on the derived category seems very natural to me, and a satisfactory answer to the question "why derived functors". If you were asking specifically about satellites, then I don't know.</p> http://mathoverflow.net/questions/29552/intuition-for-the-satellite-of-a-functor/48745#48745 Answer by Buschi Sergio for Intuition for the satellite of a functor Buschi Sergio 2010-12-09T11:01:33Z 2010-12-09T11:01:33Z <p>This is a classical reference: </p> <p>F. Ulmer, Satelliten und derivierte funktoren. I, Math. Z. 91 (1966)</p> <p><a href="http://www.digizeitschriften.de/main/dms/img/?PPN=GDZPPN00239703X" rel="nofollow">http://www.digizeitschriften.de/main/dms/img/?PPN=GDZPPN00239703X</a></p> <p>A more simple exposition can find in Barry Mitchell - Theory of Categories </p> <p><a href="http://books.google.it/books?id=hgJ3pTQSAd0C&amp;printsec=frontcover&amp;dq=mitchell,+theory+of+categories&amp;source=bl&amp;ots=erlQ8pdHiS&amp;sig=JteX6fmj-qAco7tWA2dmDe-e9_Y&amp;hl=it&amp;ei=prYATeSHJsKi4Qb1n4z0Ag&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CCUQ6AEwAA" rel="nofollow">http://books.google.it/books?id=hgJ3pTQSAd0C&amp;printsec=frontcover&amp;dq=mitchell,+theory+of+categories&amp;source=bl&amp;ots=erlQ8pdHiS&amp;sig=JteX6fmj-qAco7tWA2dmDe-e9_Y&amp;hl=it&amp;ei=prYATeSHJsKi4Qb1n4z0Ag&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CCUQ6AEwAA</a></p> <p>Or in classical Cartan Eilenberg "Homological ALgebra"</p>