# Intuition for the satellite of a functor

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.

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

$S_-^1(F)(M) = lim(ker(F(M) \rightarrow F(P)))$

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 nlab.

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.

So my question is

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?

-
I think you want an additive functor between abelian categories. –  Qiaochu Yuan Jun 26 '10 at 0:35
Yes, thanks. The different setting in which I'm learning about derived functors also does not require additivity of the functor or abelian categories. It has been changed. –  Eric A. Bunch Jun 26 '10 at 3:36
Don't you want to map from a projective object $P \to M$ and to an injective object $M \to I$? –  K.J. Moi Dec 9 '10 at 17:38

I don't know if this can be considered as "intuition", anyway

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.

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 $\pi_B \circ \mathcal{K}(F) = RF \circ \pi_A$.

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).

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 $\mathcal{A}$.

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.

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.

-
Thanks for the answer. I'll have to think about this a bit more; I've never worked with Kan extensions before. –  Eric A. Bunch Jun 29 '10 at 15:21
Well, neither have I :) in this case it is just a name, I don't know if there's more to it. –  Mattia Talpo Jun 30 '10 at 9:14
I agree with you that your answer answers the question of `why derived functors' pretty well. Thanks again for your answer, although I was hoping for something specific about the satellite. –  Eric A. Bunch Jul 5 '10 at 5:30

This is a classical reference:

F. Ulmer, Satelliten und derivierte funktoren. I, Math. Z. 91 (1966)

http://www.digizeitschriften.de/main/dms/img/?PPN=GDZPPN00239703X

A more simple exposition can find in Barry Mitchell - Theory of Categories