Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sometimes (specifically in this case I'm looking at Charles Rezk's "A Model for the Homotopy Theory of Homotopy Theory") it seems that people refer to the homotopy category of a model category as a "homotopy theory." Is this just a different name for the homotopy category of a model category, or does this somehow denote some associated object or functor in the same way that we can associate a cohomology "theory" to a spectrum? For instance, is the "homotopy theory" of the model category of spectra (and Quillen equivalent categories) somehow stable homotopy of topological spaces, and as such is there some (perhaps idealized, not really existent) model category for unstable homotopy? What would other homotopy theories be?

Thanks!

share|improve this question

1 Answer 1

up vote 5 down vote accepted

My impression is that a "homotopy theory" is more than the homotopy category: it is an $(\infty, 1)$-category. The paper you describe introduces one model for $(\infty, 1)$-categories (complete Segal spaces); the most popular one today seems to be the "quasi-categories" used by Lurie and Joyal. A slogan is that model categories are presentations for "homotopy theories" or $(\infty, 1)$-categories. To get an $(\infty, 1)$-category from a model category, you should localize at the weak equivalences, but in the higher categorical sense. Alternatively, if you have a simplicial model category (at least, a reasonably nice one, e.g. a combinatorial one), then you can describe the associated homotopy theory as a simplicial category (recall that simplicially enriched categories are another model for $(\infty, 1)$-categories) by just taking the cofibrant-fibrant objects with the given simplicial structure.

There is a lot more information in these higher categories than there is in their homotopy categories. For instance, if you take the homotopy category of a "stable" $(\infty, 1)$-category, then you get a triangulated category; a classic example is the derived category, or the homotopy category of spectra. But just working with triangles is in many ways unnatural (and caused difficulties when historically triangulated categories were used), because they aren't unique. However, if you think of them as cofibers, then they are unique (in the higher categorical sense). The analog is that the homotopy fibered product of spaces does not correspond to a strict universal property after applying $\pi_0$. The universal property is actually a higher categorical one, which doesn't work in the homotopy category alone.

Having this additional structure is especially important if you want to do algebra. For instance, if you work in the derived category (a monoidal category under the derived tensor product) of modules over some commutative ring $R$, then an algebra object in here is something like a homotopy associative, commutative H-space in topology: it's a complex of projectives $P$ together with an algebra map $P \otimes_R P \to P$ which is homotopy associative and commuative and unital. But if you have a map of "modules" (where modules are only required to be modules up to homotopy) $f: M \to N$, one can't make the cofiber of $f$ into a $P$-module. One has the same problem when working in the homotopy category of spectra with ring spectra. I believe (someone more knowledgeable will hopefully chime in), that this was historically one of the main motivations for symmetric monoidal categories of spectra such as $S$-modules and symmetric/orthogonal spectra (and before that, for various flavors of structured ring spectra).

(Note that there are model categories of spectra. See for instance this paper.)

share|improve this answer
    
Thanks. That initial comment, that a "homotopy theory" is in fact all of the higher homotopy, or sort of $\infty$ information, is very helpful. If I read it in that fashion, the name makes a lot more sense. –  Jon Beardsley Nov 14 '11 at 20:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.