Are generalized cohomology theories a homotopy category of some category of invariants?

I was taught to think of generalized cohomology theories as the homotopy category of (symmetric) spectra. But is there also a category of 'invariants', that is, some category of contravariant functors from a suitable category of topological spaces to a suitable category of algebraic objects, which has a model category structure such that the homotopy category gives the category of generalized cohomology theories, without referring to spectra?

If such a thing exists, why do people prefer to use spectra?

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1 Answer

Here is a short argument why we don't expect generalized cohomology theories to behave so well.

In the stable homotopy category, there is a generalized homology/cohomology theory represented by the sphere spectrum S, so that S_*(X) are the stable homotopy groups of X. It has a multiplication-by-2 self-map and we can use the triangulated structure to find an exact triangle S -> S -> M -> S[1]. We think of M as "the sphere mod 2", and it is called the mod-2 Moore spectrum.

In most derived categories coming from algebra, such an exact triangle would have the property that the multiplication-by-2 map on M was zero. However, we know that this is not the case here; the multiplication-by-2 map is not zero, but the multiplication-by-4 map is.

One of the problems with using a "functor" language to get at generalized cohomology theories is that given a natural transformation E -> F of generalized cohomology theories, it is not clear what the associated cofiber should be in order to produce a triangulated structure. Spectra have a natural triangulated structure and they rectify this problem. (A map between spectra also includes "phantom" data that isn't easily detected by the natural transformation between associated generalized cohomology theories.)

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When you said "the identity self-map actually has order 4", did you mean that "the multiplication-by-2 self-map has order 4"? –  Chris Schommer-Pries Oct 21 '09 at 13:12
I mean that 4 * id = 0, but 2 * id is not zero. Sorry. –  Tyler Lawson Oct 21 '09 at 13:14
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