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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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up vote 9 down vote accepted

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 $f$ and we can use the triangulated structure to find an exact triangle $S \stackrel{f}{\to} S \to M \to S[1]$ where $S[1]$ is the suspension (equivalently, the shift) of $S$. 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 $E \to F$ 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

An even more compelling reason is that stable natural transformations between generalized cohomology theories on spaces form the wrong category, i.e. are not the same as the morphisms of the stable category. Let $E,F$ be generalized cohomology theories. Let $Z_n$ be the terms of the $\Omega$-spectrum of a CW model of $E$. Then, more or less by definition, stable natural transformations from $E$ to $F$ are the inverse limit of $F^nZ_n=F^0\Sigma^{\infty-n}Z_n$. This is the correct Hom set in the stable homotopy category when $\lim^1=0$, but this may not be the case (for example when $E=S$, $F$ is the wedge of $\Sigma^n HZ/2$ over $n\in \mathbb{Z}$).

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Welcome to MO, Professor Kriz! – Todd Trimble Dec 1 '14 at 4:48

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