It is a standard consequence of the Brown Representability Theorem for $\operatorname{Ho}(\operatorname{Top}_*)$ that the category of generalized cohomology theories for spaces (pointed CW complexes, more specifically) is equivalent to the stable homotopy category $\operatorname{SHC}$ (defined as the homotopy category of the stable model structure on sequential spectra) via representability. On the other hand, we can define a cohomology theory for spectra (as in Barnes and Roitzheim's Foundations of Stable Homotopy Theory) to be a contravariant functor $E^*:\operatorname{SHC}^{op}\to\operatorname{Ab}_*$ such that

  1. Each exact triangle in $\operatorname{SHC}$ gives rise to a long exact sequence in $\operatorname{Ab}$
  2. $E^n$ preserves products for each $n$ (that is, it sends wedge sums to products of abelian groups)
  3. $E^*$ preserves suspension up to a specified natural isomorphism $E^{n+1}(\Sigma X)\cong E^n(X)$.

Then it is easy to check that $[,E]$ is always a cohomology theory for spectra. But is every cohomology theory for spectra thus represented? Given such a cohomology theory $E^*$, we obtain an associated cohomology theory for spaces by restricting to suspension spectra. Thus, by the result for spaces mentioned above, the question becomes whether a cohomology theory for spectra is determined by its restriction to spaces.

This is certainly true if we require $E^*$ to preserve sequential homotopy colimits, because any spectrum is weakly equivalent to a CW spectrum with basepoint at its unique $0$-cell, which is a sequential homotopy colimit of homotopy cofibers of coproducts of shifted sphere spectra. But without this requirement, does the result still hold? If so, why? If not, is there a standard counterexample?

  • $\begingroup$ Regarding the non-equivalence of cohomology theories and the stable homotopy category see also this mathoverflow question: mathoverflow.net/questions/117684/… -- For representability results for cohomology theories see also the very nice book 'Axiomatic Stable Homotopy Theory' by Hovey, Palmieri and Strickland. $\endgroup$ – Lennart Meier Jul 9 at 11:24

The category of cohomology theories on pointed CW-complexes is not equivalent to the stable homotopy category. The latter projects onto the former, and this projection induces a bijection on isomorphism classes, but there is a kernel, containing superphantom maps, see [Christensen, J.Daniel. “Ideals in Triangulated Categories: Phantoms, Ghosts and Skeleta.” Advances in Mathematics 136, no. 2 (June 1998): 284–339. https://doi.org/10.1006/aima.1998.1735.].

The category of cohomology theories on spectra is equivalent to the stable homotopy category. This is also called Brown representability. You essentially have to show that all cohomology theories are representable, and the rest follows from Yoneda.

By the aforementioned theorems, any cohomology theory for spectra is determined by the induced cohomology theory for spaces up to a non-canonical isomorphism.

| cite | improve this answer | |
  • $\begingroup$ I see. But why is it that cohomology theories for spectra are representable? $\endgroup$ – Doron Grossman-Naples Jul 8 at 23:00
  • 4
    $\begingroup$ @DoronGrossman-Naples as I said, that's Brown's representability theorem. You can check it in numerous references, like [Margolis, H. R. Spectra and the Steenrod Algebra. Vol. 29. North-Holland Mathematical Library. Amsterdam: North-Holland Publishing Co., 1983.] $\endgroup$ – Fernando Muro Jul 8 at 23:06
  • 1
    $\begingroup$ @DoronGrossman-Naples : it's possible that you get sequential colimits back from asking $E^n$ to respect all products, and not just finite ones $\endgroup$ – Maxime Ramzi Jul 9 at 8:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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