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I know that every generalised (Eilenberg-Steenrod) cohomology theory defines a spectrum (in the sense of Lewis-May), and vice-versa. I also know that maps between spectra are richer than maps between cohomology theories (due to the existence of phantom maps).

How unique is the/a spectrum representing a cohomology theory?

In particular, are Eilenberg-MacLane spectra unique up to weak equivalences? (An Eilenberg-MacLane spectrum for the abelian group $A$ is a spectrum representing ordinary cohomology with coefficients in $A$.)

Ben Steffan
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user09127
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    it's unique and this is just the yoneda lemma: if you have a natural isomorphism between two representable functors it is induced by an isomorphism between the representing objects. –  Jan 02 '19 at 17:42
  • Could you elaborate? I do not understand how to make this argument work on the level of spectra. – user09127 Jan 02 '19 at 18:37
  • What is unclear? That cohomology theories also define functors on spectra? That the induced map $E^(X) \to F^(X)$ is also an isomorphism for some cohomology theories $E, F$? –  Jan 02 '19 at 18:42
  • I apologize for my claims that this is easy; questions of this nature rarely are. The last point in my comment above is resolved by the Milnor lim^1 sequence for generalized cohomology of spectra, but the statement that you can upgrade some equivalence $E \to F$ of cohomology theories to a map of the corresponding $\Omega$-spectra seems subtle. You can get a map of pointed spaces $\Omega^\infty E \to \Omega^\infty F$, but you need to show that this is an infinite loop map. –  Jan 02 '19 at 23:22
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    More Yoneda shows that you can get something like "map of N-indexed spaces $f_k: \Omega^{\infty - k}E \to \Omega^{\infty - k} F$ so that $\Omega f_k$ is homotopic to $f_{k-1}$", but that is not the same as an infinite loop map. Something more needs to be shown. –  Jan 02 '19 at 23:24
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    That's precisely how my question arose. – user09127 Jan 03 '19 at 06:44
  • Maybe it's worth knowing that the uniqueness of (some) Eilenberg-MacLane spectra (as $\Omega$-spectra) is proved in an answer to another one of my questions: https://mathoverflow.net/questions/319919/why-is-pi-fh-mathbbf-p-h-mathbbf-p-the-mod-p-steenrod-algebra – user09127 Jan 03 '19 at 06:46
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    I sincerely apologize for being so blithe about it (though, in the future, it is probably good to add things like this to the post itself if you already know them - for the sake of fools like I!). I hope you get a good answer. I still think it should look like the sketch I provided, but the higher homotopies are not clear to me. –  Jan 03 '19 at 07:05
  • Well, I do not think that you are too far off (at least for EM spectra). I found a proof of their uniqueness in the Alaska notes. It boils down to proving that EM spectra define cohomology theories of spectra. Then, of course, you can use the Yoneda lemma, as you suggested! I haven't found anything on the general case though. – user09127 Jan 03 '19 at 07:18
  • Have you checked Switzer's book, Algebraic Topology? Section 9 is devoted to the representation theorems, although since I'm not too familiar with the text myself I couldn't say if it includes a discussion of exactly what you are looking for or not. – Tyrone Jan 03 '19 at 10:43
  • I can also (possibly) recommend you Rudyak's book On Thom Spectra, Orientability, and Cobordism. Section 3 is devoted to phantom maps and in particular 3.3 discusses the representability theorems. Maybe Theorem 3.20 (iii) and Theorem 3.21 are the statements you are looking for? – Tyrone Jan 03 '19 at 10:48
  • I had a discussion about this in the homotopy theory chatroom here which may be of use to you. –  Jan 03 '19 at 13:48
  • Thanks a lot, it seems to answer my question! – user09127 Jan 04 '19 at 07:45

1 Answers1

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Spectra, qua representing objects of cohomology theories, are unique up to equivalence (i.e. up to isomorphism in the stable homotopy category $\mathrm{SHC}$; what this means precisely depends on your choice of model).

In fact, there is very little to this argument (and I have to admit I do not understand the confusion in the comments, or the need to cite any literature for this): Let a cohomology theory $E^*\colon \mathrm{hTop}^\mathrm{op} \to \mathrm{grAb}$ be given where $\mathrm{hTop}$ is the homotopy category of spaces (whatever this precisely means to you) and $\mathrm{grAb}$ is the category of ($\mathbb{Z}$-)graded abelian groups. Let $\mathrm{Sp}$ denote your favourite choice for a (model or $\infty$-)category of spectra. The Brown representability theorem spits out a spectrum $E$ such that $E^* \cong [\Sigma^\infty_+ {{-}}, E]$ as functors $\mathrm{hTop}^{\mathrm{op}} \to \mathrm{grAb}$ where $\Sigma^\infty_+\colon \mathrm{Top} \to \mathrm{Sp}$ is the pointed suspension spectrum functor. This extends to a functor $\mathrm{SHC}^\mathrm{op} \to \mathrm{grAb}$ by omitting $\Sigma^\infty_+$, i.e. by sending a spectrum $X$ to $[X, E]$. But this is just a Hom-functor, so the Yoneda lemma yields that $E$ is unique up to isomorphism in $\mathrm{SHC}$, q.e.d.

"But what about phantom maps?" You say that maps between spectra are richer than maps between cohomology theories. This is not true. In fact the proof above shows that there is a bijection between maps $E^* \to F^*$ and maps $E \to F$ on representing spectra up to homotopy since the Yoneda embedding is fully faithful. What is true is that maps between spectra are richer than maps between homology theories. Recall that if $E$ is spectrum, then the $E$-homology of another spectrum $X$ is given by $E_n(X) = [\Sigma^n \mathbb{S}, X \wedge E]$ where $\mathbb{S}$ is the sphere spectrum. This isn't simply a Hom-functor, so the Yoneda argument does not apply, and in fact the functor assigning $E_*$ to $E$ is not faithful. It is however full, and moreover it is still true that the spectrum $E$ giving rise to $E_*$ is unique up to equivalence (i.e. the functor is conservative), for a reference for which and more on phantom maps I will commend you to Lurie's Chromatic Homotopy Theory Notes, Lecture 17.

Ben Steffan
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