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I have to say that I'm very new to this subject and I'm trying to get the general idea of what we're talking about.

Starting from the category of spectra, whose objects are sequences of spaces $\{X_i\}_{i\in\mathbb N}$ with structure maps $\Sigma X_n\to X_{n+1}$, we want to define the stable homotopy category as its $\mathsf{Ho}$, for a suitable model structure. But before that, what I'd like to understand is what goes wrong in defining naively (degreewise) homotopy equivalences of maps of spectra and then taking homotopy equivalence classes.

On a book I'm reading it says that a main problem is that we could have two non-homotopically equivalent ($\Omega$-)spectra $X,Y$ representing the same cohomology theory, i.e. such that the functors of maps of spectra $[-,X_i]$ and $[-,Y_i]$ (with suitable group structure) are the same cohomology theory. Is that true? Is there an easy example of that happening, maybe involving the Eilenberg MacLane spectrum?

The point I'm starting to get and which is suggested in the book is that we do not care that much about the behaviour of the sequence of spaces if not at infinity, so maybe by "truncating" (replace e.g. the first term with a point) we could find such an example? I really can't see how is possible that the zero degree cohomology doesn't change if I replace like that the space in degree zero

Thanks in advance for any suggestion.

Nikio
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1 Answers1

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The spectrum $\Sigma^\infty S^0$ and its truncation $X$, where we replace the $0$th space with a point, both represent stable cohomotopy. If these spectra were literally homotopy equivalent, the (degree 0) self maps of these two spectra would be the same. The self maps of $\Sigma^\infty S^0$ are the constant map and the identity map, coming from suspending the constant map and the identity of $S^0$.

The truncation $X$ has $\mathbb{Z}$ many self maps, classified by the degree of the second space. The difference in these two computations is that the map $\Sigma S^0 \rightarrow S^1$ is an isomorphism while the map $\Sigma * \rightarrow S^1$ is constant.

I am assuming you are taking the most naive definition of homotopy classes possible, and in particular you aren't allowed to replace by a cofinal spectrum (which is how Adams corrects these problems). If you want a more specific answer, you will have to be specific about the model of spectra you are using. You mention $\Omega$-spectra, but I was under the impression that the result is true for $\Omega$-spectra, i.e. two which represent the same cohomology are directly equivalent, perhaps a CW assumption is necessary.

Connor Malin
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  • Thank you so much for your answer. The fact that the two spectra you took represent the same algebraic object is because we are assuming to work with connected spaces (CW-complexes) X, and hence $[X,\ast]$ is trivial just as $[X,S^0]$ (here square brackets are homotopy equivalence classes of pointed topological space maps)? Or is there a more "profound" reason for which we can truncate and obtain the same cohomology theory? – Nikio Feb 10 '22 at 09:36
  • @Nikio Do you mind if I use homology? Every spectrum $E$ determines a homology theory by declaring the homology of a pointed space $X$ is given by the homotopy groups of the spectrum $E \wedge X$. The definition of the homotopy groups of a spectrum are in terms of a directed limit , and if two directed diagrams agree after some finite stage, then their directed limits will agree. Hence, if I alter some finite part of a spectrum, it will not change the homology theory associated to it. – Connor Malin Feb 10 '22 at 17:26