I'm struggling with understanding the proof that the suspension isomorphism in the homotopy category of topological spectra is equivalent to smashing with the sphere $\mathbb{S}^1$.

The proof I'm reading, which is Switzer's book proof, seems overly complicated to me.

Start with a spectrum $E$ and first construct the mapping telescope ${F}_n=E_n\wedge SE_{n-1}\wedge [n-1, n]^+\wedge...\wedge S^nE_0\wedge[0,1]^+$. Now it is clear that the obvisou inclusion $E_n$ to $F_n$ is not a function of spectra (it is not compatible with bonding maps), but using Whitehead theorem it is easy to see that these spectra are equivalent.

Next, and this is where I'm confused, he constructs a homotopy equivalence of $F\wedge S^1 \to \Sigma E$ and he uses for this a "twist" on the smashing. I really dont see why the obvious retraction map $F\wedge S^1 \to \Sigma E$ mapping everyone to the "lid" $E_{n+1}$ doesnt do the trick as it will be this time a function of spectra.

On the other hand I do realize that this "twist" must be needed because of sign reasons in cohomology. So I'm guessing the reason is that the "obvious retraction" is actually not a function of spectra but when I'm writing it, it seems to me that it obvisously is (although it is tedious to write down).

What am I missing here?

I do know that there are more modern proofs of that fact using different definitions of the homotopy category of spectra and I am willing to look at those proofs also obviously but I'd really like to understand this as I feel I am missing something essential.

Edit: Judging from what I read the subtelty might lie in the intertwining of the $\mathbb{S}^1$ factors appearing when smashing $\mathbb{S}^1\wedge S^kE_n=\mathbb{S}^1\wedge...\wedge \mathbb{S}^1\wedge E_n$... but I'm not completely sure how to take that into account, nor why the twist by the antipodal map solves this problem.



Your Answer

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

Browse other questions tagged or ask your own question.