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This might be well-known or trivial, but I could not figure out how to fill in the details: For an $S$-algebra $K$ denote its associated multiplicative cohomology theory by $h^*_K$. Suppose that I have an isomorphism of multiplicative cohomology theories: $$ h^*_{K}(X) \cong K^*(X) $$ for finite CW-complexes $X$, where $K^*(X)$ denotes the complex periodic $K$-theory of $X$. In particular, this isomorphism has to respect the products on both sides. The paper "$\Gamma$-cohomology of rings of numerical polynomials and $E_{\infty}$-structures on $K$-theory" by Baker and Richter states that the $E_{\infty}$-structure on $K$-theory is unique.

Can I use this result to show that there has to exist an equivalence of $S$-algebras between $K$ and the $S$-algebra $KU$, ie. can I "lift" the multiplicative isomorphism all the way to an isomorphism of $S$-algebras? If so, how do I start? Can I lift the multiplicative isomorphism to a map of homotopy ring spectra, which is then refined to an $E_{\infty}$-map by Baker and Richter?

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  • $\begingroup$ Can I "snaith" it? Maybe I can... $\endgroup$ Commented Oct 2, 2021 at 17:47
  • $\begingroup$ I think the answer is yes. At least, you get a map of homotopy ring spectra from your morphism of multiplicative cohomology theories by Brown representability, see mathoverflow.net/questions/387107/… . Then if I'm understanding the Baker-Richter result correctly must be an E_infty ring iso? $\endgroup$
    – kiran
    Commented Oct 7, 2021 at 0:22

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