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I have seen the Landweber exact functor theorem beeing used to retrieve cohomology theories, in particular singular cohomology and K-Theory. However the statement of the theorem itself is always homological.

But how does one deal with the fact, that the tensor product is in general not compatible with direct products. If we consider the example of singular cohomology, i.e. $\mathbb{Q}$ with the additive group law $X+Y$ and note that $\mathbb{Q}$ is not finitely presented over $MU^*(pt)=\mathbb{Z}[x_1,\ldots]$, I don't see why $$ X \mapsto MU^*(X)\otimes_{MU^*} \mathbb{Q} $$ should satisfy additivity.

So is there a cohomological version of the exact functor theorem?

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  • $\begingroup$ In abelian groups finite products and finite coproducts are the same. There is a reason why the cohomological LEFT is always stated only for finite complexes (also, from your question seems that you're unaware that a homology theory and a cohomology theory are essentially the same amount of data, so retrieving one and retrieving the other are essentially the same thing) $\endgroup$
    – Denis Nardin
    Nov 27, 2017 at 17:09
  • $\begingroup$ I am aware that they are coming from spectra, but I am not sure how to translate one into the other. Landweber writes in the original paper that it gives a homology theory on all CW-spectra and Rudyak does the same. So there are versions with non-finite complexes. Also surely one would want to be able to talk about $\mathbb{C} P^\infty$ which is an infinite complex $\endgroup$
    – Rene Recktenwald
    Nov 27, 2017 at 17:13
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    $\begingroup$ I did say cohomological LEFT :). For finite spectra you just precompose with SW duality, and you get that $E^*X=MU^*X\otimes_{MU_*}E_*$. For non-finite spectra this breaks down since cofiltered limits do not commute with the tensor product (as you noticed) $\endgroup$
    – Denis Nardin
    Nov 27, 2017 at 17:26
  • $\begingroup$ Ups my bad :) So we can get $H^*(X;\mathbb{Q})$ but only for finite CW-Complexes? In particular I don't see any of the theory of chern classes here? This is confusing to me, because I would start with the Chern classes to make $H^*(X;\mathbb{Q})$ into a $MU^*$-module in the first place. $\endgroup$
    – Rene Recktenwald
    Nov 27, 2017 at 17:36
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    $\begingroup$ You can in fact get $H^*(X;\mathbb{Q})$ from $MU$ but it is more complicated and I do not know of a nice formula from it (it's some kind of limit over the values on the finite complexes, but it is complicated by the existence of $lim^1$ terms). However you can still see the whole theory of Chern classes, at least for finite complexes, so I do not understand your objection (hint: $H^*(BU;\mathbb{Q})=\lim_{n,k} H^*(Gr_k(\mathbb{C}^n);\mathbb{Q})$) $\endgroup$
    – Denis Nardin
    Nov 27, 2017 at 17:49

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