TITLE: On the comparison map $MU^\bullet(X)\otimes_{MU^\bullet(pt)}E^\bullet(pt)\to E^\bullet(X)$ for complex oriented multiplicative cohomology theories
QUESTION [7 upvotes]: Whatever complex oriented multiplicative cohomology theories are, they come with two basic properties (among many others):
i) a complex oriented multiplicative cohomology theory is a contravariant functor form (nice) topological spaces to (graded) rings;
ii) there is an initial object $MU^\bullet$.
Properties i) and ii) alone imply that for a (nice) topological space $X$ we have two distinguished morphisms of rings,
$$
E^\bullet(pt)\to E^\bullet(X)
$$
induced by the terminal morphism $X\to pt$, and
$$
MU^\bullet(X)\to E^\bullet(X)
$$
induced by the fact $MU^\bullet$ is initial. By naturality we have the commutative diagram of rings
$\require{AMScd}$
\begin{CD}
MU^\bullet(pt) @>>> MU^\bullet(X)\\
@V V V @VV V\\
E^\bullet(pt) @>>> E^\bullet(X)
\end{CD}
Then, by the universal property of the tensor product of rings, we have a distinguished morphism of rings
$$
MU^\bullet(X)\otimes_{MU^\bullet(pt)}E^\bullet(pt)\to E^\bullet(X).
$$
What I'm unable to understand is whether this is always an isomorphism or, if it is not, which are examples where it is not an isomorphism.
Clearly, if $E^\bullet$ is Landweber exact, the above comparison map is easily seen to be an isomorphism. Yet, as far I understand it, Landweber exactness is more concerned with determining whether $MU^\bullet(X)\otimes_{MU^\bullet(pt)}R$ defines a muliplicative cohomology theory given a ring morphism $MU^\bullet(pt)\to R$ than into investigating what happens when one a priori knows $R$ is $E^\bullet(pt)$ and the morphism $MU^\bullet(pt)\to E^\bullet(pt)$ the universal one. Moreover, Landweber exact functor theorem gives a sufficient condition for $MU^\bullet(pt)\to R$ to induce a multiplicative complex oriented cohomology theory and not a necessary one.
In other words, my question is: the comparison morphism
$$
MU^\bullet(X)\otimes_{MU^\bullet(pt)}E^\bullet(pt)\to E^\bullet(X).
$$
is always defined, and there are "nice" cohomology theories, for which this is an isomorphism. Nice cohomology theories include Landweber exact theories. But what are nice theories? Are they all? If not, what is an example of a not nice theory?

REPLY [6 votes]: Firstly, even in the Landweber exact case your comparison map need not be an isomorphism unless $X$ is a finite complex.  It is more often the case that $E^*(X)$ agrees with the completed tensor product $MU^*(X)\widehat{\otimes}_{MU^*}E^*$.  However, I am still not sure whether that works for all $X$; exactness properties of completed tensor products are quite delicate.
On the other hand, the Landweber exactness criterion is actually equivalent to $MU^*(-)\otimes_{MU^*}R^*$ being a cohomology theory on finite complexes.  One direction is Landweber's theorem, the other follows easily from the existence of generalized Moore spectra $S/(v_0^{i_0},\dotsc,v_n^{i_n})$ (which is a highly nontrivial theorem, of course).