A question regarding generalized cohomology and spectra : proof of $E^{\ast}(S)\otimes\mathbb{R} = H^{\ast}(S;\pi_{\ast}E\otimes \mathbb{R})$ I asked a question on m.se about generalised cohomology and spectra. Not having received any specific answer I attempted to draw more attention by offering a bounty. But I still could not get any help. If it does not violate this site's policies, I would like to reproduce my question below. I would really appreciate any help. Thanks 
Question
I am reading the paper Quadratic functions in geometry, topology and M theory by M.J.Hopkins and I.M.Singer, and in section 4.8 they say : 

'Recall that for any compact $S$, and for any cohomology theory $E$ there is a canonical isomorphism 
  $E^{\ast}(S)\otimes\mathbb{R} = H^{\ast}(S;\pi_{\ast}E\otimes \mathbb{R})$.When $E$ is $K$-theory, this isomorphism is given by the Chern character.
  The isomorphism arises from a universal cohomology class $
i_{E}\in H^{0}(E;\pi_{\ast}E\otimes\mathbb{R}) = \varprojlim
H^{n}(E_{n};\pi_{\ast+n}E\otimes\mathbb{R})$
  and associates to a map $f:S\to E_{n}$ representing an element of
  $E^{n}(S)$ the cohomology class $f^{\ast}i_{n}$, where $i_{n}$
  is the projection of $i_{E}$ to $H^{n}(E_{n};\pi_{\ast+n}E)$.

I do not understand most of the above paragraph. My specific questions are : 

1.) Where can I find a proof of the isomorphism mentioned ? I am familiar with the basic definitions of spectra and generalized cohomology but can not quite understand the isomorphism. Is this some kind of Brown representability ?
2.) I do not understand the system of abelian groups over which the limit is taken. 

Of course, it is unreasonable to ask for detailed explanations for my doubts. I would really appreciate some hints, and pointers to good texts where I can find a discussion of the above issues and educate myself. Thanks so much !
 A: For any spectra $X$ and $Y$ put 
$$ M(X,Y)_* = \text{Hom}(H_*(X;\mathbb{Q}),H_*(Y;\mathbb{Q})) 
    = \prod_{i\in\mathbb{Z}} H^i(X;H_{*-i}(Y;\mathbb{Q}))
$$
There is an evident natural map
$$ h_{XY}\colon [X,Y]_* \to M(X,Y)_*, $$
which induces a map $h'_{XY}\colon [X,Y]_*\otimes\mathbb{Q}\to M(X,Y)_*$.  The main point is that $h'_{XY}$ is an isomorphism whenever $X$ is a finite spectrum.  To prove this, we first need to know that $h'_{S^p,S^q}$ is an isomorphism.  This is essentially the fact that $\pi_0(S^0)=\mathbb{Z}$, with $\pi_i(S^0)$ being finite for $i\neq 0$.  Now put 
$$ \mathcal{A} = \{Y : h'_{S^0,Y} \text{ is an isomorphism }\}. $$
This is easily seen to be closed under suspension and desuspension and under arbitrary coproducts.  If we have a cofibre sequence $Y_0\to Y_1\to Y_2\to\Sigma Y_0$, then we get long exact sequences of homotopy and homology groups.  Thus, if two of the spectra $Y_i$ lie in $\mathcal{A}$, then we can use the five lemma to show that the third does also.  It follows by cellular induction that all spectra lie in $\mathcal{A}$, so $h'_{S^0,Y}$ is always an isomorphism.  In particular, this means that $\pi_*(Y)\otimes\mathbb{Q}=H_*(Y;\mathbb{Q})$ so we can rewrite $M(X,Y)$ as
$$ M(X,Y) = \prod_iH^i(X;\pi_{*-i}(Y)\otimes\mathbb{Q}) $$
if desired.
Now put
$$ \mathcal{B} = \{X:h'_{XY} \text{ is an isomorphism for all } Y\}. $$
We have seen that $S^0\in\mathcal{B}$.  It is also easy to see that $\mathcal{B}$ is closed under suspension and desuspension and under finite coproducts.  Moreover, if we have a cofibre sequence $X_0\to X_1\to X_2\to\Sigma X_0$ then we again get long exact sequences for $[X_i,Y]_*\otimes\mathbb{Q}$ and $M(X_i,Y)_*$.  Thus, we can again use the five-lemma to see that if two of the spectra $X_i$ are in $\mathcal{B}$, then the third is also.  It follows by cellular induction that all finite spectra lie in $\mathcal{B}$, so $h'_{XY}$ is an isomorphism whenever $X$ is finite.  The same kind of argument also shows that when $X$ is finite we have $H^*(X;\pi_*(Y)\otimes\mathbb{Q})=H^*(X;\pi_*(Y))\otimes\mathbb{Q}$.
To reconcile this with your notation, take $Y=E$.  The case $X=S^0$ gives $H_*(E;\mathbb{Q})=\pi_*(E)\otimes\mathbb{Q}$, so 
$$ E^*(X)\otimes\mathbb{Q} = H^*(X;\pi_*(E)\otimes\mathbb{Q}) 
                           = H^*(X;\pi_*(E))\otimes\mathbb{Q}.
$$
You can now tensor everything with $\mathbb{R}$ if you like.
