Cohomology ring of BG Let $G$ be a compact Lie group, let $T$ be a maximal torus, and let $W$ be the Weyl group.  My main question is as follows: 


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*How does one prove that $H^\ast(BG,\mathbb{Q})$ is isomorphic to the $W$-invariant part of $H^\ast(BT,\mathbb{Q}) \cong \mathbb{Q}[[x_1, \ldots, x_n]]$?  This is apparently basic knowledge in algebraic topology, because I keep reading "recall that..." followed by some version of this statement and no references.  But I can't find a proof in any of my textbooks.


I would ideally like a reference which also addresses the following secondary question:


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*When is the natural map $H^\ast(BG,\mathbb{Z}) \to H^\ast(BT,\mathbb{Z})^W$ an isomorphism, and what can one say about the integral cohomology ring of $BG$ when it is not?  Note the fact that the map above is an isomorphism for $G = U(n)$ is equivalent to the statement that the Chern classes are integral.


Thanks!
 A: Notice that there is a sequence of homomorphisms $T \to N \to G$, where $N$ is the maximal torus normaliser (so $W = N/T$).  $W$ acts on $BT$ (because it acts on $T$ by conjugation through group homomorphisms), and there is an equivalence from the classifying space of $N$ to the Borel construction for this action:
$$BN \simeq EW \times_W BT.$$
Consequently, we can compute the cohomology of $BN$ from the Leray-Serre spectral sequence
$$H^\ast(W; H^\ast(BT)) \implies H^\ast(BN).$$
Taking rational cohomology, this spectral sequence is concentrated in group-cohomological degree $0$, since $W$ is a finite group.  Therefore the spectral sequence collapses at $E_2$, which is $H^0(W, H^\ast BT) = H^\ast(BT)^W$.
It therefore suffices to show that the map $BN \to BG$ is an isomorphism in rational cohomology.  If we write $BN$ as $EG / N$, this map is a fibre bundle with fibre $G / N$, so it's enough to show that $G/N$ has the rational homology of a point.  
For instance, if $G = SU(2)$, $N = \mathbb{Z} / 2 \ltimes T$, and $T = S^1$.  Then $G/T = \mathbb{C} P^1$, and the action of $\mathbb{Z} / 2$ is antipodal, giving $G / N = \mathbb{R} P^2$, which is indeed rationally a point.  I don't remember the argument in general, but I think this is always true.
Hopefully this indicates how the corresponding integral statement can fail - there can be torsion contributions from the higher group cohomology of $W$, which needs to be exactly cancelled (via a differential in the second spectral sequence above) with a torsion cohomology class from $G/N$.
A: Q1: Let me first note, that the statement
$$H^\ast(BG,\mathbb{Q}) \cong H^\ast(BT,\mathbb{Q})^W\tag{$\ast$}$$
made in the question requires in addition that $G$ is connected. However, the general case can be reduced to the connected case since
$$H^\ast(BG,\mathbb{Q}) \cong H^\ast(BG_0,\mathbb{Q})^{G/G_0}$$
where $G_0$ is the identity component of $G$.
A text book reference for $(\ast)$ can be found in Hsiang, Cohomology theory of topological transformation groups, Chapter III, §1, Lemma 1.1. The results of the book that are relevant for your question can also be found in the following paper: Richard Gonzales, Localization in equivariant cohomology and GKM theory (cf. Remark 9, Lemma 5). Other approaches and more information can be found in this answer.
Q2: First note that the kernel of the restriction map
$$\rho^\ast: H^\ast(BG;\mathbb{Z}) \to H^\ast(BT;\mathbb{Z})$$
is the tosion subgroup $Tors$ of $H^\ast(BG;\mathbb{Z})$. So, $\rho^\ast$ is injective iff $H^\ast(BG;\mathbb{Z})$ is torsion-free (there is a lot of literature about torsion
of $H^\ast(BG;\mathbb{Z})$ so I won't discuss it here).
A short paper of Feshbach

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*The image of $H^\ast(BG;\mathbb{Z})$ in $H^\ast(BT;\mathbb{Z})$ for $G$ a compact Lie group with maximal torus $T$. Topology 20(1981) 93-95 doi:10.1016/0040-9383(81)90015-X,

characterizes when the induced map
$$\bar{\rho}^\ast: H^\ast(BG;\mathbb{Z})/Tors \to H^\ast(BT;\mathbb{Z})$$
is an isomorphism:

$\bar{\rho}^\ast$ is an isomorphism iff $H^\ast(BG;\mathbb{Z})/Tors \otimes \mathbb{F}_p$ is an integral domain for each prime $p$.

There is also a counter-example, $\mathrm{Spin}(12)$, that shows that $\bar{\rho}^\ast$ is not always an isomorphism.
A: I don't remember where I heard the following proof/sketch:
Using the fibering $G/T\to BT\to BG$ and the fact that the Euler class of $G/T$ is nonzero, we have that $H^\ast(BG)$ embeds into $H^\ast(BT)$ (it composes with the transfer map to be multiplication by the Euler class); and the desired isomorphism comes from the fact that $W$ acts on $H^*(G/T)$ as the regular representation. 
This is actually a special case of equivariant cohomology, where we instead use the Borel construction and the fibering $G/T\to M_T\to M_G$.
