Is Lie group cohomology determined by restriction to finite subgroups? Consider the restriction of the ordinary group cohomology $H^*(BG,\mathbb{Z})$, where $G$ is a compact Lie group and $BG$ is its classifying space, to finite subgroups $F < G$. If we consider the product of all such restrictions
$$H^*(BG,\mathbb{Z}) \to \prod_F H^*(BF,\mathbb{Z}),$$
is this map injective?
EDIT: I asked this question at mathoverflow, and Tim Campion provided an argument for the torsion elements, which together with Qiaochu's answer below one has a complete solution to the question, so yes the map is injective.
 A: If the cohomology of $BG$ is torsion-free (which happens e.g. when $G = U(n)$ but not when $G = O(n)$) then it injects into the cohomology of $BT$ where $T \to G$ is any maximal torus (and in fact rationally $H^{\bullet}(BG, \mathbb{Q})$ is precisely the Weyl group invariants $H^{\bullet}(BT, \mathbb{Q})^W$). So in this case it suffices to know that this is true for a torus. The question for tori further reduces to the case of a single circle $S^1$, and now we want to know whether the map
$$H^{\bullet}(BS^1, \mathbb{Z}) \to \prod_{n \ge 1} H^{\bullet}(B \mathbb{Z}/n, \mathbb{Z})$$
is injective. Now, I believe but don't know how to prove that in positive degree the restriction
$$\mathbb{Z} \cong H^{2k}(BS^1, \mathbb{Z}) \to H^{2k}(B\mathbb{Z}/n, \mathbb{Z}) \cong \mathbb{Z}/n$$
is just reduction $\bmod n$, which if true would mean the answer is yes in the torsion-free case. In the general case we at least get that the kernel of the map consists at worst of torsion classes. Hopefully someone who actually knows group cohomology can say more from here.
