Is Lie group cohomology determined by restriction to finite subgroups? Consider the restriction of the group cohomology $H^*(BG,\mathbb{Z})$, where $G$ is a compact Lie group and $BG$ is its classifying space, to finite subgroups $F \le G$. If we consider the product of all such restrictions
$$H^*(BG,\mathbb{Z}) \to \prod_F H^*(BF,\mathbb{Z}),$$
is this map injective?
Note, according to McClure - Restriction maps in equivariant K-theory a similar result holds in equivariant K-theory. Perhaps there is a way to derive the above from McClure's theorem?
I asked this question on stackexchange and didn't manage to get a (complete) answer nor solve it myself. However, Qiaochu Yuan offered a proof for the non-torsion elements.
 A: $\DeclareMathOperator\Image{Image}$Here's the most general result I think I can muster. I've split it out into a second answer in order to keep the answer to the original question more self-contained.

Theorem 1: Let $G$ be a compact Lie group, let $X$ be a $G$-space, and let $E$ be a spectrum. Then the following hold, where $F$ ranges over finite subgroups of $G$:

*

*The image of $\bigoplus_F E_\ast(X_{hF}) \to E_\ast(X_{hG})$ contains all of the torsion;


*The kernel of $E^\ast(X_{hG}) \to \prod_F E^\ast(X_{hF})$ is contained in the subgroup of divisible elements.

This follows from the following two more precise theorems:

Theorem 2: Let $G$ be a compact Lie group, and let $X$ be a $G$-space. Let $N \subseteq G$ be the normalizer of a maximal torus $T \subseteq G$, and let $W = N / T$ be the Weyl group. Then $\Sigma^\infty_+ X_{hG}$ splits off of $\Sigma^\infty_+ X_{hN}$.
Proof: The splitting is given by the Becker-Gottlieb transfer: the fiber of $X_{hN} \to X_{hG}$ is $G/N$, the same as the fiber of $BN \to BG$, which has Euler characteristic 1.

Theorem 3:
Let $N$ be an extension of a finite group $W$ by a torus $T$, and let $E$ be a spectrum and $m \in \mathbb N_{\geq 2}$. Then the following hold, where $F$ ranges over finite subgroups of $N$:

*

*$\varinjlim_F (E/m)_\ast(X_{hF}) \to (E/m)_\ast(X_{hN})$ is an isomorphism;


*$(E/m)^\ast(X_{hN}) \to \varprojlim_F (E/m)^\ast(X_{hF})$ is an isomorphism.

Proof of Theorem 1 from Theorems 2 and 3: By Theorem 2, it will suffice to consider the case where $G = N$ is an extension of a finite group by a torus. Theorem 3 establishes Theorem 1 for $E/m$. Then (1) follows by considering the natural short exact sequence $0 \to E_{\ast}(-)/m \to (E/m)_\ast(-) \to E_{\ast-1}(-)^{\text{$m$-tor}} \to 0$, and the argument for (2) uses a similar exact sequence.

The proof of Theorem 3 will follow from a series of lemmas. For the remainder, we let $U(1)^n \to N \to W$ be an extension of a finite group by a torus, and we let $(\mathbb Q/\mathbb Z)^n \to N_\ast \to W$ and $(C_q)^n \to N_q \to W$ be the subextensions which exist by the analysis in my other answer. We fix a spectrum $E$, $m \in \mathbb N_{\geq 2}$, and an $N$-space $X$.

Lemma 4: The fiber of $X_{hN_\ast} \to X_{hN}$ is $B\mathbb Q^n$, and in particular this map is an $(E/m)_\ast$ and $(E/m)^\ast$ equivalence.
Proof: This comes via a diagram chase from the fiber sequence $B\mathbb Q^n \to B(\mathbb Q/\mathbb Z)^n \to BU(1)^n$.

Lemma 5: We have $(E/m)_\ast(X_{hN_\ast}) \cong \varinjlim_q (E/m)_\ast(X_{hN_q})$ canonically, and a canonical short exact sequence $0 \to \varprojlim^1 (E/m)^{\ast+1}(X_{hN_q}) \to (E/m)^\ast(X_{hN_\ast}) \to \varprojlim (E/m)^\ast(X_{hN_q}) \to 0$.
Proof: By the analysis in my other answer, we have $N_\ast = \varinjlim N_q$. Therefore $BN_\ast = \varinjlim BN_q$, and it follows that $X_{hN_\ast} = \varinjlim X_{hN_q}$. The lemma follows by the usual formulas for homology and cohomology of a filtered colimit.

Lemma 6: If $E$ is $m$-torsion, then $\varprojlim^1 E^\ast(X_{hN_q}) = 0$.

Proof of Theorem 3: This follows from Lemmas 4, 5, and 6, once we note that $E/m$ is $m^2$-torsion.

It remains to prove Lemma 6.

Lemma 7: Let $q,r \in \mathbb Z$, and consider the inclusion $C_{qr}^n \xrightarrow i (\mathbb Q/\mathbb Z)^n$. Consider also the inclusion $C_q^n \xrightarrow j C_{qr}^n$ with quotient $C_r^n$. Let $A$ be an $r$-torsion and $q$-torsion abelian group. Then $H^\ast(ij;A)$ is injective and $\Image(H^\ast(ij;A)) = \Image(H^\ast(j;A))$.
Proof: Direct calculation. More precisely, $H^\ast(BC_q;A)$ and $H^\ast(BC_{qr};A)$ both have $A$ in all degrees; the inclusion of $H^\ast(B(\mathbb Q/Z);A)$ is an isomorphism onto the even degrees, and $H^\ast(j;A)$ kills the odd classes while being an isomorphism on even classes. Then one extends this analysis to $n > 1$.

Lemma 8: Let $A \xrightarrow i B \xrightarrow j C$ be maps of chain complexes. Suppose that $ji$ is injective and $\Image(ji) = \Image(i)$. Then the sequence of homologies $H_\ast(A) \xrightarrow{i_\ast} H_\ast(B) \xrightarrow{j_\ast} H_\ast(C)$ has $i_\ast$ injective and $Image(j_\ast i_\ast) = Image(i_\ast)$.
Proof: Diagram chase.

Corollary 9: Fix $s \in \mathbb Z$, and consider the maps $H^s(BN_\ast;A) \xrightarrow i H^s(BN_{qr};A) \xrightarrow j H^s(BN_q;A)$. For $q$, $r$ sufficiently divisible by $m$ and $A$ $m$-torsion, we have that $ji$ is injective and $\Image(ji) = \Image(j)$.
Proof: Using Lemma 7 as a base case, use Lemma 8 to induct through the pages of the Serre spectral sequences for the fibrations over $BW$. This is a first-quadrant spectral sequence, so for fixed $s$ it stabilizes at a finite page. The statement can be tested on associated gradeds, so there are no extension problems.

Corollary 10: Assume that $E$ is bounded below and $m$-torsion, and fix $s \in \mathbb Z$. Consider the maps $E^s(X_{hN_\ast}) \xrightarrow i E^s(X_{hN_{qr}}) \xrightarrow j E^s(X_{hN_q})$. For $q,r$ sufficiently divisible by $m$, we have that $ji$ is injective and $\Image(ji) = \Image(j)$.
Proof: Using Corollary 9 as a base case, use Lemma 8 inductively to walk through the Atiyah–Hirzebruch spectral sequences for the fibrations over $BN_\ast$, $BN_{qr}$, and $BN_q$ respectively (which all have fiber $X$). Since we are assuming that $E$ is bounded below, this is essentially a first-quadrant spectral sequence so the argument goes in the same way as before.

Proof of Lemma 6: That this follows from Corollary 10 in the case where $E$ is bounded below can be seen in two ways — either from the eventual injectivity of $E^\ast(X_{hN_\ast}) \to E^\ast(X_{hN_q})$, or from the fact that sequence is Mittag–Leffler. When $E$ is not bounded below, we simply pass to a suitable connective cover of $E$, since we are always taking the cohomology of a suspension spectrum, which is bounded below.
A: After the heavy lifting done by people on MSE and in the comments, I think it's not too bad to finish off the proof that the answer is yes.
As argued by Ben Wieland in the comments, we reduce to showing that for any short exact sequence of topological groups
$$U(1)^n \to G \to W $$
where $W$ is finite, we have that $H^\ast(BG;\mathbb Z)$ injects into the product of $H^\ast(BF;\mathbb Z)$ over all finite subgroups $F \subseteq G$. The jist of the argument is going to be replacing $U(1)^n$ with $(Q/\mathbb Z)^n$, and then arguing that every finitely-generated subgroup of the resulting extension $G_\ast$ is finite.
The first thing to note is that for some $m \in \mathbb Z$, there exist maps of exact sequences
$$\require{AMScd} \begin{CD} (\mathbb Z/m)^n @>>> G_m @>>> W \\ @VVV @VVV @VV{=}V\\(\mathbb Q /\mathbb Z)^n @>>> G_\ast @>>> W \\ @VVV @VVV @VV{=}V\\ U(1)^n @>>> G @>>> W \end{CD}$$
The middle line exists and maps to the bottom line because (1) $\mathbb Q/\mathbb Z$ is the torsion subgroup of $U(1)$, and so must be preserved by the action of $W$, and (2) the discrete group quotient $U(1)^\delta / (\mathbb Q/\mathbb Z)$ is a rational vector space, so no matter the action of $W$, the cohomology of $W$ with values in this quotient vanishes by Maschke's theorem. Thus $H^\ast(BW; \underline{(\mathbb Q/\mathbb Z)^n}) \to H^\ast(BW; \underline{U(1)^{\delta,n}})$ is an isomorphism and in particular the class classifying the extension is hit.
The top line exists and maps to the middle line because when we choose a cocycle $W \times W \to (\mathbb Q/\mathbb Z)^n$, we see that because $W$ is finite, the cocyle has finite image, and every finitely-generated subgroup of $(\mathbb Q/\mathbb Z)^n$ is finite — thus the cocycle lives in some finite, $W$-invariant subgroup $(\mathbb Z/m)^n \subseteq (\mathbb Q / \mathbb Z)^n$ (using that $(\mathbb Z/m)^n \subseteq (\mathbb Q/\mathbb Z)^n$ is the $m$-torsion subgroup and so must be $W$-invariant).
By similar reasoning, we see that every finitely-generated subgroup of $G_\ast$ is finite. Therefore, because homology commutes with filtered colimits, we have $H_\ast(BG_\ast) = \varinjlim_{G' \subseteq G_\ast} H_\ast(BG')$, where the colimit is over finite subgroups (or even just those finite subgroups $G' = G_m$ of the form given above), and we use any constant coefficients.
Now, because $B(\mathbb Q/\mathbb Z)^n \to BU(1)^n$ is a homology isomorphism with any finite coefficients (this can be checked in several ways), we see that $BG_\ast \to BG$ is likewise a homology isomorphism with finite coefficients by the Serre spectral sequence. So the composite map $\varinjlim H_\ast(BG_m) \to H_\ast(BG_\ast) \to H_\ast(BG)$ is an isomorphism with finite coefficients. By the universal coefficient theorem, the map $H^\ast(BG) \to \prod H^\ast(BG_m)$ is injective with finite coefficients. Since Qiaochu has already shown that it is an injection on nontorsion elements, it follows that this map is an injection on integral cohomology (noting that these things are sufficiently finite to be safe).

Note that most of the above boiled down to facts about the extension $U(1)^n \to G \to W$, not really facts about (co)homology — the only thing we really needed was the isomorphism $H_\ast(B(\mathbb Q/\mathbb Z)) \cong H_\ast(BU(1))$ with finite coefficients.
In fact, Theorem 5.7 of Becker and Gottlieb's original paper The transfer map and fiber bundles (DOI) is actually stated for general cohomology theory, and implies by Ben Wieland's argument that $\Sigma^\infty BG$ splits off of $\Sigma^\infty BN(T)$, so the reduction to extensions $T \to N \to W$ with $T$ a torus and $W$ finite holds for an arbitrary homology or cohomology theory $E$.
It's not hard to show that if $E$ is a spectrum with trivial rationalization, then $B(\mathbb Q/\mathbb Z)^n \to BU(1)^n$ is and $E$-homology or cohomology equivalence. So the above argument shows that in this case, we have that $N_\ast \to N$ is an $E$-homology and $E$-cohomology equivalence, where $N_\ast$ fits in the extension $(\mathbb Q/\mathbb Z)^n \to N_\ast \to W$ as above. Moreover, since $BN_\ast = \varinjlim BN'$ where the colimit is over finite subgroups, and since this is a homotopy colimit, we have $E_\ast(BN) = \varinjlim E_\ast(BN')$, so that $\bigoplus E_\ast(BN') \to E_\ast(BN)$ is surjective. For cohomology, there are potential $\varprojlim^1$ issues.
Thus the statement we get is:
Theorem: Let $E$ be a spectrum with trivial rationalization, and let $G$ be a compact Lie group. Then $\bigoplus_{F \subseteq G} E_\ast(BF) \to E_\ast(BG)$ is surjective, where the sum is over finite subgroups $F \subseteq G$.
It would be nice if this could be upgraded to a statement about all spectra by considering also the rationalization, but that seems unpromising because of Maschke's theorem — Qiaochu's argument for nontorsion classes is more subtle, it seems.
Probably also some statement about cohomology is possible ….

In fact, it's not hard to extend the statement to arbitrary $G$-spaces. That is:
Theorem: Let $E$ be a spectrum with trivial rationalization, let $G$ be a compact Lie group, and let $X$ be a $G$-space. Then $\bigoplus_{F \subseteq G} E_\ast(X_{hF}) \to E_\ast(X_{hG})$ is surjective, where the sum is over finite subgroups $F \subseteq G$.
