# Cohomology groups of quotient by finite group

I know there are already lots of questions about (co)homology groups of a quotient manifold, but please let me ask one more question.

Let $G$ be a finite group acting on a manifold $M$ without fixed point. The standard spectral sequence argument shows that $$H^k(M,\mathbb{Q})^G\cong H^k(M/G,\mathbb{Q}).$$ This in particular shows that rank $H^k(M,\mathbb{Z})^G$=rank $H^k(M/G,\mathbb{Z})$.

We also have a natural map $\pi^{*}:H^k(M/G,\mathbb{Z})\rightarrow H^k(M,\mathbb{Z})^G$ via the quotient map $\pi:M\rightarrow M/G$.

Is it true that $\pi^{*}$ is injection mod torsion (and hence that $H^k(M/G,\mathbb{Z})\subset H^k(M,\mathbb{Z})^G$ (mod torsions) is of finite index)? Or are there any relation between $\mathbb{Z}$-coefficient (co)homology groups?

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It is poor form to post this question on MO while the original question you asked sits on math.SE: math.stackexchange.com/questions/184529/… –  Steve D Aug 20 '12 at 21:13

Yes. Consider the commutative diagram involving the maps $H^k(M/G,\mathbb Z) \to H^k(M,\mathbb Z)^G \to H^k(M,\mathbb Q)^G$ and $H^k(M/G,\mathbb Z) \to H^k(M/G,\mathbb Q) \to H^k(M,\mathbb Q)^G$
The kernel of the map $\pi^*$ is contained in the kernel of the second sequence. Since the second map in the second sequence is an isomorphism, the kernel of the second sequence is the kernel of its first map, which is a tensoring with $\mathbb Q$, so its kernel is torsion.