# Cohomology group of free quotient.

Let $G$ be a finite group acting on a manifold $M$ without fixed point. The standard Leray-Cartan-Serre spectral sequence argument shows that $$H^k(M,\mathbb{Q})^G\cong H^k(M/G,\mathbb{Q}).$$ This in particular means 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 (and hence that $H^k(M/G,\mathbb{Z})\subset H^k(M,\mathbb{Z})^G$ is of finite index)?

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No. Take, for example, $M=S^n$, $G=\mathbb Z/2$ (acting by antipodal involution). Now all cohomology groups $H^k(M)$ for $0<k<n$ are trivial but half of $H^k(M/G=\mathbb RP^n)$ are not.
But the statement becomes true for n-fold coverings if one adjoins 1/n to coefficients: since (for n-fold covering) $\pi_*\pi^*$ equals multiplication by n, $f^*$ is injective whenever n is invertible; or to put it another way: $\ker \pi^*\subset \ker(x\mapsto nx)$.