I have a finite group $G$ acting freely on a manifold and $\pi:M\to M/G$ is the covering map. I need to show that the induced (pullback) map $\pi^*:H^*(M/G)\to H^*(M)$ is injective, where $H^*(M)=\oplus_{k=0}^n H^k(M).$
I have some questions. The first one is how to justify that if $\dim M=n$ then $\dim M/G=n$ (is that always true?)
In such a case, the induced map $\pi^*:H^0(M/G)\oplus\cdots\oplus H^n(M/G)\to H^0(M)\oplus\cdots\oplus H^n(M)$ would be defined as $$\pi^*(a_0+\ldots+a_n)=\pi_0^*(a_0)+\ldots+\pi_n^*(a_n),$$ where $\pi_k^*:H^k(M/G)\to H^k(M)$ is the pullback of a class of $k-$forms in $H^k(M/G)$ to $H^k(M).$
Now I try to prove injectivity. If I assume that $$\pi^*(a_0+\ldots+a_n)=\pi^*(b_0+\ldots+b_n),$$ then $$\pi_0^*(a_0)+\ldots+\pi_n^*(a_n)=\pi_0^*(b_0)+\ldots+\pi_n^*(b_n)\in H^*(M).$$ By uniqueness in the direct sum, we have $\pi_j^*(a_j)=\pi_j^*(b_j)$ for all $j=0,1,\ldots,n.$ This means that, for every $j,$ there exists $\gamma_j\in\Omega^{j-1}(M/G)$ such that $\pi_j^*(a_j)=\pi_j^*(b_j)+d\gamma_j.$ Since $\pi_j^*$ is a linear map, then $\pi_j^*(a_j-b_j)=d\gamma_j.$
This implies that $d(\pi_j^*(a_j-b_j))=d(d\gamma_j))=0$ and using the fact that derivative and pullback commute, then $\pi_j^*(da_j-db_j)=0,$ so $da_j-db_j\in\ker\pi_j^*.$
I don't know how to proceed because nothing ensures that $\pi_j^*$ is an injective map.
Corrections and suggestions are appreciated.