What is the map $\pi^*:H^{\dim X/G}(X/G,Z)\rightarrow H^{\dim X}(X,Z)$?

Question

Let $\pi:X\rightarrow X/G$ be a free quotient map by a finite group $G$. Assume that both $X$ and $X/G$ are oriented. We know that $\pi_*$ maps the fundamental class $[X]$ to $|G|[X/G]$.

What about cohomology groups? Assume that $\alpha_X$ and $\alpha_{X/G}$ are the generators of top cohomology groups (determined up to sign). Can we tell where $\alpha_{X/G}$ is mapped by $\pi^*$? I think $\pi^*(\alpha_{X/G})=\alpha_{X}/|G|$. Am I right?

Answer 1

Aren't $\pi^*$ and $\pi_*$ adjoint so that $\langle\pi^*(\alpha_{X/G}),[X]\rangle=\langle\alpha_{X/G},\pi_*([X])\rangle$? That would mean you should multiply, not divide, by $|G|$ in your proposed formula. And that would have the big advantage that the answer is still a cohomology class with integer (not merely rational) coefficients.