Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.