Let $X$ be a manifold or scheme with a finite group $G$ acting on it freely. Let $\pi:X\rightarrow X/G$ be the natural projection. We have $\pi_{\*}\pi^{\*}=|G|id$ on $H^*(X/G,\mathbb{Z})$. Can we say anything about the map $\pi^{\*}\pi_{\*}$?
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2$\begingroup$ It maps a cochain to its orbit under the group action. $\endgroup$– HJRWCommented Jan 7, 2013 at 14:41
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4$\begingroup$ For $\pi_{\ast}\pi^{\ast}$ one can say much more than the limited statement you make about integral cohomology, and such refinements are useful. Please think for yourself about more interesting things for $\pi_{\ast}\pi^{\ast}$. One can "say things" about $\pi^{\ast}\pi_{\ast}$ (perhaps $\mathcal{F} \rightarrow (\pi^{\ast}\pi_{\ast}\mathcal{F})^G$ being an isomorphism is what you want?), but what is the goal? This question is too vague. $\endgroup$– user30379Commented Jan 7, 2013 at 15:08
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Consider the product $X\times G$. There are two maps, the projection $p:X\times G \to X$ and the action $a:X\times G \to X$. One can check that $\pi^*\pi_* \cong a_*p^*$. In other words, consider the image of $X\times G$ in $X\times X$ (under the map $(p,a)$) as a correspondence. Then $\pi^*\pi_*$ is equal to the map given by that. In particular, if you can describe the class of $X\times G$ in $H^*(X\times X,\mathbb{Z})$, you can describe the map explicitly.
