Let $f\colon X\to Y$ be a (finite sheeted) cover of connected (nice) spaces, $V\to X$ a vector bundle over $X$, and $p\colon Z\to Y$ a (finite) regular cover of $Y$ which factors through $f$, i.e., $Y=Z/G$ for some (finite) group $G$. Then $$ f_*(V)=\left(\bigoplus_{h\in \operatorname{Hom}(Z,X)}h^*(V)\right)\bigg/G, $$ where $\operatorname{Hom}(Z,X)=\{h\colon Z\to X\mid fh=p\}$ and $G$ acts in the obvious way. What is this lemma called? (Apologies in advance for ag tag but maybe it's appropriate given that I don't really care about the category.)
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$\begingroup$ How do you define $f_*(V)$? $\endgroup$– abxCommented Oct 13, 2017 at 14:36
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$\begingroup$ It's just the direct image of V, so $f_*(V)_y=\oplus_{x\in f^{-1}(y)}V_x$. $\endgroup$– seldom seenCommented Oct 13, 2017 at 14:38
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$\begingroup$ Err, push forward, I guess. $\endgroup$– seldom seenCommented Oct 13, 2017 at 14:44
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$\begingroup$ if $f$ is regular cover as well (group $H$), the fibre prod. $X \times_Y Z$ is trivial $H$-cover over Z (since by assumption has a section over $Z$). It's then a disjoint union of copies of $Z$ indexed by the maps $h$ above, mapping as identity to $Z$ and by $h$ to $X$. In this case, the formula follows since pushing forward by $f$ and pulling back to $Z$ is the same thing as pulling back to $X \times_Y Z$ and pushing forward to $Z$ AND using that pullback under $Z \to Y$ is equivalence of sheaves on $Y$ with $G$-equivariant sheaves on $Z$ (with inverse=quotient / $G$ as in your formula). $\endgroup$– JoSCommented Oct 22, 2017 at 18:19
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$\begingroup$ I can do basic covering space theory, thanks. $\endgroup$– seldom seenCommented Nov 9, 2017 at 21:52
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