I am currently reading D. Mumford´s Abelian Varieties and it came up the following question: let $X$ be an algebraic variety over an algebraically closed field $k$ and $G$ a finite group acting on $X$. Assume we are in a situation that there exists a quotient $(Y, \pi:X \rightarrow Y)$, he then proves a proposition that there is a one-to-one correspondence between coherent $\mathcal{O}\_{Y}$-modules and $G$-equivariant coherent $\mathcal{O}\_{X}$-modules. In the course of the proof he shows that for such a $G$-sheaf $\mathfrak{F}$ on $X$ the natural morphism $\pi^{-1}((\pi\_{\*}\mathfrak{F})^{G}) \rightarrow \mathfrak{F}$ is an isomorphism. 


What about other kind of sheaves, especially locally constant sheaves (regarding whether this natural morphism is an isomorphism)?



Group action on sheaves
Hi Peter! 


I just checked in Mumford's book and there is an assumption in the theorem that *G* acts freely. If you don't assume this, it might be easier to instead work with the stack quotient $[X/G]$. Then the statement you are trying to show fits into the general framework of descent along torsors. See for instance Vistoli's notes from *Fundamental algebraic geometry* for a nice treatment. All of this amounts to a very general set-up for showing that *G*-equivariant sheaves on *X* are equivalent to sheaves on the quotient. This will for instance work for locally constant sheaves in the étale topology. Maybe you can also do it directly without invoking any general theory by imitating Mumford's proof: he passes to a completion to assume that the covering is trivial, but it seems that you could take an étale cover instead.


Depending on what you want to do with your sheaf, having a sheaf on the quotient stack might be just as good as having a sheaf on the scheme quotient. For instance, if you want to compute its cohomology you can apply Leray to the projection to the coarse moduli space. In any nice situation all higher derived functors of the pushforward will vanish so you get the cohomology on the scheme quotient as well.


Hope this helps.


