If a finite group $G$ acts on a complex variety $X$, then it is elementary to show that $G/X$ is a complex variety.  Taking quotients like this preserves nice properties like being projective or quasiprojective.  I was wondering whether you could go in the other direction.

Here is a precise question.  Let $Y$ be a complex manifold that can be given the structure of a complex variety $Y'$.  Let $\pi:\tilde{Y} \rightarrow Y$ be a finite, unramified cover of $Y$.  Can $\tilde{Y}$ be given the structure of a complex variety $\tilde{Y}'$ such that there is a finite map $\pi' : \tilde{Y}' \rightarrow Y'$ making the obvious diagram commute?  If the answer is yes, then can we take $\tilde{Y}'$ to be projective/quasiprojective if $Y'$ is projective/quasiprojective?

This kind of thing is true for Riemann surfaces, but even there I don't know how to prove it except by going through the whole machinery showing that all compact Riemann surfaces are projective varieties.  Since such things are not available in higher dimensions, I'm stuck.

Thanks!