Finite covers of complex varieties (all but two questions answered!) EDIT: Thanks to several people I almost have a complete answer.  BCnrd pointed out that the generalized Riemann Existence Theorem shows that $\tilde{Y}$ can be uniquely given the structure of a complex variety $\tilde{Y}'$.  Also, Georges pointed out that the Kodaira embedding theorem implies that $\tilde{Y}'$ is projective if $Y'$ is projective.
This leaves two parts open to my original question.  If $Y'$ is quasiprojective, then must $\tilde{Y}'$ be quasiprojective?  Also, if $Y'$ is affine, then must $\tilde{Y}'$ be affine?
The impression I get from reading the comments is that the answers are "yes" and that everyone but me is able to easily prove them given what has already been said.  Can anyone give me a hint or a reference as to how to proceed?
Thanks to everyone for all your help so far.

ORIGINAL 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/affine if $Y'$ is projective/quasiprojective/affine?
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.
I should maybe remark that I don't even know how to do the above for affine varieties.
Thanks!
 A: Georges' answer is very nice so I will just add some comments.
In the curve case the assumption "unramified" is not necessary; in fact, every finite cover 
of a Riemann surface is still a Riemann surface (this is essentially Riemann Existence Theorem).
In higher dimension the situation is more complicate and "unramified" is definitely necessary. In fact, Donaldson and Auroux proved that every real, symplectic 4-manifold $(X, \omega)$ can be realized as a finite branched cover of $\mathbb{CP}^2$, and moreover such a cover $f \colon X \to \mathbb{CP}^2$ is 
completely determined, up to symplectomorphisms, by 


*

*the branch curve $D \subset \mathbb{CP}^2$ and

*the monodromy representation $\theta \colon \pi_1(\mathbb{CP}^2 -D) \to S_{N}$, where $N:= \deg f$.


Finally, $X$ is complex - projective  if and only if $D$ is isotopic to a complex curve. 
A: A finite unramified map mapping to an algebraic variety is finite in the algebraic sense and hence affine, so the preimage of an affine subset is affine. (I suppose this assumes that by affine you mean affine algebraic. I am not entirely sure what happens if the target is an affine analytic subset of $\mathbb C^n$. I assume there may be an analytic equivalent of a finite map being affine, but I don't know). 
A finite map is also projective, hence if the target is quasi-projective, then the source is projective over something quasi-projective so itself is quasi-projective.
A: Dear Nikita, Kodaira has proved the following fantastic embedding  theorem for complex manifolds (conjectured by Hodge in 1950):
A compact complex manifold X is projective algebraic if and only if it has a closed positive (1,1) form whose cohomology class is rational.
Succinct explanation : The positivity is a complex differential geometric notion: locally a (1,1) form can be written $ i \Sigma h_{jk} dz_j d\bar {z_k}$ and the matrix $(h_{jk})$ is required to be positive definite.This closed form represents a class in singular cohomology by De Rham's theorem and this class should actually be in $H^2(X,\mathbb Q)$
It follows easily from this that
Given a holomorphic finite unramified covering   $\tilde X \to X$ of compact holomorphic manifolds, the manifold $X$ is projective algebraic if and only if $\tilde X$ is 
You can read the proofs in Griffiths-Harris's Principles of Algebraic Geometry. 
