Let $X$ be a smooth quasi-projective variety (so irreducible) over $\mathbf{C}$. We may think of $X$ as a complex manifold which we denote by $X^{an}$. Of course the topology on $X^{an}$ is finer than the Zarisiki topology on $X$. Now let us suppose that we have a surjective finite unramified analytic cover

$f:Y\rightarrow X^{an}$.

Now for the sake of simplicity (I'm quite sure that one may relax considerably these assumptions) we will assume that there exists a normal projective variety $\overline{X}\supseteq X$ (as an open subset in the Z-topology) and that there exists a normal compact analytic variety $\overline{Y}\supseteq Y$ ( as an open subset in the analytic topology) and a finite ramified analytic covering map

$\overline{f}:\overline{Y}\rightarrow\overline{X}^{an}$

which extends the map $f$.

Then one may look at the analytic coherent sheaf $O_{\overline{Y}}$ push it forward by $f_{*}$ and obtain the following analytic coherent sheaf on $\overline{X}^{an}$:

$\mathcal{F}^{an}:=f_{*}{\mathcal{O}}_{\overline{Y}}$.

Now by GAGA we now that there exists a unique algebraic coherent sheaf $\mathcal{F}$ on $\overline{X}$ such that the

(1) The "analytification" of $\mathcal{F}$ is equal to $\mathcal{F}^{an}$.

By definition of coherence of $\mathcal{F}$ we know that

(2) For evey $x\in X$ there exists a Zariski open set $U$ of $x$ such that the sequence of algebraic sheaves

$({O_{\overline{X}}|U})^n\ \rightarrow ({O_{\overline{X}}|U})^m\rightarrow\mathcal{F}|U\rightarrow 0$

is exact for some integers $m,n\in\mathbf{Z}_{\geq 0}$ (which may depend on $x$).

Now using $(1)$ and $(2)$ is there a simple way to deduce that $\overline{Y}$ is projective. Note that once we know that $\overline{Y}$ is projective then $\overline{Y}\backslash Y$ is analytically closed and therefore Zariski closed which implies that $Y$ is quasi-projective.

The conclusion that I was interested in was $Y$ is quasi-projective. So it seems that one may find a proof that $\overline{Y}$ is projective in Chap 12 of SGA1, but I'm sure that there must be a direct and easy way to deduce the algebraicity of $\overline{Y}$ using $(1)$ and $(2)$.