So coming back to what I have saying see here are two ways of showing that a compact Riemann surface is algebraic. First approach: Using harmonic analysis, show that there ()is enough meromorphic functions which separate points and tangents. Now using () and some basic cohomology prove Riemann-Roch. Now using Riemann-Roch you may fin a very ample line bundle to embedd your Riemann surface in a projective space.

You wrote <<Everything uses coh>> Well I'm not sure that I would agree that coherent sheaves are ubiquitous to mathematics but for sure this is a key finiteness notion which appear at many places. Nevertheless, I still think you can come up with a better answer and point me out the key places in the argument where the finiteness condition of coherence is used :) In any case, thanks a lot for your comments, I appreciate.

As you know in complex dimension > 1 this is precisely the lack of meromorphic functions (or differentials) which prevent an compact complex manifold from being algebraic. So here from a single analytic map $f:Y\rightarrow X$ we get the existence of meromorphic functions on $Y$ which separate points and tangents. You see, there is no harmonic analysis involved in this argument. It seems to be purely some "homological algebras" and some probably clever commutative algebra.

Hi BCnrd, you wrote <<Your amazement about many rat'l fns must already occur for the proof of projectivity of compact Riemann surfaces (crux is to make one non-constant mero. fn), >> That is exactly it! You see in general I know how to prove that a compact Riemann is projective. First using some harmonic analysis you prove that you have () enough functions which separate points and tangent and from there you prove Riemann-Roch using only () as the input. See for example Miranda's book on Comapct Riemann surfaces and algebraic.

And by many rational functions I mean enough functions so that you can separate points and tangents. This is kind of fascinating since a priori I don't quite see how to use the mere existence of $f$ to construct a meromorphic function $g:Y\rightarrow\mathbf{C}$ such that $g(P)=0$ and $g(Q)=1$ where $P$ and $Q$ are 2 points in the same fiber of the map $f:Y\rightarrow X$.

So you see the thing which I find fascinating about the algebraicity of $Y$ is the following. So we use the same notation as in the question. So from the existence of this analytic map $f:Y\rightarrow X$ we get that $Y$ is quasi-projective which implies that $Y$ has a lot of rational functions (so meromorphic)! So from the existence of only one analytic map we get the existence of many rational functions (even many regular functions if $X$ is affine for example)! So I'm trying to understand the heuristic behind that!

Are we somehow also using in the course of the argument the fact that on a Stein manifold $W$ a coherent sheaf of $O_W$-module is (1) generated by global sections and (2) has trivial cohomology groups in degrees larger than $0$.

Hi BCnrd, thanks a lot for Houzel's references. So in total Houzel's 4 papers "Geometrie analytique locale" consists of $12+22+25+15=74$ pages which is quite a lot of pages to read. So you made a very good remark by emphasizing the fact that $\mathcal{F}$ is not only a coherent sheaf of modules but of algebras. This remark is quite important. Nevertheless I would like to see at what places do we use the exact sequence which appears in (2) of my question.

Dear BCnrd, <<So apply GAGA to get coherent sheaf of algebras on $\overline{X}$>> Ok so far so good. Then you wrote: <<then finite $\overline{X}$-scheme which must recover $\overline{Y}$>> Could you explain this in greater details. This I think is what I was looking for. So how do you get this finite $\overline{X}$-scheme structure on $\overline{Y}$ using as the only input data the coherent sheaf $\mathcal{F}$ on $\overline{X}$? If you could make this statement precise so that I can see (2) is used I guess this would answer my question :)

It would be nice to have some kind of "hands on" description of this coherent sheaf $\mathcal{F}^{an}$ since after all it comes from the push forward of a map which is finite unramified outside some analytic divisor.

<<Anyway, the equality $\pi_1(X) = \pi_1(X^{\rm{an}})$ is very deep>> Well you see here I take for granted the existence of the map $\overline{f}$ so there is no need to use Grauert and Remmert constructions which I think guarantee the existence of such a map $\overline{f}$ compatible with $f$. Personally I think that the existence of the map $\overline{f}$ is deeper than what I'm asking for. So here the whole point is I take the existence of this map as granted! I think that from there the argument should be "clever homological algebra" but I cannot figure it out by myself!

Hi BCnrd, <<And do you want that in such cases the alg. structure is also unique?>> Yes but I think that this follows from my setup. Once you know that $\overline{Y}$ is projective and by this I mean of course that this projective structure is compatible with the analyitc structure which is given on $\overline{Y}$ then we know that this projective structure is unique, this follows from the corollary on p. 30 of Serre's GAGA paper. <<you must intend to be a finite analytic morphism: proper with finite fibers.>> Yes of course I want that! I will add it