Ok I got it. So by taking the fiber product you can bring it back to the case where you only have to deal with one arrow (modulo the simple but important observation that the connected components in the analytic and algebraic category agree). Many thanks!

I don't see it. So if $X$ and $Y$ are algebraic varieties and $f:X\rightarrow Y$ is analytic, I agree that $f$ will be regular iff the graph $\Gamma_f\subseteq X\times Y$ is Zariski locally closed but I dont see how this answer my question...

So using Grauert-Remmert and GAGA we have: for every finite analytic unramified covering $f:X_1^{an}\rightarrow X_2$ with $X_2$ algebraic, there exists a unique algebraic structure un $X_1$ such that $f$ is finite etale. But In my question I'm also requiring that this regular map $f$ is compatible with the algebraic structure of $Z$.

So I never read Grauert und Remmert, but are they proving the existence of a finite analytic ramified covering over suitable compactifications of a single arrow $f:X\rightarrow Y$ or they also do it for a commutative triangle like in my question?

So taking for granted Grauer-Remmert, I would you prove the Riemann Existence theorem?

Dear Matthew, so when you say that abelian varieties are rigid do you mean that you are looking for complex deformations that preserve their endomorphism ring? If yes, then how does this fit into classical deformation theory "a la Kodaira-Spencer"?

Donu, Is there a way using (only) deformation theory to see that an abelian variety with complex multiplication can be defined over number field? When $X$ is a smooth projective variety over $\amthbf{C}$ such that $H^1(X,\Theta_X)=0$ then since the Kodaira-Spencer map is trivial we see readily that $X$ can be defined over a number field. I know about the classical proof for elliptic curves which uses the $j$-invariant but I'm wondering if there exists some refinement of the deformation theory argument that I have just explained.

Thanks Donu, so you are using the Dolbeault's isomorphism $H^1(A,\mathcal{O}_A)≃H_{\overline{\partial}}^{0,1}(A)$.

Of course intuitively I see why we should have $g^2$ since one needs to choose $2g$ $R$-linearly vectors in $C^g$ so we see that the moduli space of $g$-dimensional complex tori should be something like $GL_{2g}(Z)\backspace M_{2g\times 2g}(R)/GL_g(C)$ so the complex dimension is $g^2$ but this is just a heuristic. But at the end one needs to invoke something in order to compute the dimension of this $H^1$ since it boils down to solve some system of differential equations.