Let $f:S \to T$ be a surjective, unramified, holomorphic map between connected Riemann surfaces. If $S$ is not compact is it always true that $f$ is a covering?


This is of course true if $S$ is compact or, more generally, if $f$ is proper. However, I can not see why this should be true in general.



Unramified map of Riemann surfaces
The simplest "non-trivial" example is $$z\mapsto \int\_0^ze^{-\zeta^2}d\zeta:\quad C\to C.$$
It is surjective, and not ramified. But it is certainly not a covering because every covering over a simply connected surface is a homeomorphism.


You can make the target surface compact if you wish. Consider the map from $C$ to
$S$, where $S$ is the Riemann sphere, $f(z)=y\_1/y\_0$, where
$y\_j$ are two linearly independent solutions of the Airy equation
$$y''=zy.$$ It is also surjective and unramified: $f'=-W(y\_1,y\_0)/y\_0^2$,
where $W$ is the Wronskian determinant which is constant.


