Unramified map of Riemann surfaces

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.

2 Answers
2

The simplest "non-trivial" example is $$zmapsto int_0^ze^-zeta^2dzeta:quad Cto C.$$
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.

Let $C to D$ be a nontrivial finite étale cover and $p in C$ a point. Then $C setminus p to D$ is still surjective and unramified, but it's not a covering.

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