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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.

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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.

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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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  • $\begingroup$ Yes, of course, I did not consider this trivial case. I shall see if I can somehow reformulate my question. Thanks, Dan. $\endgroup$ – Chitrabhanu Sep 2 '18 at 10:35

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