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.
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.
Here is a more "topological" counterexample.
Let $C = \mathbb{C}^\times$ be the punctured plane. Let $D = \{ z \in \mathbb{C}^\times \mid |z| < 1 \}$ be the punctured disk. Define $\iota \colon D \to C$ to be the usual inclusion. Let $\rho \colon C \to T = (\mathbb{C} - 0)/ (z \sim 2z)$ be the quotient map. Then $\rho \circ \iota \colon D \to T$ is the desired counterexample. (You can pass to the universal cover $S$ of $D$ if you like, to make $S$ simply connected.)
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.
Unlike in the example of Dan Petersen, all my surfaces are simply connected.
