Quasi-finite surjective holomorphic map is proper? Let $X$ and $Y$ be connected complex manifolds. Let
$f:X \to Y$ be a surjective holomorphic map such that pre-image of every $y \in Y$ is a finite set.
Then can we say that $f$ is a proper map? I feel that this is not true but I am unable to think of a counterexample.
If not counterexample, hints towards $f$ is proper are welcome!
 A: Such a map need not be proper. A simple example is given by the map $f \colon z \mapsto z^2$ with $Y = \mathbb{C}\setminus \{0\}$ and $X$ any connected open subset of $Y$ such that $f(X) = Y$. Standard choices for $X$ are $\mathbb{C} \setminus (-\infty,0]$ or a three-quarter plane. A maximal $X$ is for example $Y \setminus \{1\}$.
Then if $z_0$ is a boundary point of $X$ in $Y$, the preimage of a compact neighbourhood of $f(z_0)$ isn't compact.
A: Without the surjectivity assumption, a counterexample is the map $f: \bar{B}^c \to B$ given by $z \mapsto 1/z$ where $B$ is the open unit disk in the complex plane.
A: Recently, I found a very interesting counter-example of such mapping in higher dimensions: $f: \mathbb{C}^2 \to \mathbb{C}^2$
defined as
$$
f (x, y) = (x + x^2y, y).
$$
Note that $f$ is surjective and pre-image of any point contains at most two points.
But, $f$ is not proper because for any compact neighborhood $K$ of $(0,0)$, the set $f^{−1}(K)$ is never compact.
Interestingly, we cannot have surjective finite to one holomorphic self-map of $\mathbb{C}$ that is not proper.
