Proper Holomorphic Mapping implies Branched Covering

Suppose $D_1$ and $D_2$ are two bounded domains in $\mathbb{C}^n$ and $\text{ }f: D_1 \to D_2$ is a surjective proper holomorphic map. Then why is $f$ a branched cover of finite order?

• What happens when $n=1$ ? – reuns Jun 20 '17 at 0:08
• Not sure. I want to consider $f^{-1}\{0\}$ and show that on $D_1$ minus this set that $f$ is a covering map. I believe that I need to show that $f^{-1}\{0\}$ is a finite set, but I am not sure if that is true. – Nick Jun 20 '17 at 1:24
• I'm not sure of your definition, but I would think asking $f$ to be continuous on the boundary (assumed to be smooth..). By the open mapping we would get that $f$ is continuous $\partial D_1 \to \partial D_2$ so it is a finite covering $\partial D_1 \to \partial D_2$ and hence $D_1 \to D_2$. – reuns Jun 20 '17 at 1:28
• I don't have that f extends continuously to the boundary. But if it did, why does it imply that $\partial D_1 \to \partial D_2$ is a finite cover and not an infinite cover? – Nick Jun 20 '17 at 1:42
• If $n=1$, I'd say since $\partial D_1$ is compact, if $|f^{-1}(p)| = \infty$ then the $z_i, f(z_i) = p$ accumulate at some $z$. Also between $z_i,z_{i+1}$ there is always a $s_i$ such that $f(s_i) = p_2$, thus $f$ is discontinuous at $z$. I wonder if it works with $n=2$. – reuns Jun 20 '17 at 2:16