It is easy to see that there cannot be a proper holomorphic map from the punctured unit disk to the unit disk in the complex plane. What about the other direction; does there exist a proper holomorphic map from the unit disk to the punctured unit disk?
$\begingroup$ $\endgroup$
Add a comment |
2 Answers
$\begingroup$ $\endgroup$
I don't think so. A map $\mathbb D \to \mathbb D^\ast$ would lift to a map $\mathbb D\to \mathbb H$, where the upper halfplane $\mathbb H$ is seen as the universal cover of $\mathbb D^\ast$. As $\mathbb D \to \mathbb D^\ast$ is proper, so is $\mathbb D\to \mathbb H$. In particular it has closed image but by the open mapping theorem it also has open image and is hence equal to $\mathbb H$. That means that the inverse under $\mathbb D \to \mathbb D^\ast$ of a point is the disjoint countable topological union of non-empty sets which contradicts properness.
$\begingroup$ $\endgroup$
1 Question is somewhat not clear: take $i\colon D^*\rightarrow D$ to be identity map, which is holomorphic. Then you can take $f_n\colon D^{*}\rightarrow D^{*}$, $f(z)=z^n$, for $n\in \mathbb{Z}$,and compose with $i$ to get different holomorphic maps from punctured unit disc to unit disc.
- 1$\begingroup$ In my question, proper means that the inverse image of a compact set is compact. $\endgroup$Jaikrishnan– Jaikrishnan2011-02-08 12:33:50 +00:00Commented Feb 8, 2011 at 12:33