# Non Existence of a proper holomorphic map from the punctured unit disc to an Annulus

Show that there is no proper holomorphic map from the punctured unit disc to an annulus $A_r=\{z \in \mathbb C:1 <|z| < r \}$.

Def:A map $f: X \to Y$ is called proper if $f^{-1}(K)$ is compact for every compact set $K$ in Y.

please give some hints/ideas to prove this.Can someone please give a reference for reading about construction of proper maps between different domains in $\mathbb C$ ?

• Not every continuous map is proper. – Jesse Madnick May 30 '15 at 6:01
• @John what if the domain is bounded ? [ in $\mathbb R^n$ only] – Arpit Kansal May 30 '15 at 6:18
• @John I mean as $f$ is continuous therefore $f^{-1}(K)$ is closed and bounded(due to punctured disc) in domain punctured disc.Am I missing something? – Arpit Kansal May 30 '15 at 6:23
• @John but any such map must be surjective because it an open and closed map. – Arpit Kansal May 30 '15 at 18:35

Sketch: Suppose $f:\mathbb {D}\setminus \{0\}\to A_r$ is holomorphic and proper. Let $z_n \to 0$ within $\mathbb {D}\setminus \{0\}.$ Show then that the distance from $f(z_n)$ to $\partial A_r$ goes to $0;$ this follows from $f$ being proper. Now $f$ is bounded and holomorphic in $\mathbb {D}\setminus \{0\},$ so the isolated singularity of $f$ at $0$ is removable. We then arrive at a nonconstant holomorphic map $F : \mathbb {D} \to \overline {A_r}$ with $F(0) \in \partial A_r.$ This can't happen.
• I think there is no proper holomorphic map in the other direction. Otherwise it seems to me that would imply the map is identically $0$ on one of the circles making up $\partial A_r.$ That would then imply the map is identically $0,$ contradiction. Maybe there's an easier way … – zhw. May 31 '15 at 3:43