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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$ ?

Arpit Kansal
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1 Answers1

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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.

zhw.
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  • Dear, Thank you for your response.First claim follows from :http://math.stackexchange.com/questions/1299051/classification-of-proper-maps-between-domains-in-mathbbrn and the last contradiction comes with open mapping theorem,right? – Arpit Kansal May 30 '15 at 20:14
  • @zhw +1,nice proof.Sorry..I am also interested in knowing the map in the reverse direction.I believe again there is no map in the reverse direction too and I think it will follow by using a covering map.Can you please give some idea? – Dontknowanything May 31 '15 at 02:58
  • 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 03:43
  • @zhw Can you please tell why is map zero on circle? – Arpit Kansal May 31 '15 at 06:15