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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$ ?

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  • $\begingroup$ Not every continuous map is proper. $\endgroup$ – Jesse Madnick May 30 '15 at 6:01
  • $\begingroup$ @John what if the domain is bounded ? [ in $ \mathbb R^n$ only] $\endgroup$ – Arpit Kansal May 30 '15 at 6:18
  • $\begingroup$ @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? $\endgroup$ – Arpit Kansal May 30 '15 at 6:23
  • $\begingroup$ @John but any such map must be surjective because it an open and closed map. $\endgroup$ – Arpit Kansal May 30 '15 at 18:35
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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.

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  • $\begingroup$ Dear, Thank you for your response.First claim follows from :math.stackexchange.com/questions/1299051/… and the last contradiction comes with open mapping theorem,right? $\endgroup$ – Arpit Kansal May 30 '15 at 20:14
  • $\begingroup$ @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? $\endgroup$ – Dontknowanything May 31 '15 at 2:58
  • $\begingroup$ 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 … $\endgroup$ – zhw. May 31 '15 at 3:43
  • $\begingroup$ @zhw Can you please tell why is map zero on circle? $\endgroup$ – Arpit Kansal May 31 '15 at 6:15

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