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It is well known that there is no proper holomorphic map from complex plane onto disc by Liouville's theorem.Does there exist a proper holomorphic map $f$ from the unit disc onto the complex plane?I believe that such map does not exists but I'm unable to prove this.Please Help!

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

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  • $\begingroup$ Any holomorphic map from the unit disk to the complex plane is automatically proper, so I think you can remove the properness assumption. The real question is whether there is a holomorphic map from the unit disk ONTO the complex plane. $\endgroup$ – Alex Fok May 29 '15 at 11:03
  • $\begingroup$ @AlexFok This is not true. For example, the map described in this question is not proper. $\endgroup$ – Jim Belk May 30 '15 at 21:07
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There is no such map. For convenience, I will use the upper half-plane $\mathbb{H}$ instead of the unit disk.

Suppose $f\colon \mathbb{H}\to\mathbb{C}$ is a proper holomorphic map.

Note first that, for each sequence $\{z_n\}$ in $\mathbb{H}$ that converges to a point $p$ on the real axis, the image sequence $\{f(z_n)\}$ must must converge to $\infty$ on the Riemann sphere. To prove this, observe that if $\overline{D}_R$ is the closed disk of radius $R$ in $\mathbb{C}$ centered at the origin, then $\overline{D}_R$ is compact, and hence $f^{-1}(\overline{D}_R)$ is a compact subset of $\mathbb{H}$. Then $\mathbb{C}-f^{-1}(\overline{D}_R)$ is an open neighborhood of $p$, so $z_n$ lies in the complement of $f^{-1}(\overline{D}_R)$ for all but finitely many $n$. This holds for all $R>0$, which proves the claim.

Now, since $f$ is a nonzero holomorphic function, the zeroes of $f$ form a discrete subset of $\mathbb{H}$. Since $f$ is proper, the set of zeroes must be compact, and therefore there are only finitely many zeroes. Let $D$ be any open disk centered on the real axis that does not contain any zeroes of $f$, and consider the holomorphic function $g(z) = 1/f(z)$ on $D\cap\mathbb{H}$.

Now, $g$ has the property that $g(z_n) \to 0$ for any sequence $z_n \in D\cap\mathbb{H}$ converging to a point on the real axis. Thus $g$ extends continuously to $D\cap\overline{\mathbb{H}}$ by setting $g(x) = 0$ for $x \in D\cap\mathbb{R}$. By the Schwarz reflection principle, $g$ now extends to a holomorphic function on all of $D$, which is nonsense since $g$ is zero on $D\cap\mathbb{R}$.

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    $\begingroup$ Sorry - why does properness imply $f(z_n)\to \infty$ for $Im(z_n)\to 0$? $\endgroup$ – Neal May 30 '15 at 22:18
  • $\begingroup$ @Neal For a proper map, the image of any sequence that escapes to infinity must escape to infinity. See the Wikipedia article. In this case, the argument would be that the preimage of the closed disk of radius $R$ in the complex plane must be a compact subset of $\mathbb{H}$, and therefore contains only finitely many $z_n$. $\endgroup$ – Jim Belk May 30 '15 at 22:25
  • $\begingroup$ In this case, would it be more precise to say the image of any sequence with no convergent subsequence must be a sequence with no convergent subsequence? For instance, $1 + i/n$ doesn't escape to infinity as $n\to \infty$. $\endgroup$ – Neal May 30 '15 at 23:14
  • $\begingroup$ @Neal Here "escape to infinity" is being used in the technical sense -- a sequence $\{x_n\}$ "escapes to infinity" in a topological space $X$ if, for every compact subset $K$ of $X$, the sequence has only finitely many terms in $K$. Thus the sequence $\{1+i/n\}$ does escape to infinity in $\mathbb{H}$. $\endgroup$ – Jim Belk May 31 '15 at 0:04
  • $\begingroup$ @Neal I've added an argument about this to clarify my answer. $\endgroup$ – Jim Belk May 31 '15 at 0:18

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