Let $f:\mathbb{C}\cup \{\infty\} \to \mathbb{C}\cup \{\infty\}$ be a holomorphic, non-constant map and $G\subseteq \mathbb{C}\cup \{\infty\}$ non-empty, connected, $\infty\in G$ and $G$ satisfies: $G\cap \mathbb{C}\subseteq \mathbb{C}$ is open and $\varphi(G)\cap \mathbb{C}\subseteq \mathbb{C}$ is open, where $\varphi: \mathbb{C}\cup \{\infty\} \to \mathbb{C}\cup \{\infty\}$, $\varphi(z)=\frac{1}{z}$.

I need for a proof, that $f(G)\cap \mathbb{C}\subseteq \mathbb{C}$ and $\varphi(f(G))\cap \mathbb{C}\subseteq \mathbb{C}$ is open, but I don't see why this should be satisfied. Are $f(G)\cap \mathbb{C}\subseteq \mathbb{C}$ and $\varphi(f(G))\cap \mathbb{C}\subseteq \mathbb{C}$ open?

  • $\begingroup$ @Tanuj No, it must not. $f$ can be any (non constant) rational function. $\endgroup$ Jan 19, 2016 at 21:53
  • $\begingroup$ I thought $f$ is holomorphic? $\endgroup$
    – user301452
    Jan 19, 2016 at 21:59
  • 1
    $\begingroup$ rational functions are holomorphic on $\mathbb{C}\cup \{\infty \}$, or am I wrong? $\endgroup$
    – dgl
    Jan 19, 2016 at 22:06
  • $\begingroup$ @Tanuj And $f(z)=(z-42)^{-1}$ is a rational function with a pole at $z=42$, but not in $z=\infty$. As I said: Poles can be anywhere. Compactness does not change anything about that. $\endgroup$ Jan 20, 2016 at 19:11
  • $\begingroup$ @Tanuj: That's not true. $z=\infty$ isn't a pole in the example I gave you and still the function is not bounded, because $z=42$ is a pole. This has nothing to do with Liouville's theorem. You seem to think that $\infty$ not being a pole is the same as not being in the image. That's not the case as you can clearly see in my example. The value $\infty$ is attained in the point $z=42$. $\endgroup$ Jan 21, 2016 at 19:16

1 Answer 1


To finally answer the question: This is a special case of the open mapping theorem for Riemannian surfaces. The theorem states that every holomorphic function $f:X\to Y$ between two Riemannian surfaces is an open mapping. If you apply it for $X=Y=\mathbb{C}\cup\{\infty\}$ you get that $f(G)$ is open in $Y$ for all open $G\subseteq X$. In particular the intersection $f(G)\cap\mathbb{C}$ is open because $\mathbb{C}$ is open in $Y$.

First note that openess is a local property: $f$ is open iff for all $x\in X$, $f$ maps (small enough) neighbourhoods of $x$ to neighbournoods of $f(x)$.

Now note that we can actually choose a small enough disc $U$ around $x$ such that $f(U)$ is contained in a disc $V$ around $f(x)$. Therefore we only have to show that holomorphic maps $\{z : |z|<r\} \to \{z : |z|<R\}$ that send $0$ to $0$ also map neighbourhoods of $0$ to neighbourhoods of $0$. Now this is either a consequence of the open mapping theorem for functions on the Gaussian plane if one has proven that previously or it is a relatively simple argument:

Write $f(z)=z^k \cdot g(z)$ for some $k\in\mathbb{N}_{\geq 1}$ and holomorphic $g$ with $g(0)\neq 0$.

First we show that we can reduce to the case $k=1$. By continuity $g(z)\neq 0$ for all $z$ in a sufficiently small neighbourhood of $0$. We can therefore write $g(z)=\exp(h(z))$ for some holomorphic function $h$. Now that means that $g$ has a holomorphic $k$th root: $g^{1/k}(z):=\exp(\frac{1}{k}h(z))$. Therefore we have $f(z)=(z\cdot g^{1/k}(z))^k$. Now the power map $w\mapsto w^k$ certainly maps open neighbourhoods of $0$ to open neighbourhoods of $0$. Therefore $f$ has this property if the inner function has it.

But in the case $k=1$ we have $f'(0) = 1\cdot g(0) + 0 \cdot g'(0)=g(0)\neq 0$ so that $f$ is locally a diffeomorphism around $0$ by virtue of the inverse function theorem. In particular it maps small enough open neighbourhoods of $0$ homeomorphically to open neighbourhoods of $0$. QED.


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