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Like in Riemann's mapping theorem, we have a conformal mapping $f:\Omega\rightarrow\mathbb{D}$ (so $f$ is bijective and holomorphic), where $\mathbb{D}=\{|z|<1,\ z \in\mathbb{C}\}$ is the set of the open unit disks.

Why does it follow from Liouville's theorem, that $\Omega$ is not the whole $\mathbb{C}$, so $\Omega \neq \mathbb{C}$?

Liouville's theorem states that if $f$ is holomorphic on $\mathbb{C}$ and bounded, the $f$ is constant.

I really don't see how this determines $\Omega \neq \mathbb{C}$.

It would be so nice, if someone could walk me through this.

Thank you for your time,


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Well, $\mathbb{D}$ is bounded. If the image of $f$ is in there, $f$ is bounded.... – user20266 Jun 17 '12 at 14:21
up vote 2 down vote accepted

If $\Omega=\Bbb C$, since $f$ gets it's values in the unit disk, it's a bounded function. So it can't be bijective.

Note that Liouville theorem doesn't work if the open set is not $\Bbb C$. For example, when $\Omega$ is missing a ball $B(z_0,r)$, the map $f(z):=\frac 1{z-z_0}$ is bounded, holomorphic but not constant.

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Ahh, the bijective property does not uphold, I see. – Chris Jun 17 '12 at 14:32
You probably mean $\Omega$ is missing a ball $B(z_0,r)$. – Fan Zheng Dec 14 '14 at 3:15
@FanZheng You are right. Fixed now. – Davide Giraudo Dec 14 '14 at 10:26

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