By Liouville's theorem, there is no non-constant holomorphic function from the complex plane to the unit disc. I wonder what the converse is like--surely there are holomorphic functions on the open unit disc into the complex plane, but are there any bijective ones?
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2I don't understand the question. You know that there are no holomorphic bijections from $\mathbb C$ onto a disk, but you ask if there are holomorphic bijections from a disk onto $\mathbb C$? – Feb 08 '13 at 01:23
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4@5PM The OP just forgot that holomorphic + bijective implies that the inverse is automatically holomorphic. – Julien Feb 08 '13 at 01:25
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A bijective holomorphic function has a holomorphic inverse, so certainly there are no bijective holomorphic functions from $\mathbb D$ to $\mathbb C$.
For an example of a function form $\mathbb D$ to $\mathbb C$ that's unbounded, consider $\frac{1}{1-x}$ or $e^{\frac{1}{1-x}}$.
JSchlather
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More generally: there is a holomorphic bijection from the open unit disk onto a region $U$ if and only if $U$ is simply connected and the complement of $U$ (in the Riemann sphere) has at least two points.
GEdgar
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For reference this is The Riemann Mapping theorem which may be stated as $U\subseteq\mathbb{C}$ is open and simply connected implies that $U$ is biholomorphic to the unit disc. – Alex Youcis Feb 08 '13 at 01:57
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