- Is the unit disc \( \displaystyle \mathbb{D} = \left\{ z \in \mathbb{C} \, \Big| \, |z| < 1 \right\} \) biholomorphic to \( \mathbb{C} \)?
- Is the punctured unit disc \( \displaystyle \mathbb{D}^{*} = \left\{ z \in \mathbb{C} \, \Big| \, 0< |z| < 1 \right\} \) biholomorphic to \( \mathbb{C} \smallsetminus \left\{0\right\} \)?
Are These Sets Biholomorphic?
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Are These Sets Biholomorphic?
Last edited by Tsakanikas Nickos on Mon Nov 23, 2015 1:54 am, edited 1 time in total.
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Re: Are These Sets Biholomorphic?
1.Answer
The unit disc \(\displaystyle{D=\left\{z\in\mathbb{C}: \left|z\right|<1\right\}}\) is not biholomorphic
to \(\displaystyle{\mathbb{C}}\) .
Indeed, suppose that the unit disc \(\displaystyle{D}\) is biholomorphic to \(\displaystyle{\mathbb{C}}\) .
Then, there exists a holomorphic function \(\displaystyle{f:\mathbb{C}\longrightarrow D}\) which is
one to one, onto \(\displaystyle{D}\) and the function \(\displaystyle{f^{-1}}\) is biholomorphic.
Since \(\displaystyle{D}\) is a bounded subset of \(\displaystyle{\mathbb{C}}\), we have that the
function \(\displaystyle{f}\) is bounded. According to \(\displaystyle{\rm{Liouville's}}\) theorem,
the function \(\displaystyle{f}\) is constant, a contradiction since it is one to one.
On the other hand, the disc \(\displaystyle{D}\) is homeomorphic to \(\displaystyle{C}\) since the
function \(\displaystyle{g:\mathbb{C}\longrightarrow D\,,g(z)=\dfrac{z}{1+\left|z\right|}}\)
is one to one, onto \(\displaystyle{D}\), it is continuous, and the inverse function
\(\displaystyle{g^{-1}:D\longrightarrow \mathbb{C}\,,g^{-1}(z)=\dfrac{z}{1-|z|}}\) is continuous.
The unit disc \(\displaystyle{D=\left\{z\in\mathbb{C}: \left|z\right|<1\right\}}\) is not biholomorphic
to \(\displaystyle{\mathbb{C}}\) .
Indeed, suppose that the unit disc \(\displaystyle{D}\) is biholomorphic to \(\displaystyle{\mathbb{C}}\) .
Then, there exists a holomorphic function \(\displaystyle{f:\mathbb{C}\longrightarrow D}\) which is
one to one, onto \(\displaystyle{D}\) and the function \(\displaystyle{f^{-1}}\) is biholomorphic.
Since \(\displaystyle{D}\) is a bounded subset of \(\displaystyle{\mathbb{C}}\), we have that the
function \(\displaystyle{f}\) is bounded. According to \(\displaystyle{\rm{Liouville's}}\) theorem,
the function \(\displaystyle{f}\) is constant, a contradiction since it is one to one.
On the other hand, the disc \(\displaystyle{D}\) is homeomorphic to \(\displaystyle{C}\) since the
function \(\displaystyle{g:\mathbb{C}\longrightarrow D\,,g(z)=\dfrac{z}{1+\left|z\right|}}\)
is one to one, onto \(\displaystyle{D}\), it is continuous, and the inverse function
\(\displaystyle{g^{-1}:D\longrightarrow \mathbb{C}\,,g^{-1}(z)=\dfrac{z}{1-|z|}}\) is continuous.
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Re: Are These Sets Biholomorphic?
Suppose that \( \displaystyle f : \mathbb{C} \smallsetminus \{ 0 \} \longrightarrow \mathbb{D}^{*} \) is a biholomorphic map. Observe that \( \displaystyle f \) is a holomorphic function (into \( \mathbb{C} \)), bounded on a neighborhood of \( 0 \) - which is a singular point of \( \displaystyle f \). By Riemann's Theorem on Removable Singularities, we conclude that \( 0 \) is a removable singularity of \( \displaystyle f \), which means that \( \displaystyle f \) can be defined at \( 0 \) in such a way that \( \displaystyle f \) becomes holomorphic on all of \( \mathbb{C} \). But now \( \displaystyle f : \mathbb{C} \longrightarrow \mathbb{D}^{*} \) is a bounded entire function, so \( \displaystyle f \) is constant by Liouville's Theorem. However, this is a contradiction to the fact that \( \displaystyle f \) is injective. Hence, the given sets are not biholomorphic.
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