Is there a holomorphic and bijective function of the unit disc onto the complex plane?
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Is there a holomorphic and bijective function between the open unit ball of $mathbbC$ and $mathbbC$?
The usual homeomorphisms $psi(z):=fracz$ and his composition with the conjugate map are not holomorphic on $mathbbC$.
complex-analysis holomorphic-functions
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up vote
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Is there a holomorphic and bijective function between the open unit ball of $mathbbC$ and $mathbbC$?
The usual homeomorphisms $psi(z):=fracz$ and his composition with the conjugate map are not holomorphic on $mathbbC$.
complex-analysis holomorphic-functions
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is there a holomorphic and bijective function between the open unit ball of $mathbbC$ and $mathbbC$?
The usual homeomorphisms $psi(z):=fracz$ and his composition with the conjugate map are not holomorphic on $mathbbC$.
complex-analysis holomorphic-functions
Is there a holomorphic and bijective function between the open unit ball of $mathbbC$ and $mathbbC$?
The usual homeomorphisms $psi(z):=fracz$ and his composition with the conjugate map are not holomorphic on $mathbbC$.
complex-analysis holomorphic-functions
complex-analysis holomorphic-functions
edited 16 mins ago


Giuseppe Negro
16.5k328119
16.5k328119
asked 18 mins ago
Federico Fallucca
1,42618
1,42618
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add a comment |Â
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No, because its inverse would be a holomorphic map from $mathbb C$ to the open unit disk on $mathbb C$. Therefore, it would be bounded and then, by Liouville's theorem, it would be a constant map.
Sorry. I asked a stupid question.
– Federico Fallucca
7 mins ago
I don't think so.
– José Carlos Santos
4 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
No, because its inverse would be a holomorphic map from $mathbb C$ to the open unit disk on $mathbb C$. Therefore, it would be bounded and then, by Liouville's theorem, it would be a constant map.
Sorry. I asked a stupid question.
– Federico Fallucca
7 mins ago
I don't think so.
– José Carlos Santos
4 mins ago
add a comment |Â
up vote
4
down vote
No, because its inverse would be a holomorphic map from $mathbb C$ to the open unit disk on $mathbb C$. Therefore, it would be bounded and then, by Liouville's theorem, it would be a constant map.
Sorry. I asked a stupid question.
– Federico Fallucca
7 mins ago
I don't think so.
– José Carlos Santos
4 mins ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
No, because its inverse would be a holomorphic map from $mathbb C$ to the open unit disk on $mathbb C$. Therefore, it would be bounded and then, by Liouville's theorem, it would be a constant map.
No, because its inverse would be a holomorphic map from $mathbb C$ to the open unit disk on $mathbb C$. Therefore, it would be bounded and then, by Liouville's theorem, it would be a constant map.
answered 16 mins ago


José Carlos Santos
125k17101188
125k17101188
Sorry. I asked a stupid question.
– Federico Fallucca
7 mins ago
I don't think so.
– José Carlos Santos
4 mins ago
add a comment |Â
Sorry. I asked a stupid question.
– Federico Fallucca
7 mins ago
I don't think so.
– José Carlos Santos
4 mins ago
Sorry. I asked a stupid question.
– Federico Fallucca
7 mins ago
Sorry. I asked a stupid question.
– Federico Fallucca
7 mins ago
I don't think so.
– José Carlos Santos
4 mins ago
I don't think so.
– José Carlos Santos
4 mins ago
add a comment |Â
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