Riemann mapping theorem with pathological boundary












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From wikipedia: In complex analysis, the Riemann mapping theorem states that if $U$ is a non-empty simply connected open subset of the complex number plane $mathbb{C}$ which is not all of $mathbb{C}$, then there exists a biholomorphic mapping $f$ (i.e. a bijective holomorphic mapping whose inverse is also holomorphic) from $U$ onto the open unit disk.



Is this also true for the boundary is not smooth? Even the boundary is a Jordan curve? Because we have to transform holomorphically from that to a smooth boundary which for me is not very possible. Or there are some condition omitted in the statement?










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  • The wikipedia statement is true in all generality. Boundary smoothness has nothing to do with it.
    – zhw.
    Nov 30 at 23:54
















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From wikipedia: In complex analysis, the Riemann mapping theorem states that if $U$ is a non-empty simply connected open subset of the complex number plane $mathbb{C}$ which is not all of $mathbb{C}$, then there exists a biholomorphic mapping $f$ (i.e. a bijective holomorphic mapping whose inverse is also holomorphic) from $U$ onto the open unit disk.



Is this also true for the boundary is not smooth? Even the boundary is a Jordan curve? Because we have to transform holomorphically from that to a smooth boundary which for me is not very possible. Or there are some condition omitted in the statement?










share|cite|improve this question






















  • The wikipedia statement is true in all generality. Boundary smoothness has nothing to do with it.
    – zhw.
    Nov 30 at 23:54














0












0








0







From wikipedia: In complex analysis, the Riemann mapping theorem states that if $U$ is a non-empty simply connected open subset of the complex number plane $mathbb{C}$ which is not all of $mathbb{C}$, then there exists a biholomorphic mapping $f$ (i.e. a bijective holomorphic mapping whose inverse is also holomorphic) from $U$ onto the open unit disk.



Is this also true for the boundary is not smooth? Even the boundary is a Jordan curve? Because we have to transform holomorphically from that to a smooth boundary which for me is not very possible. Or there are some condition omitted in the statement?










share|cite|improve this question













From wikipedia: In complex analysis, the Riemann mapping theorem states that if $U$ is a non-empty simply connected open subset of the complex number plane $mathbb{C}$ which is not all of $mathbb{C}$, then there exists a biholomorphic mapping $f$ (i.e. a bijective holomorphic mapping whose inverse is also holomorphic) from $U$ onto the open unit disk.



Is this also true for the boundary is not smooth? Even the boundary is a Jordan curve? Because we have to transform holomorphically from that to a smooth boundary which for me is not very possible. Or there are some condition omitted in the statement?







complex-analysis riemann-surfaces






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asked Nov 30 at 23:26









CO2

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  • The wikipedia statement is true in all generality. Boundary smoothness has nothing to do with it.
    – zhw.
    Nov 30 at 23:54


















  • The wikipedia statement is true in all generality. Boundary smoothness has nothing to do with it.
    – zhw.
    Nov 30 at 23:54
















The wikipedia statement is true in all generality. Boundary smoothness has nothing to do with it.
– zhw.
Nov 30 at 23:54




The wikipedia statement is true in all generality. Boundary smoothness has nothing to do with it.
– zhw.
Nov 30 at 23:54










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It is true without any assumption on the boundary. It is another question whether the biholomorphic $h : U to U_1(0)$ extends to homeomorphism $bar{h} : overline{U} to overline{U_1(0)}$. See for example https://arxiv.org/pdf/1307.0439.pdf.






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    It is true without any assumption on the boundary. It is another question whether the biholomorphic $h : U to U_1(0)$ extends to homeomorphism $bar{h} : overline{U} to overline{U_1(0)}$. See for example https://arxiv.org/pdf/1307.0439.pdf.






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      It is true without any assumption on the boundary. It is another question whether the biholomorphic $h : U to U_1(0)$ extends to homeomorphism $bar{h} : overline{U} to overline{U_1(0)}$. See for example https://arxiv.org/pdf/1307.0439.pdf.






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        It is true without any assumption on the boundary. It is another question whether the biholomorphic $h : U to U_1(0)$ extends to homeomorphism $bar{h} : overline{U} to overline{U_1(0)}$. See for example https://arxiv.org/pdf/1307.0439.pdf.






        share|cite|improve this answer














        It is true without any assumption on the boundary. It is another question whether the biholomorphic $h : U to U_1(0)$ extends to homeomorphism $bar{h} : overline{U} to overline{U_1(0)}$. See for example https://arxiv.org/pdf/1307.0439.pdf.







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        answered Nov 30 at 23:59


























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