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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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  • $\begingroup$ The wikipedia statement is true in all generality. Boundary smoothness has nothing to do with it. $\endgroup$ – zhw. Nov 30 '18 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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