Let $U$ be a simply connected open set in $\mathbb{R}^2$. Is it true that $U$ is homeomorphic to an open ball?
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4Yes. This follows from the Riemann mapping theorem. – Mark Aug 02 '11 at 11:27
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I'm asking more general question. – J-Holomorph Aug 02 '11 at 11:45
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You have changed the question after correct answers have been posted. I think it'd be better to ask a separate question for general $n$. – lhf Aug 02 '11 at 11:47
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Since you now re-asked your follow-up question I rolled back to the previous version. – t.b. Aug 02 '11 at 11:58
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3BUT, isn't the fact in the question MUCH EASIER TO PROVE than the Riemann mapping theorem? (A snipe: is the empty set simply connected?) – GEdgar Aug 02 '11 at 14:39
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4@GEdgar: Proving this without the Riemann mapping theorem was the subject of this MathOverflow question. – Jonas Meyer Aug 02 '11 at 16:26
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see also Pedro's answer https://math.stackexchange.com/questions/1064066/is-every-path-connected-open-subset-of-mathbbr2-homeomorphic-to-mathbbr/1064068#1064068 – D.R. Aug 06 '23 at 20:51
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Yes, this is the Riemann mapping theorem. You get much more than a homeomorphism: you get a biholomorphic map.
lhf
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Yes. In fact, more can be said... The Riemann Mapping Theorem states that the homeomorphism can be taken to be biholomorphic (as a complex map), if $U \neq \mathbb{C}$. See this link for a much more detailed treatment and proof.
Hope this helps!
Shaun Ault
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According to the Riemann mapping theorem that's true iff U is a simply connected nonempty open set in $\mathbb{R}^2$ which is a strict subset. That is, $U\subsetneq \mathbb{R}^2$.
user864940
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