The complement of a simple closed curve in the Riemann sphere has two connected components (Jordan). Schoenflies theorem implies that each of these components is homeomorphic to a disk and hence each has trivial fundamental group. Is there a direct (and simple) way to see that each component has trivial fundamental group without invoking Schoenflies?
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1$\begingroup$ The Wikipedia article lists four "elementary proofs." Have you looked at these? en.wikipedia.org/wiki/Schoenflies_problem $\endgroup$– Christian RemlingCommented Sep 3, 2016 at 19:12
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$\begingroup$ @Christian Remming: Thanks for your question. "Elementary" is not the same as "direct" or "simple". I have read the Thomassen proof and have looked at the Moise and Bing proofs.They are elementary but I wouldn't call them "simple". In fact, this question was motivated by a desire to flesh out the proof that uses the Carath\'eodory extension theorem. The missing link for me is between the output of the Jordan curve theorem and the input of the Riemann mapping theorem. $\endgroup$– Chris JudgeCommented Sep 3, 2016 at 21:57
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$\begingroup$ (continued) I think that I now know how to bridge that gap, but it's not simple: Jordan-Brouwer implies that first homology of a connected component is trivial and then since each curve is null-homologous one can construct a complex logarithm function. Then the usual proof (e.g. Ahlfors) of the Riemann mapping goes through. $\endgroup$– Chris JudgeCommented Sep 3, 2016 at 21:57
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$\begingroup$ Chris, you are essentially using Schoenflies in your proof, as you are giving a pretty standard proof of the Schoenflies theorem with that argument. $\endgroup$– Ryan BudneyCommented Sep 4, 2016 at 3:57
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1$\begingroup$ If you'll grant that Schoenflies is easy for a piecewise linear Jordan curve, try this: Show that each component of the complement of any compact connected set in the plane has trivial fundamental group. Do this by reducing to the case when the compact set is a piecewise linear $2$-manifold. (Any given loop in the complement of the compact set is in the complement of some neighborhood, therefore in the complement of some PL $2$-manifold neighborhood.) $\endgroup$– Tom GoodwillieCommented Sep 4, 2016 at 12:48
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