Timeline for Why is a component of complement of Jordan curve 1-connected w/o Schoenflies?
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14 events
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| Sep 7, 2016 at 16:58 | comment | added | Chris Judge | @Misha. Yes, but not really simple or direct.For example, on mathoverflow.net/questions/18454/… one learns of techinical combinatorial approaches including triangulations. Existence of triangulations is my ultimate aim...In the end, I chose to define the complex logarithm in the proof of surjectivity of Riemann mapping theorem via integrating piecewise vertical/horizontal curves and using closedness of the form $(f'(z)/f(z))dz$. Integrating over any piecewise vertical/horizontal loop in the complementary region gives zero. | |
| Sep 5, 2016 at 15:23 | comment | added | Misha | Actually, this follows directly from Alexander duality that each component of the complement is acylic. Now, use the fact that each 2-dimensional manifold is homotopy-equivalent to a graph and hence has free fundamental group, see mathoverflow.net/questions/18454/…. | |
| Sep 4, 2016 at 13:32 | comment | added | Chris Judge | @ Tom. Thanks! That seems like a good way to go as I agree that PL-Schoenflies is not difficult. | |
| Sep 4, 2016 at 13:28 | comment | added | Chris Judge | @ Ryan Budney. Yes, I am trying to flesh out the "pretty standard" proof of Jordan-Schoenflies using Carath\'eodory's theorem. A reference to complete details would be appreciated. | |
| Sep 4, 2016 at 12:48 | comment | added | Tom Goodwillie | 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.) | |
| Sep 4, 2016 at 6:23 | comment | added | Włodzimierz Holsztyński | Oops, here is a better link: ams.org/journals/proc/1971-027-03/S0002-9939-1971-0271949-X/… | |
| Sep 4, 2016 at 5:31 | comment | added | Włodzimierz Holsztyński | Perhaps Borsuk's separation theorem + jstor.org/stable/2036506?seq=1#page_scan_tab_contents should lead to the required proof. The two mentioned starter theorems are simple/easy when compared to Schoenflies. I hope that you give it a try. | |
| Sep 4, 2016 at 3:57 | comment | added | Ryan Budney | Chris, you are essentially using Schoenflies in your proof, as you are giving a pretty standard proof of the Schoenflies theorem with that argument. | |
| Sep 4, 2016 at 3:07 | history | edited | Michael Albanese | CC BY-SA 3.0 |
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| Sep 3, 2016 at 21:57 | comment | added | Chris Judge | (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. | |
| Sep 3, 2016 at 21:57 | comment | added | Chris Judge | @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. | |
| Sep 3, 2016 at 19:12 | comment | added | Christian Remling | The Wikipedia article lists four "elementary proofs." Have you looked at these? en.wikipedia.org/wiki/Schoenflies_problem | |
| Sep 3, 2016 at 14:19 | history | edited | Chris Judge | CC BY-SA 3.0 |
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| Sep 3, 2016 at 13:18 | history | asked | Chris Judge | CC BY-SA 3.0 |