most recent 30 from http://mathoverflow.net 2013-05-23T11:36:13Z http://mathoverflow.net/feeds/question/11182 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/11182/riemann-mapping-for-doubly-connected-regions Anweshi 2010-01-08T20:33:01Z 2010-01-08T20:56:05Z

<p>Remove the closure of simply connected region from the interior of a simply connected region. Is it true that the resulting domain can be mapped conformally to some annulus?</p>

http://mathoverflow.net/questions/11182/riemann-mapping-for-doubly-connected-regions/11185#11185 Mariano Suárez-Alvarez 2010-01-08T20:47:31Z 2010-01-08T20:47:31Z

<p>The answer is yes. This is a special case of theorem 10 in Ahlfors' <em>Complex Analysis</em>, section 5, chapter 6. (Special in that the theorem more generally says that if the complement of the domain has $n$ connected components not reduced to points in the extended plane, then the domain is equivalent to an annulus from which $n-2$ concentric slits have been removed. In your case $n=2$.)</p>

http://mathoverflow.net/questions/11182/riemann-mapping-for-doubly-connected-regions/11186#11186 S. Carnahan 2010-01-08T20:48:20Z 2010-01-08T20:48:20Z

<p><a href="http://en.wikipedia.org/wiki/Riemann%5Fmapping%5Ftheorem#Why%5Fis%5Fthis%5Ftheorem%5Fimpressive.3F" rel="nofollow">See Wikipedia</a>. The third entry of the list gives an affirmative answer.</p>

http://mathoverflow.net/questions/11182/riemann-mapping-for-doubly-connected-regions/11187#11187 Harald Hanche-Olsen 2010-01-08T20:56:05Z 2010-01-08T20:56:05Z

<p>No. The resulting set need not even be connected. And if it is, it need not be doubly connected, as the interior region may have boundary points in common with the original region. Aside from these crude objections, however, the answers of Mariano and Scott are OK.</p>