Jordan curve theorem for cylinders - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:32:32Z http://mathoverflow.net/feeds/question/100555 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100555/jordan-curve-theorem-for-cylinders Jordan curve theorem for cylinders B Leraut 2012-06-25T00:24:24Z 2012-08-04T13:14:28Z <p>Hello, I would like to know if the following result is true: </p> <p>Let $A,B$ be two embedded circles in $S^2$ which do not intersect and let $C$ be the $\textit{closed}$ region bounded by $A$ and $B$ (this region is well-defined by the Jordan Curve Theorem). Then $C$ is homeomorphic to $S^1\times [0,1]$.</p> <p>If so, is there an easy proof, or does this result have a name?</p> http://mathoverflow.net/questions/100555/jordan-curve-theorem-for-cylinders/100556#100556 Answer by tweetie-bird for Jordan curve theorem for cylinders tweetie-bird 2012-06-25T00:32:37Z 2012-06-25T00:57:49Z <p>This is the annulus theorem, and is indeed true for circles in $S^2$ without need for further hypotheses (which are needed in higher dimensions). </p> <p>See <a href="http://en.wikipedia.org/wiki/Annulus_theorem" rel="nofollow">en.wikipedia.org/wiki/Annulus_theorem</a></p> http://mathoverflow.net/questions/100555/jordan-curve-theorem-for-cylinders/103922#103922 Answer by Alexandre Eremenko for Jordan curve theorem for cylinders Alexandre Eremenko 2012-08-04T08:49:37Z 2012-08-04T08:49:37Z <p>A more general result is true. Let F and G be two disjoint connected compact subsets of the sphere. Let D be the component of the complement whose boundary intersects both. Then D is conformally equivalent to a standard ring (or a punctured disc, or the punctured plane). This theorem can be easily derived from the Riemann mapping theorem in the simply connected case. See, for example the book by Goluzin, Geometric theory of functions of complex variable. </p> <p>In your case, then F and G are Jordan curves, one also needs to extend this homeomorphism to the boundary. This can be done with the same technique as for the Riemann case. The boundary extension is a local question.</p> <p>Of course, a purely topological proof, not using complex variables must also exist.</p> http://mathoverflow.net/questions/100555/jordan-curve-theorem-for-cylinders/103943#103943 Answer by Lee Mosher for Jordan curve theorem for cylinders Lee Mosher 2012-08-04T13:14:28Z 2012-08-04T13:14:28Z <p>If you are willing to quote the <a href="http://en.wikipedia.org/wiki/Jordan%25E2%2580%2593Sch%25C3%25B6nflies_theorem%20%22Sch%5C%5C%22onflies%20Theorem%22" rel="nofollow">Schoenflies</a> theorem and the classification of surfaces, a quick proof of this result is a standard exercise. First, using algebraic topology (as in the proof of the Jordan curve theorem itself), one shows that $S^1 - (A \cup B)$ is a union of three components, whose closures are three compact subsets $S_A$, $S_{AB}$, $S_B$ with frontiers $A$, $A \cup B$, $B$ respectively. Applying the Schoenflies theorem, each of $S_A$, $S_{AB}$, $S_B$ is a compact surface with boundary, and in fact $S_A$, $S_B$ are homeomorphic to a disc. Next one has $\chi(S_A) + \chi(S_{AB}) + \chi(S_B) = \chi(S^2)$, which is a standard result about gluing together a bunch of compact surfaces-with-boundary by pairwise identifying boundary circles. So that equation becomes $1 + \chi(S_{AB}) + 1 = 2 \implies \chi(S_{AB})=0$. Using the classification of surfaces, and ruling out all compact surfaces of Euler characteristic zero with other than two boundary components, $S_{AB}$ is an annulus. </p> <p>The same technique carried out in general allows one to take any compact surface $S$ with boundary (specified by its Euler characteristic, orientability, and number of boundary components), any integer $n$, and any subset $\mathcal{C} \subset S$ consisting of $n$ disjointly embedded circles, and enumerate all of the finitely many possibilities for the topological types of the components of $S-\mathcal{C}$.</p>