Does tetrahedral complex in R^3 with "2-spherical" boundary have to be a 3d ball? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T02:34:48Z http://mathoverflow.net/feeds/question/95287 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95287/does-tetrahedral-complex-in-r3-with-2-spherical-boundary-have-to-be-a-3d-ball Does tetrahedral complex in R^3 with "2-spherical" boundary have to be a 3d ball? IL. 2012-04-26T19:12:44Z 2012-04-26T19:12:44Z <p>Let $S$ be a finite set of tetrahedrons in $\mathbb{R}^3$. Let $S$ be tetrahedral complex in a sense that if two tetrahedrons intersect, the intersection is a face of both. In what follows we view tetrahedrons and their faces as <em>closed</em> sets in $\mathbb{R}^3$. Let the underlying topological space $|S|$ (domain occupied by all tetrahedrons with the topology inherited from $\mathbb{R}^3$) be manifold with boundary.</p> <p>Let $T$ be arbitrary subset of tetrahedrons from $S$. Let $\partial T$ consist of those 2-simplices, which are faces of exactly one tetrahedron from $T$. Let $|\partial T|$ be the underlying topological space. </p> <p>Suppose $|\partial T|$ is homeomorphic to a 2-sphere. Does it follow that $T$ is homeomorphic to a 3d ball? If not, what would be a counter-example?</p>