# Does tetrahedral complex in R^3 with “2-spherical” boundary have to be a 3d ball?

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 closed 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.

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.

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?

• Yes, this is Alexander's theorem. This answer was given as part of an answer to your previous question as well. – Ryan Budney Apr 26 '12 at 20:18
• I am voting to close as "spam". – Igor Rivin Apr 26 '12 at 20:37
• In that answer there was additional requirement that $|T|$ is a manifold with boundary. I asked in comments if this requirement could be dropped, and somebody answered "no" without any explanation. – IL. Apr 26 '12 at 20:44
• I would have thought that if a collection of tetrahedra in $\mathbb R^3$ has a manifold boundary, then it is itself a manifold with boundary, since all non-manifold points would have to involve the boundary. – Jim Conant Apr 26 '12 at 21:37
• Jim, yes, that's why I asked if the requirement on $|T|$ could be dropped. – IL. Apr 26 '12 at 21:46