# A question on PL-topology and polytopal complex

Question : $C$ is a pure, full-dimensional polytopal complex(a special case of a regular cell complex) in $\mathbb{R}^d$. I know that the boundary of the underlying set is a PL-sphere. Is it true that $C$ is a PL-ball?

Definitions for this question:

1. Polytopal complex is a finite nonempty collection of convex polytopes in $\mathbb{R}^d$ that contains all faces of its polytopes, and such that the intersection of two of its polytopes is a face of each of them.
2. Dimension of the complex is the largest dimension of a polytope in the complex.
3. A complex $C$ is pure if each of its faces is contained in a face of dimension $dim(C)$.
4. Underlying set is the union of its faces.
5. PL stands for piecewise-linear
6. PL-k-ball is something that is PL-homemorphic to a simplex of $k$ dimensions.
7. PL-(k-1)-sphere is something that is PL-homeomorphic to a boundary of a simplex of $k$ dimension.

(If the answer is yes for the above question) Further Question : Now we look at a non-full dimensional pure polytopal complex $C$. We define the "boundary" of $C$ to be the set of points where there does not exist a neighborhood that is PL-homemorphic to $\mathbb{R}^{dim(C)}$. Then if the "boundary" of $C$ is a PL-(dim(C)-1)-sphere, is $C$ a PL-dim(C)-ball?

P.S. Please give a comment if some parts of the question is not suitably defined. Also, I have no idea how hard this is, so please comment if you know it is obvious or is really hard.

9/19 : Edited so it includes the dimension restriction.

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First question. Your question is equivalent to the Schoenflies problem: does a locally flat PL $(d-1)$-sphere in $S^d$ (and hence in $\mathbb R^d$) bound a PL-disc?
The answer is "yes" for $d \neq 4$; for a proof when $d > 5$ see the book Rourke C.P., Sanderson B.J. Introduction to piecewise-linear topology (Springer, 1972). It is a direct consequence of the Cobordism theorem and the Generalized Poincaré Conjecture in the PL setting. The answer in dimension $d=4$ is however unknown.
Second question. The answer here is "no": take a triangulated surface in $\mathbb R^3$ with one boundary component which is not a disc (for instance, a holed torus). Note also that one can choose a non-manifold example: for instance, a sphere with a disc attached to its equator.
Thank you for the answer. 1. I need a bit of help with the answer for the second question: "The triangulated surface in $\mathbb{R}^3$ with one boundary component which is not a disc". Since the complexes I'm looking at are compact, I guess this would look something like a two-dimensional doghnut in $\mathbb{R}^3$. Then doesn't the boundary consist of the inner/outer circles and is not connected? 2. I want to try to modify the second question to make it work, so I hope you won't mind me not accepting the answer yet. –  Suho Oh Sep 19 '10 at 13:43
By a "holed torus" I mean a standard torus surface in $\mathbb R^3$ (a 2-dimensional object) with one small open disc removed: it is a compact surface with (1-dimensional) connected boundary. Another example of compact surface in 3-space with connected boundary which is not a disc is a Mobius strip. In fact, every compact surface with boundary embeds in 3-space. –  Bruno Martelli Sep 19 '10 at 19:01