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Let $T$ be finite set of tetrahedrons in $\mathbb{R}^3$. Let $T$ be tetrahedral complex in a sense that if two tetrahedrons intersect, the intersection is a face of both. Let $\partial T$ consist of those 2-simplices, which are faces of exactly one tetrahedron from $T$. In what follows we view $\partial T$ as 2-dimensional simplicial complex. Suppose $\partial T$ satisfies the following conditions:

A. Each edge (1-simplex) in $\partial T$ is a face of exactly two 2-simplices from $\partial T$.

B. $\partial T$ is ``edge-connected''. That is, dual graph of $\partial T$ (where 2-simplices play the role of vertices and two vertices are connected iff the corresponding 2-simplices share 1-simplex) has one connected component.

C. For each vertex (0-simplex) $v$ in $\partial T$ the 'neighborhood' (all 2-simplices from $\partial T$ that have $v$ as their face) is 'edge-connected' in a sense defined above.

D. The Euler number of $\partial T$ is that of 2-sphere.

Let $|\partial T|$ be the underlying topological space. Given all the conditions outlined above, does $|\partial T|$ have to be homeomorphic to a 2-sphere? If not, what would be a counter-example? If yes, how would the proof go and what do I need to reference?

The motivation for the question is as follows: I need to learn or develop fast algorithm to determine if a given set of tetrahedrons spans a domain homeomorphic to a ball.

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Have you looked at the classification of 2-manifolds? –  Ryan Budney Aug 27 '12 at 6:38
Yes I looked at that. My understanding is that the classification suggests conditions B and D above (but conditions A and C are also necessary). –  user30061 Aug 30 '12 at 20:13

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