Minimum valence of a vertex in a simple triangular mesh with no boundaries Suppose M is a simple (manifold, orientable and connected) triangular mesh with no boundaries (in other words, a mesh homeomorphic to a sphere).
I have the conjecture that the following statement is true:

The valence (the number of connecting edges) of
  each vertex is greater or equal than 3.

My current definition of manifoldness is that each edge is incident to only one or two faces and the faces incident to a vertex form a closed or an open fan.
Valence is a natural number by definition, so:


*

*Valence = 0 breaks connectivity

*Valence = 1 breaks manifoldness because the only edge associated to the vertex would not be able to be incident to any face.


However I am having problems proving that Valence = 2 should break the manifoldness condition.
Can someone help me to better understand those conditions so I can prove if the statement is either false or true?

Okay. It took me the entire day to realize. It probably deserves a more formal proof, but at least it is an argument:
If a vertex has valence 2, the 2 neighboring vertices must have 2 edges connecting the same 2 vertices. This would make the edges in the star of v boundaries, as they would be homeomorphic to the closed upper half-plane.
So, in this case, it is guaranteed that each vertex in a 2-manifold mesh with no boundaries has valence greater or equal to 3.

