Elementary proposition on triangulations I have a question on triangulations.
Let T be a triangulation of a d-dimensional cross-polytope.
Let s be a (d-1)simplex that does not lie on the boundary of the cross-polytope.
 How can we show that s is a face of exactly two (d)simplices ? Am I wrong with the hypothesis ?
Thanks in advance.

 A: That is correct. This is one of those statements that is easy to "see" but tricky to write down a proof. But I tried anyway (maybe someone can give a better explanation):
Let $p$ be a point in the interior of a (d-1)-simplex (Say $S$). Then since $p$ does not lie on the boundary of the cross polytope (say $P$) there is a neighbourhood of $p$ in $P$ homeomorphic to $\mathbb{R}^{d}$, and $S \subset P$ is (locally) homoemorphic to $\mathbb{R}^{d-1} \subset \mathbb{R}^{d}$. Note that $ \mathbb{R}^{d} \setminus \mathbb{R}^{d-1}$ has two connected components, i.e. $S$ has two "sides " in P. So any $d$-simplex containing $S$ will locally look be equal to one of these two connected components, hence there at most two of them. 
A: Thank you for your answer, I get the idea.
I came up with a more elementary proof that does not use the notion of neighborhoods.
Let us show the following result : a (d-1)-simplex S that lies on the boundary of a d dimensional polytope P is a face of exactly one d-simplex.
First notice that the reunion of the d-simplices is equal to P. Indeed the reunion of all simplices is equal to P, and a "void" in the reunion of the d-simplices can't be filled with the finite reunion of simplices of dimension lower than d-1. So we can always find a d-simplex that contains any point.
If d=1, then S={x}, and we can find a d-simplex R that contains x. Moreover, if another d-simplex R' contains x, since x is on the boundary, then the intersection of R and R' is of dimension d, then $R' \cap R$ is a face of R so it is R so R'=R.
If d>=1, then the interior of S is not void, and let x a point from the interior. Again we can find a d-simplex R that contains x. Since the intersection S and R is a face of S containing an interior point, it is S, so S is a face of R. If another d-simplex R' contains S, then since it is on the boundary, $R' \cap R$ is of dimension d, and it is a face of R so it is R.
Now by looking at the half spaces (say E1,E2) induced by S (which is a hyperplane), we find the two d-simplices in $E1 \cap P$ and $E2 \cap P$.
Is this proof valid and is it efficient ? Do I implicitely use the notion of neighborhoods when I say that $R' \cap R$ is of dimension d ?
Thank you.
