It's well-known that any compact polyhedron $P$ in $\mathbb{R}^n$ (we talk about piecewise-linear setting there, i.e. $P$ is a finite union of compact convex polytopes) can be triangulated into (geometric) simplices, although sometimes it is necessary to add "extra points" in $P$ to serve as vertices of simplices in the triangulation $\mathcal{T}$. E.g.  [Schönhardt polyhedron](http://en.wikipedia.org/wiki/Sch%C3%B6nhardt_polyhedron) requires such extra points. (Here By $\mathcal{T}$ we mean a partition of $P$ into finitely many simplices $T\in\mathcal{T}$ --- more precisely, the interiors $int(T)$ of $T$'s do not intersect, and the closure of $\cup\_{T\in\mathcal{T}}int(T)$ equals $P$.
The vertices of $T$'s that are not vertices of $P$ are these "extra points" we talk about.)


It looks correct that one can always construct such a $\mathcal{T}$ so that each extra point in it is "non-rigid", i.e. it can be continuously moved inside an open subset of the face of minimal dimension it is inserted into, so that after such a deformation $\mathcal{T}$ remains a triangulation of $P$.
Is this indeed correct, and can anyone point out a reference? 


**Added:** A *weaker* form of the question: show that each extra poing in $\mathcal{T} $ is not prescribed, i.e. for any vertex $y$ of $\mathcal{T}$ which is not a vertex of $P$ there exists another triangulation of $P$ which does not have $y$ as a vertex. [This still suffices for our purpose, of showing that a part of certain kind of moment generating functions, for moments of a uniform measure supported on $P$, does not depend upon $\mathcal{T}$ ]. 
This is easy to see that $y$ lying in the interior of $P$ is not prescribed---one can directly construct a new triangulation not using $y$, by choosing the points of intersection of the edges on $y$ with a sufficiently small sphere around $y$ and re-triangulating new convex pieces without using $y$. But it is not obvious for $y$ lying in a proper face of $P$. 



non-rigidity of interior points in polyhedral triangulations?
The answer for arbitrary polyhedra is no. If a 4-dimensional polyhedron has a 3-dimensional Schönhardt polyhedron as one of its faces, there will need to be a new vertex added somewhere within that face, which will not be free to move in an open set.


I believe that the answer is yes in 3d and yes to higher-dimensional polyhedra all of whose faces are already simplices, but I don't have a proof handy.


