A *triangulation* of a convex polytope $P\subset\Bbb R^n$ is a partition of $P$ into $n$-simplices $\{\Delta\_1,...,\Delta\_m\}$ each of which has all its vertices among the vertices of $P$. A polytope may have many different triangulations.



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> **Question I:** do all combinatorially equivalent polytopes have the same triangulations?
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More precisely, let $P,Q\subset\Bbb R^n$ be combinatorially equivalent, $\phi:\mathcal F(P)\to\mathcal F(Q)$ be a face-lattice isomorphism, and $\{\Delta\_1,...,\Delta\_m\}$ a triangulation of $P$.
If $\Delta\_i$ has vertices $p\_1,...,p\_{n+1}\in\mathcal F\_0(P)$, then let $\phi(\Delta\_i)\subset Q$ be the simplex with vertices $\phi(p\_1),...,\phi(p\_{n+1})$. Do the simplices $\phi(\Delta\_1),...,\phi(\Delta\_m)$ form a triangulation of $Q$? That is, do they have disjoint interiors and cover all of $Q$?



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> **Question II:** if not, is there always a **universal triangulation**? That is, a special triangulation for $P$ that carries over to every combinatorially equivalent polytope in the way described above?
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Do combinatorially equivalent polytopes have the same triangulations?
Just to mark this question as answered, the comment by Tobias Fritz is spot on. In [this answer](https://math.stackexchange.com/a/2122958/79593) to a Math Stack Exchange question, Francisco Santos completely resolves your questions. On the one hand, the answer to Question I is no: combinatorially equivalent polytopes can have different triangulations. On the other hand, the answer to Question II is: yes, there is at least one triangulation they all share (the so-called "pulling triangulations").


