Let $X$ be a triangulable topological space, i.e, a topological space which is homeomorphic with a finite simplicial complex. Suppose $X$ can be embedded in $\mathbb{R}^d$ for some $d$. Is it possible to find a simplicial complex inside $\mathbb{R}^d$ that triangulates $X$?
As a motivation for this question: Fáry's theorem states that any simple planar graph can be drawn without crossings so that its edges are straight line segments. That is, the ability to draw graph edges as curves instead of as straight line segments does not allow a larger class of graphs to be drawn
Best Answer
As it was mentioned in the question any planar graph has a planar drawing where all edges are straight segments. But, a higher-dimensional analog of this assertion fails! Brehm and Sarkaria showed that for every $d\geq 2$ and every $k$, $d+1\leq k\leq 2d$, there exist a finite $d$-dimensional simplicial complexes $K$ that can be embedded in $\mathbb{R}^k$ but not linearly. For more information, please see