On embedding of a triangulation

Question

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

Answer 1

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

"Brehm, Ulrich, and Karanbir S. Sarkaria. "Linear vs. piecewise-linear embeddability of simplicial complexes." MPI-[Ber.]/Max-Planck-Inst. f@: ur Mathematik; 92-52 (1992)."

Answer 2

I think what you are looking for is the Simplicial Approximation Theorem:

If $K$ is a finite simplicial complex and $L$ is an arbitrary simplicial complex, then any map $f :K\to L$ is homotopic to a map that is simplicial with respect to some iterated barycentric subdivision of $K$. You can find the proof in Hatcher Algebraic Topology book.

The proof makes it clear that the simplicial approximation $g$ can be chosen not just homotopic to $f$ but also close to $f$ (If you Choose a metric on $K$ that restricts to the standard Euclidean metric on each simplex of $K$) if we allow subdivisions of $L$ as well as $K$.