Triangulation of a manifold adapted to a submanifold I am not extremely proficient in topology, and am concerned with the following question :
Given a compact manifold $X$ and a submanifold $Y \subset X$, is it always possible to find a triangulation of $X$ which restricts to a triangulation on $Y$ ? Ie. such that the simplices which meet $Y$ define a triangulation on it. 
If this is not possible, does it become true if $X$ is a compact complex manifold and $Y$ a smooth analytic subset ? 
 A: Provided your submanifold is neatly embedded the answer is yes.  Neat embedding means that the boundary of the submanifold (if it exists) is in the boundary of the ambient manifold, intersecting transversely.  It also requires that compact sets in the ambient manifold intersect the submanifold in compact sets, so things like $(0,1) \times \{0\} \subset \mathbb R^2$ are not allowed. 
The proof is just a variation on the proof that smooth manifolds have triangulations.  You repeat the same proof (of Whitehead's) but keeping track of the submanifold in the process.  If you don't know Whitehead's proof that smooth manifolds have triangulations, the basic idea is to embed the manifold in any manifold that you know has a triangulation (like a high-dimensional euclidean space or sphere).  By refining the ambient triangulation you can ensure the embedding appears locally linear in the simplices of the ambient triangulation.  The stratification of the ambient triangulation pulls-back to a stratification of the embedded manifold, which you subdivide into a triangulation.  
