Let $K$ be a finite simplicial complex with the underlying topological space $|K|=\cup K$. If $|K|$ is also a topological manifold with boundary, does it hold that some subcomplex of $K$ triangulates exactly the boundary $\partial |K|$?
Intuitively, I would say it is obvious. It clearly holds for combinatorial manifolds, but as far as I know, not every "simplicial topological manifold" is so nice.
Answer is positive and is a corollary of the following general "homogeneity lemma":
Let X be a simplicial complex and x, y are points in the same open face. Then there is a homeomorphism $X\to X$ sending x to y.
(Let me know if you have trouble proving this lemma.)
Once you have this lemma, you see that if a the interior of a simplex in your triangulated manifold intersects the boundary then the entire simplex is contained in the boundary. It follows that the boundary is a subcomplex, hence, is triangulated.