2

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.

Peter Franek
  • 11,522

1 Answers1

1

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.

Moishe Kohan
  • 97,719
  • Concerning your "lemma", you find an isotopy $h_t$ of a simplex that contains $x,y$ in the interior s.t. $h_0$ is the identity and $h_1(x)=y$, and use the isotopy extension theorem, is that right? – Peter Franek Jul 04 '14 at 08:01
  • @PeterFranek: Yes, in the case when this simplex $s$ is top-dimensional. However, in general, $s$ is not top-dimensional; then you consider the interior $U$ of the union of all simplices containing $s$ and use the fact that $U$ (near the segment in $s$ connecting $x$ to $y$) is a product of an open subset of $s$ and a certain simplicial complex. Now, you run the same isotopy argument in this product neighborhood. – Moishe Kohan Jul 04 '14 at 15:18