How do we express the union of closed neighborhoods of a cover with pairwise boundary intersection for a manifold $M$?

Question

Consider local charts $(U_{\alpha},\varphi_{\alpha})$ of an $n$-dimensional manifold $M$ with closed neighborhoods $U_{\alpha}$, $\varphi_{\alpha}:U\to V\subset\mathbb{R}^n$, and $\alpha\in A$ an element of the index set $A$. Then we choose an atlas $\mathcal{A}=\{(U_{\alpha},\varphi_{\alpha})|\alpha\in A\}$ in such a way that for any two distinct $i,j\in A$ in the atlas, the intersection of $U_i$ and $U_j$ is their shared boundary (a subset of $\partial U_i$ and $\partial U_j$), i.e. $U_i\cap U_j=\partial U_i\cap\partial U_j\ne\emptyset$ and the union of the closed neighborhoods still covers $M$.

What can the union of the neighborhoods be expressed as, besides the tautological $\bigcup_{\alpha\in A}U_{\alpha}$? That is, we want a covering of $M$ where the weak topology is such that the neighborhoods only intersect on boundaries, which is $\bigcup_{\alpha\in A}\text{Int}(U_{\alpha})\cup{\text{boundary intersections}}$.

Any help would be much appreciated! Thanks in advance.

Answer 1

You do not mention what kind of manifolds you are interested in (topological, smooth, $C^r$?), but let us ignore that point.

A chart $\varphi : U \to V \subset \mathbb{R}^n$ for $M$ is always defined on an open $U \subset M$. The boundary of $U$ can be understood as the topological boundary of the subspace $U$ of $M$. But then $\varphi$ is not defined on the boundary and you do not get something like "a chart on $\overline{U}$". The only interpretation that could make sense is this: For each closed ball $B \subset V$ consider the pair $(\varphi,B)$. Then $\varphi^B = \varphi \mid_{\varphi^{-1}(B)} : \varphi^{-1}(B) \to B$ may be regarded as a closed chart for $M$. It is essential that it is the restriction of an ordinary chart. Instead of closed balls you may also take closed $n$-simplices, closed $n$-cubes etc.

If you consider topological manifolds, then all these objects $B$ and $\varphi^{-1}(B)$ are manifolds with boundary. In the smooth case this is true for balls $B$, for simplices or cubes we get a manifold with corners (see Definition of "a topological manifold with corners".).

There are theorems saying that manifolds can be triangulated under suitable assumptions. See https://en.wikipedia.org/wiki/Triangulation_(topology) and https://mathoverflow.net/q/44021. This gives a positive answer to your question. Note, however, that this is highly non-trivial.