Decomposition of a topological manifold into sets with low-dimensional intersections

Question

Let $f:M\rightarrow N$ be a continuous function between connected topological manifolds, modelled respectively on $\mathbb{R}^m$ and $\mathbb{R}^n$; $n,m\in \mathbb{Z}^+$. Can we always find a cover $\{C_i\}_{i \in I}$ of $N$ such that:

• $I$ is finite,
• The $C_i$ are regular closed subsets, i.e. $\overline{\text{int}(C_i)}=C_i$,
• $\text{int}(C_i)\cong \mathbb{R}^n$,
• $\text{dim}(C_i\cap C_j)<n$? and $C_i\cap C_j$ is a Borel set?

When $N$ is compact, this is clear, since we just cover $N$ with locally Euclidean neighbourhoods, take their closures, and reduce to a finite set using compactness. However, what about in general?

Answer 1

I will ignore $M$ and $f$ since they play no role in the question. Here is what I know about the compact case:

1. If $N$ admits a triangulation or, more generally, a handle decomposition, then the finite collection of subsets $C_i$ does exist.

2. Every topological manifold of dimension $\le 3$ admits a triangulation.

3. Every topological manifold of dimension $> 4$ admits a handle decomposition.

It is not known if compact topological 4-manifolds admit structure of CW complexes.

Edit. I just realized that the answer to your question is positive for all connected manifolds. Even two subsets $C_1, C_2$ will suffice. It is an application of Berlanga-Brown theorem which states that every connected topological n-manifold contains an open and dense subset homeomorphic to the open n-ball.

Here are some details:

Berlanga in

R.Berlanga "A mapping theorem for topological sigma-compact manifolds", Compositio Math, 1987, vol. 63, 209-216.

generalizes an earlier work of Morton Brown (in the case of compact manifolds) proves that every connected $n$-dimensional topological manifold $N$ contains an open and dense subset $U$ homeomorphic to $R^n$. I will consider the case $n\ge 2$ since the situation with $n=1$ is clear.

Let $A:= N - U$. Pick a sequence $x_i\in U$ whose accumulation set in $N$ equals $C$. Since $U$ is homeomorphic to $R^n$, there exists a hypersurface $H\subset U$ homeomorphic to $R^{n-1}$, containing the sequence $(x_i)$ and separating $U$ in two open subsets $V_1, V_2$ each homeomorphic to $R^n$. Then the closure $C_i$ of $V_i$ in $N$ will be regular (see below) and the intersection $B=C_1\cap C_2$ has empty interior in $N$. Thus, $\dim(B)=n-1$. (In general, each closed subset with empty interior in an $n$-dimensional manifold has covering dimension $\le n-1$, this is Menger-Urysohn theorem. But in our case $B$ contains $H$, so $\dim(B)=n-1$.)

To see regularity of $C_i, i=1, 2$ note that the boundary of $C_i$ equals $A\cup H$ and, by the construction, each point of $A\cup H$ is a boundary point of both $V_1$ and $V_2$. Thus, $int C_i= V_i$, while $C_i=cl(V_i)$, $i=1, 2$.