# On partitioning a manifold into submanifolds of the same dimension

Let $$M$$ be an $$n$$ dimensional topological manifold with boundary. Let $$M_1,M_2,...,M_k$$ be $$n$$ dimensional submanifolds of $$M$$ such that:

1. $$M=\cup_{i=1}^k M_i$$
2. $$\{Int(M_i)\}_{i}$$ is a family of pairwise disjoint sets . (Where $$Int$$ denotes the interior of a topological manifold)

Does it follow that for any nonempty $$A\subseteq\{1,2,..,k\}$$ that $$\cap_{a\in A}M_a$$ is either empty or a manifold (with boundary) of dimension $$n+1-|A|$$ and that we have $$\partial M\cap \cap_{a\in A}M_a$$ is either empty or a manifold with boundary of dimension $$n-|A|$$ ?

I admit I have not spent much time thinking about the problem as I don't have knowledge of the usual machinery needed to deal with topological manifolds as opposed to smooth manifolds. I guess it might be enough to prove the statement for $$k=2$$ ? The mathematical situation I described reminds me of phase diagrams in thermodynamics/materials science

I would be grateful if someone can point out a reference or a counterexample.

No, already for $$n=2$$. Take any compact triangulated surface and take $$M_i$$'s to be the 2-dimensional simplices of the triangulation. Now, take two simplices $$M_1, M_2$$ which meet at a vertex, so $$A=\{1,2\}$$ but the intersection does not satisfy your dimension count. You can relax the dimension count condition and simply ask for intersections to be manifolds with boundary of some dimension. This will also fail when $$n=3$$, but examples are a bit complicated.
• @Amr: Then the intersection is a submanifold, the common boundary of $M_1, M_2$. Jun 1 at 13:25