# Geometric argument for “easy” part of Jordan-Brouwer separation theorem without local flatness

Let $M^n \subset \mathbb{R}^{n+1}$ be an $n$-dimensional compact connected topological submanifold. The Jordan-Brouwer separation theorem says that $\mathbb{R}^{n+1} \setminus M^n$ contains two connected components. This for instance is immediate from Alexander duality.

For more geometric proofs, there are two steps: the "hard" step proves that there are at least $2$ components, and the "easy" step shows that there are at most $2$ components.

Now, the "easy" step is indeed easy if you assume that $M^n$ is locally flat. Just fix two points $p$ and $q$ that are on opposite sides of a local product neighborhood of $M^n$. You can then connect any point in the complement of $M^n$ to either $p$ or $q$ by first moving close to $M^n$ and then using the local product neighborhoods to "follow" $M^n$ to either $p$ or $q$ (depending on what side you end up on). But I don't see how to do this if $M^n$ is not locally flat. Does anyone know a reasonably geometric argument for this?

EDIT: Since no one has answered, let me ask a somewhat more focused question. We certainly do not need the full strength of local flatness to make the above argument go through. All we need would be a positive answer to the following question:

Question 1 Let $M^n \subset \mathbb{R}^{n+1}$ be an $n$-dimensional compact connected topological submanifold, not necessarily locally flat, and let $p \in M^n$. Does there necessarily exist an open neighborhood $U$ of $p$ in $\mathbb{R}^{n+1}$ such that the set $U \setminus (U \cap M^n)$ has at most two connected components?

One very natural way of obtaining a positive answer to question 1 would be by giving a positive answer to the following more precise question:

Question 2 Let $M^n \subset \mathbb{R}^{n+1}$ be an $n$-dimensional compact connected topological submanifold, not necessarily locally flat, and let $p \in M^n$. Does there necessarily exist a continuous map $\phi: (-1,1)^{n+1} \rightarrow \mathbb{R}^{n+1}$ with the following two properties:

1. The image of $\phi$ is an open set containing $p$, and
2. The set $\{\text{$p \in (-1,1)^{n+1}|\phi(p) \in M^n$}\}$ consists of all points whose last coordinate is $0$.

Here $(-1,1)^{n+1}$ is the cartesian product of $(n+1)$ copes of the open interval $(-1,1) \subset \mathbb{R}$.

Both of these are obviously true in the locally flat setting; indeed, in that setting you can assume that the map $\phi$ in question 2 is a topological embedding

• Have you checked whether this is true for some known example of non-locally flat submanifold, say the Alexander horned sphere? – ThiKu Sep 26 '15 at 1:58