I'm familiar with a general Jordan-Brouwer Separation theorem for a compact connected smooth manifold of codimension $1$ in $\mathbb{R}^N$ (i.e. not just hyperspheres). My ultimate goal is to better understand how the analog (in a proper sense) of this theorem carries over to topological manifolds.

This, however, seems to be a very deep result that requires very sophisticated machinery (for a non-expert in graduate-level algebraic topology and differential geometry etc..) with very steep entry barrier.

While I gather more knowledge on the subject (that might take many months), I hope to make a smaller (immediate progress) in this general direction, with the help of those two questions:

The primary question:

Is there a simpler definition/notion of the "inside" and "outside" of $\mathbb{R}^n \setminus Y$, for topological compact connected codimension-1 manifold $Y\subset \mathbb{R}^N$ (with the topology induced by the Euclidian metric)?

Secondary question:

Let $X\subset \mathbb{R}^N$ be compact, connected smooth manifold of codimension-$1$ (a hypersurface). So it has an easy notion of "inside" and "outside" (see in the elaboration below). Map $X$ by a (non-smooth) homeomorphism $h:\mathbb{R}^N \to \mathbb{R}^N$:

(a) Does the image $Y= h(X)$ have a simple definition of the "inside"?

(b) if the answer to (a) is negative, what are some of the mild conditions that need to be imposed on $h$ so that its image does have a simple definition of "inside"?

Any comments, references or intuition why those are still hard questions (or not) $-$ especially anything that helps better understanding the big picture $-$ are welcome.


Here is a quote of Jordan-Brouwer Separation theorem from lecture notes:

Theorem 20.1 [Jordan- Brouwer Separation theorem]: Let $X$ be compact, connected (smooth) manifold of codimension-$1$ in $\mathbb{R}^N$(hypersurface). The complement of $X$ in $\mathbb{R}^N$ consists of two connected open sets, the "outside" $D_0$ and the "inside" $D_1$. Moreover, $\overline{D}_1$ is compact manifold with boundary $X =\partial \overline{D}_1$.

The "inside" is, roughly, all points in $\mathbb{R}^N\setminus X$ (the complement of $X$ in $\mathbb{R}^N$) for which if we send a ray [in general position] from this point[$^*$] and then counts the number of the intersections of this ray with $X$, we have an odd number of intersections then the point is in the "inside" (conversely, if we have even number of cuts the point is in the "outside").

With the above in mind, my primary question is: suppose I have $Y$, compact connected topological manifold. So I lose the notion of transversality etc. Is there still a (relatively) simple notion of the "inside" and "outside" of $\mathbb{R}^n \setminus Y$, if yes how those are defined? Or is this already a "deep question" that requires to dive into algebraic topology and etc...

Possible duplicates

I've searched extensively to find an explanation on StackExchange or other sources: by looking up "inside topological manifold", "separation theorem for topological manifolds", "Simple Alexander duality for non-spheres" etc... but I failed to piece together any answer to my question. I find a lot of sources (here, here, here, here, here, here, the closest to my question is this one) concerned with smooth manifolds or non-smooth cases but concerned only with hyperspheres, or concerned about topology of the compliments (e.g. if it is simply connected, the counter-example of Alexander horned sphere, Alexander duality is stated in terms of hyperspheres, and here I would like to consider not just hyperspheres but compact, connected codimension-$1$ manifold)), or it is stated in a way that (with my current knowledge) is too cryptic to decipher (like here). I hope the way I'm trying to attack this question is simpler and I can make some progress.

[$^*$] A ray is going in a general direction so the ray cuts the manifold $X$ transversally, such rays are dense so we can always find one.

  • 1
    $\begingroup$ Well, the outside is the unique unbounded connected component of the complement, which clearly exists. The inside will be the bounded connected components of the complement, but it is not obvious that there is such a component, nor then that it is unique. $\endgroup$ – Lord Shark the Unknown May 1 at 18:54
  • 1
    $\begingroup$ The invariant you have, the parity of the transversal intersection number, can be extended in the topological category by using the Poincare pairing of $H^{n-1}(Y; Z_2)$ with the relative 1-st homology of neighborhoods of $Y$ in $R^n$ (relative the boundary). However, this is well beyond the tools that you currently have. My suggestion is: Instead of trying to find a non-existing black cat in a dark room, pick up an algebraic topology book and start reading. Hatcher is the standard reference these days. If this is too hard, think about something else. $\endgroup$ – Moishe Kohan May 1 at 19:54

This is the kind of situation where Alexander Duality says a lot, but establishing the theorem is admittedly no small undertaking and it's not always obvious how you should use it.

The statement of Alexander Duality for singular (co)homology concerns any subspace $X\subset S^n$ which is compact and locally contractible (which is true of a compact topological submanifold) and it gives us an isomorphism

$$\tilde{H}_q(X) \cong \tilde{H}^{n-q-1}(S^n\setminus X) $$

for every $q\geq 0$. In particular if $X$ is a closed (as in compact and without boundary) and oriented $(n-1)$-dimensional topological sub-manifold then $$\mathbb{Z} \cong \tilde{H}_{n-1}(X)\cong \tilde{H}^0(S^n \setminus X)$$

But the rank of the reduced degree $0$ homology group is one less than the number of path components, so $S^n \setminus X$ has exactly two components. They are both open in $S^n$ since $X$ is compact and hence (topologically) closed in $S^n$.

It follows that if $X\subset \mathbb{R}^n$ is a closed, connected, oriented codimension-$1$ topological submanifold then $\mathbb{R}^n\setminus X$ also has two disjoint open components, say $U_1$ and $U_2$ (by considering $S^n$ as the one-point compactification of $\mathbb{R}^n$). Since $X$ is compact one of these components is definitely unbounded, say $U_1$, so it remains to show $U_2$ is not. Let $r >0$ be large enough so that $X\subset B_r(0)$, the ball of radius $r$ centred at the origin and consider $Y:=\mathbb{R}^n \setminus B_r(0)$. Since $Y$ is connected and $Y\cap U_1 \neq \emptyset$ it follows that $U_2\cap Y = \emptyset$ (and hence $U_2$ is bounded), since $Y = (Y\cap U_1) \cup (U_2 \cap Y)$. Hence we could consider the bounded component as being "inside".

In fact, if $X$ satisfies all the above conditions but is not connected, and has say $k$ components, then $\tilde{H}_{n-1}(X)\cong \mathbb{Z}^k$ and so $\mathbb{R}^n\setminus X$ will have $k+1$ components, precisely one of which is unbounded.

(For a similar application of Alexander duality, check out my answer to this question )

I think your second question should follow from the first, since the image $h(Y)$ would be a compact topological submanifold of $\mathbb{R}^n$.

  • $\begingroup$ Thanks a lot! this helps to see the big picture. Is there an easy intuition of what the inside and outside mean in this case? For the smooth case I get the intuition by considering the number of cuts mod 2 that a ray from any point in the "inside" has with the manifold that "encloses", is there a similar intuition for the non-smooth case? $\endgroup$ – them May 5 at 16:14

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