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More precisely, let $M$ be a subspace $\mathbb R^n$ with the following properties:

  • $M$ is a topological manifold of dimension $n-1$.
  • M is compact.

Does there exist a homological characterization of when the following happens:

  • $\mathbb R^n \backslash M$ has two components, the bounded one being "inside" and the other one "outside". Both are $n$-dimensional manifolds.

If the above is not possible, is there a different formulation of the question which would allow a nice characterization?

The motivation of this question is of course the realization that the solution for $n = 3$ seems to be that $M$ is an oriented surface.

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    $\begingroup$ The answer is yes. See the proof of the Jordan-Brouwer Separation theorem in a text like Bredon's "Geometry and Topology" $\endgroup$
    – Ryan Budney
    May 16, 2010 at 14:51
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    $\begingroup$ To elaborate on Ryan's comment: Since the question is about more general manifolds than spheres, the relevant theorem is Alexander duality. This implies that the complement of $M$ has two components exactly when $M$ is connected. Alexander duality also implies, incidentally, that $M$ must be orientable. $\endgroup$
    – Allen Hatcher
    May 16, 2010 at 16:47

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More genreally, the number (finite or not) of the connected components of the complement set of a compact subset $M\subset {\mathbb R}^n$, which is the rank of $H_0({\mathbb R}^n\setminus M)$, is a homotopic invariant for compact subspaces of ${\mathbb R}^n$, by duality in homology.

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