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If a hypersurface in a manifold separates the ambient space into two disconnected pieces, is the surface necessarily orientable? This seems to be true when one considers the Jordan Brouwer theorem which implies the sphere $S^n$ embedded in $\mathbb{R}^n$ separates space into two disconnected components. But does the requirement of orientability extend to hypersurfaces in any manifold? A counter-example would show a non-orientable hypersurface separating the ambient space into two disconnected regions.

(edit: statement on Jordan Brouwer theorem refined per George Lowther's comment)

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    The line "...which implies that the hypersurface must be homeomorphic to $S^n$ embedded in $\mathbb{R}^n$ to guarantee separation" does not seem correct. A torus separates $\mathbb{R}^3$ for example.2011-09-29

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A nonorientable surface can certainly separate a nonorientable manifold. Here is an example. Note that the Klein bottle $K$ is the boundary of a $3$-manifold $M$, which is gotten by identifying the two ends of a solid cylinder in an orientation reversing way. Now consider the $3$-manifold $(K\times [0,1])\cup M_0\cup M_1$ where $M_i$ is homeomorphic to $M$, and $\partial M_i$ is identified with $K\times\{i\}$. This is a nonorientable $3$-manifold. The middle slice $K\times\{1/2\}\subset K\times [0,1]$ separates the manifold into two pieces.

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    @measure_noob: Sorry, my statement should have included "compact." My mistake.2011-09-29