TITLE: Generalized Jordan-Brower separation theorem
QUESTION [8 upvotes]: Suppose $M^{n+1}$ is a closed connected smooth manifold and $N^n$ is a closed connected smooth embedded submanifold of $M^{n+1}$. 
What's the weakest condition under which the Jordan-Brower separation theorem hold? 
Namely, $M^{n+1}\backslash N^n$ has exactly two connected components, that share $N^n$ as their common boundary. I think a simple connectedness condition suffices. But does a weaker set of conditions work? How about $H_1(M^{n+1},Z)=0$? And could you give me some reference? Many thanks!

REPLY [2 votes]: A properly embedded connected compact submanifold $N \hookrightarrow M$ gives you a map $H_1(M;Z) \to Z/2$ by counting mod 2 intersections. By counting those you see (as explained by Mike) that we have equivalent conditions: $N$ seperates $M$ into exactly 2 components (as we have a trivial normal bundle) $\Leftrightarrow$ the induced homomorphism $H_1(M;Z) \to Z/2$ is trivial.
Now this characterizes those submanifolds pretty well. In particular we see that if $H_1(M;Z)=0$ every such homomorphism induced by any submanifold is necessarily trivial. Or if $H_1(N;Z) \to H_1(M;Z)$ is an isomorphism as we see that every homology class can be represented as a curve on $N$ which can be pushed off $N$, which makes the intersection trivial.