Higher dimensional Jordan-Brower separation theorem Is there an established higher dimensional version of the Jordan-Brower theorem? I would like some statement like this:
Let $M$ be an $n$-dimensional closed, compact and connected variety and suppose that there exists an embedding $M\hookrightarrow{\Bbb R}^{n+1}$. Then ${\Bbb R}^{n+1}-M$ has exactly two connected components.
Mind that I do not want to assume $M$ orientable in the first place.
If such a statement is true, then one can show that $M$ is orientable as follows. If not, let $\gamma$ be a closed path in $M$ starting at $P$ which realizes the non-trivial monodromy of the normal line bundle. By compactness, there is a tubular neighborhood $T$ of $\gamma$ such that $T\cap M\simeq\gamma\times[-1,1]$. Starting from $P$ choose compatibly (i.e. continuously) a normal vector $\vec{n}_Q$ contained in $T$ at each point $Q$ of $\gamma$. 
The endpoint of $\vec{n}_Q$ describes a curve in $T$ which never meets $M$. It can be closed to a loop $\gamma^\prime$ moving along the normal line at $P$. The closed curve $\gamma^\prime$ meets $M$ only at $P$ and they intersect transversally there.
This would contradict the higher dimensional Jordan-Brower theorem, since a closed curve intersecting $M$ transversally must intersect an even number of times.
Motivation : I'm looking for an "easy" (maybe to be read: "highly intuitive") proof of the fact that the projective plane cannot be embedded in ${\Bbb R}^3$ which can be appreciated by a person without a solid background in mathematics (but she has a fairly good understanding of what the projective plane is). In this case, of course, the curve $\gamma$ can be taken as the "mid section" of a Möbius band in the projective plane.
 A: In the neighbourhood of a point of $M$, $M$ becomes nothing but $\mathbb R^n$ separating $\mathbb R^{n+1}$ in two half-spaces. Take two points $x_-$ and $x_+$, one in each of those “local connected components” of $\mathbb R^{n+1} \setminus M$. It is relatively easy to see that $\mathbb R^{n+1} \setminus M$ is connected if you can join $x_-$ and $x_+$ through a path avoiding $M$, and that it has 2 pathwise connected component otherwise. 
So now, assume $\mathbb R^{n+1} \setminus M$ is pathwise connected, so that there exists such a path $\gamma : I \to \mathbb R^{n+1} \setminus M$. You can assume that $\gamma$ is injective (either by “generic position” arguments or because you can easily prove that pathwise connected components and arcwise connected components are the same). I will also assume $\gamma$ to be smooth.
I can also take the segment that goes from $x_-$ to $x_+$ while remaining in the neighbourhood I started with (in which $M$ looks like $\mathbb R^n$): it crosses $M$ transversely in exactly one point. Now I can glue together $\gamma$ and this arc. By still more (easy) general position arguments, what I get can be assumed to be an embedded circle $\widehat \gamma$ in $\mathbb R^{n+1}$ intersecting $M$ transversely in exactly one point.
I've thus obtained homology classes $[\widehat \gamma] \in H_1(\mathbb R^{n+1}), [M] \in H_n(\mathbb R^{n+1})$ such that $[\widehat \gamma] \cdot [M] = \pm 1$. This is absurd, because $H_1(\mathbb R^{n+1}) =  H_n(\mathbb R^{n+1}) = 0$.
