Can someone please remind me how this goes?
Here's the idea of proof I'm trying to recall: let $S$ be a closed surface (connected, compact, without boundary) embedded in $\mathbb{R}^3$. Then one can define the "outward-pointing normal unit vector" to $S$ at any point, and subsequently an orientation of the surface.
One would like to define this vector by saying that it points towards exterior points to $S$. So we need some kind of generalization of the Jordan curve theorem saying that the surface cuts $\mathbb{R}^3$ into two pieces (interior and exterior). What is this theorem exactly?
Also, I apologize if this is silly, but is there an obvious argument that a piece of the surface cuts a small tubular neighborhood of it into interior and exterior points (this seems necessary to define the outward normal vector properly)?
Is there a "cleaner" approach to prove this fact? Thanks in advance.
Another approach might be to show directly that the normal bundle must be trivial
For instance if the Klein bottle were embedded in $\mathbb{R}^3$, then its normal bundle could not be trivial since it is an unorientable manifold. If its normal bundle were non-trivial then the boundary of a tubular neighborhood would be an oriented two fold cover and therefore a torus. Further the entire tubular neighborhood would have the homotopy type of the Klein bottle.
It is not hard to show using a Meyer-Vietoris sequence that this is impossible.
$$H_2(\mathbb{S}^3)\longrightarrow H_1(\mathbb{T^2}) \longrightarrow H_1(\hbox{Tubular neighborhood}) \oplus H_1(\hbox{Tube neighborhood complement}) \longrightarrow H_1(\mathbb{S}^3)$$
Which is
$$0 \longrightarrow \mathbb{Z} \oplus \mathbb{Z} \longrightarrow (\mathbb{Z}_2 \oplus \mathbb{Z} ) \oplus H_1(\hbox{Tube neighborhood complement})\longrightarrow 0$$
A similar arguement works for the projective plane $\mathbb{P}^2$.