TITLE: Does having a codimension-1 embedding of a closed manifold $M^n \subset \mathbb{R}^{n+1}$ require $M$ to be orientable?
QUESTION [11 upvotes]: I'm trying to follow a proof about immersing/embedding $\mathbb{RP}^n$ into $\mathbb{R}^{n+1}$, which goes roughly as follows:
Write $\tau=T\mathbb{RP}^n$.  The normal bundle $\nu$ has rank 1, so its Steifel-Whitney class is $w(\nu)=1$ or $w(\nu)=1+x$.  In every case, we need $w(\nu)\cdot w(\tau) = w(\nu \oplus \tau) = w(\epsilon^{n+1})=1$.  If $w(\nu)=1$, then $w(\tau)=(1+x)^{n+1}=1$, so $n+1=2^r$.  If $w(\nu)=1+x$, then similarly $(1+x)^{n+2}=1$ so $n+2=2^r$.  If the immersion is an embedding, the former case must hold.
Why is this true?  I feel like there should be an easy reason, but none of the people I talked with were able to nail down anything precise.  This could be wrong, but it seems like this is tacitly saying that a codimension-1 embedding of a closed manifold must be in fact of an orientable manifold, which is the same as saying that the the normal line bundle has trivial $w_1$ (since line bundles are totally classified by their orientability, i.e. by $w_1$).  Is this true?

REPLY [5 votes]: Here's a nice solution I just thought of, which may in fact be logically equivalent to Jim's (if it's even correct!).  I welcome comments addressing that.
Compactify $\mathbb{R}^{n+1}$ to $S^{n+1}$, so we consider $M \subset S^{n+1}$.  By Alexander duality, $$\tilde{H}_0(S^{n+1} \backslash M ; \mathbb{Z}/2) \cong \tilde{H}^{(n+1)-0-1}(M;\mathbb{Z}/2) = \tilde{H}^n(M;\mathbb{Z}/2)=\mathbb{Z}/2,$$ so $H_0(S^{n+1}\backslash M;\mathbb{Z}/2)=\mathbb{Z}/2 \oplus \mathbb{Z}/2$.  Hence $M$ separates $S^{n+1}$.