show that $R^n -X$ has at most 2 connected components.
Theprem: Suppose that $X$ is the boundary of $D$, a compact manifold with boundary and let $F:D \to R^n$ be a smooth map extending $f$; supposed that $z$ is a linear regular value of $F$ that doesn't belong to the image of $f$. Then $F^{-1}(z)$ is a finite set , and $W_2(f,z) =$ #$F^{-1}(z) mod 2$.
I know that $X$ is a compact, connected hypersurface in $R^n$. $X$ is winding around $R^n$, then $dim(X)=n-1$. Let $z\in R^n$ but not on $X$, then $W_2 (X,z)$ must be $0$ or $1$ depend on whether $z$ lies inside or outside of $X$.
If $z$ lie on the inside of $X$ then $R^n -X$ yield 2 components, otherwise $R^n-X$ still remain as one component.
I'm not sure if I understand this problem correctly, any help will be much appreciated.
If $X$ is an $n$-dimensional compact connected submanifold of $\mathbb{R}^{n+1}$ without boundary, its complement has 2 connected components (this follows from alexander duality).