# $f(x)=\frac{x-x_0}{||x-x_0||}$ is a zero degree map

Let $S\subset\mathbb{R}^n$ be a compact, connected, embedded hypersurface. For $x_0\notin S$, define: \begin{align*} f_{x_0}:S &\to \mathbb{S}^{n-1}\subset \mathbb{R}^n\\ x&\mapsto \frac{x-x_0}{||x-x_0||} \end{align*}

By Jordan's separation theorem, $\mathbb{R}^n\setminus S$ has two connected components $U$ (bounded) and $V$ (unbounded). Prove that the degree of $f_{x_0}$ is zero for all $x_0\in V$.

For the case $n=2$, I can easily see this is true by looking at the winding number. But I'm totally stuck with the general case.

Any ideas?

First, note that $\mathrm{deg}(f_{x_0})$ does not depend on the point $x_0 \in V$ (to see this, build an homotopy between maps with differents $x_0$'s).
Now suppose w.l.o.g that $0 \in U$. Let $x_0$ be outside of a compact big ball containing $S$. We claim that $\frac{x_0}{\|x_0\|}$ is not on the image of $f_{x_0}$. Indeed, suppose it were. Then there would be an $x \in S$ such that $$f_{x_0}(x)=\frac{x_0}{\Vert x_0\|},$$ which tells us that $$x-x_0=\Vert x-x_0\| \frac{1}{\Vert x_0\|}x_0.$$ Letting $k:=\Vert x-x_0\| \frac{1}{\Vert x_0\|}$, we see that $k>0$ and that $$x=(1+k)x_0,$$ which is an absurd, since $x$ was supposed to be of lesser length than $x_0$.
It follows that $f_{x_0}$ is not surjective. Hence, it must have degree zero.
• I know the definitions of degree by $\int_S f^*[\omega]=\deg(f)\int_{\mathbb{S}^{n-1}}$ and by $\deg(f)=\sum_{p\in f^{-1}(q), q \text{ regular}}sign_p(f)$. How does non-surjectivity imply degree zero? – rmdmc89 Nov 25 '17 at 21:36