A vector field which is nowhere $0$ have outward and inward pointing vectors

Question

I am reviewing Algebraic Topology for my Qual. This is a problem from the past.

"Let $V$ be a continuous vector field on the unit ball $B^n\subset \mathbb{R}^n$ which is nowhere $0$. Prove that there are points $x,y\in S^{n-1} = \partial B^n$ and positive numbers $a,b>0$ s.t. $V(x)=ax,V(y)=-by$."

I am confused since I never even learned about vector field, so the language is kinda weird for me. Anyway, I looked it up and it seems like I have a map $V:B^n\to \mathbb{R}^n$ and I need $a,b,x,y$ like above. This looks like Brouwer fixed point theorem, but I don't know where to start. Finding a point $x$ s.t. $V(x)=ax$ has the same direction seems hard already for me.

Answer 1

$V$ is nonzero, so normalise it: $W(x)=V(x)/\|V(x)\|$. Then $W$ is a continuous map from $B^n$ to $S^{n-1}$. Now $V(x)=ax$ on $S^{n-1}$ means $W(x)=x$ and $V(y)=-by$ on $S^{n-1}$ means $W(y)=-y$. The restriction of $W$ to $S^{n-1}$ is homotopic to a constant map, since it extends to $B^n$. So $W|_{S^{n-1}}$ has degree $0$.

If $W(y)\ne -y$ on $S^{n-1}$ then $W|_{S^{n-1}}$ is homotopic to the identity map from $S^{n-1}$ to $S^{n-1}$ (take the "straight line" homotopy in $\Bbb R^n$ from the identity to $W|_{S^{n-1}}$ and project from the origin to the sphere). Therefore $W|_{S^{n-1}}$ has degree $1$ a contradiction. The case where $W(x)\ne x$ is similar.