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

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.

1 Answer

$$V$$ 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.