Showing there is some point interor to the unit disk, where V=(0,0) using covering maps Let $B$ be the closed unit disk in $\mathbb{R}^2$
and suppose that $V = (p(x, y), q(x, y))$
is a vector field (p,q are continuous functions) defined on B. The boundary
of $B$ is the unit circle $S^1$
. Show that if at every point of $S^
1$
, $V$ is tangent
to $S^1$
and not equal to $(0,0)$, then there is some point in the interior of B
where $V = (0,0)$.
I tried to prove this by contradiction. Suppose that the hypothesis is not true. Then at every $(x,y) \in B$, we can divide $V$ by its length. Taking at each point a vector perpendicular to this unit vector, we can define a continuous map from from $B$ to $S^1$. This is a covering map. I'm not quite too sure where to go from here...

The map $V/|V|$ had better not be a covering map because the disk is two-dimensional and the circle is one-dimensional. 
Here's a guide to a solution using a little bit of homotopy theory. First, observe that the map $V/|V|$ restricted to the boundary circle of $B$ is a degree-one self-map of the circle. Now answer this question: If a map $f:S^1\to S^1$ extends to an entire map $B\to S^1$, what does that imply about the degree of $f$? Finish the argument by showing that if $V$ were everywhere nonzero, you would reach a contradiction.
