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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...

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  • $\begingroup$ Because it is a covering map does it mean that there no point on $S^1$ where $S^1$ is equal to $V$ is not equal to $(0,0)$ Any help? $\endgroup$ Dec 2, 2013 at 3:30

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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.

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  • $\begingroup$ So this is not a covering map? $\endgroup$ Dec 2, 2013 at 4:07
  • $\begingroup$ @KangHoonYou I should hope not -- are there any open sets in $B$ homeomorphic to an open interval? $\endgroup$
    – Neal
    Dec 2, 2013 at 17:43

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