Let \(f:\mathbb R^{n+1}\to\mathbb R^{n+1}\) be continuous, and suppose \(f\) does not vanish on the unit sphere \(S^n\). Define \[ F:S^n\to S^n, \qquad F(x)=\frac{f(x)}{|f(x)|}. \] Prove that if \(F\) has nonzero degree, then \(f\) has a zero in the unit ball.
Solution
1. Assume, for contradiction, that \(f\) has no zero in the closed unit ball \(B^{n+1}\).
Under this assumption the normalization \[ G:B^{n+1}\longrightarrow S^n, \qquad G(x)=\frac{f(x)}{|f(x)|}, \] is continuous. On the boundary \(S^n=\partial B^{n+1}\) it agrees with \(F\), so \(G\) is an extension of \(F\) over the ball.
2. Any map \(S^n\to S^n\) that extends continuously over \(B^{n+1}\) is null-homotopic.
Using the extension \(G\) from 1, define \[ H:S^n\times[0,1]\longrightarrow S^n, \qquad H(x,t)=G((1-t)x). \] Then \[ H(x,0)=G(x)=F(x), \qquad H(x,1)=G(0), \] so \(H\) is a homotopy from \(F\) to the constant map with value \(G(0)\).
3. The assumption in 1 contradicts \(\deg(F)\neq0\).
Degree is invariant under homotopy, while a constant map \(S^n\to S^n\) has degree \(0\). By 2, \(F\) is homotopic to a constant map, hence \[ \deg(F)=0, \] contrary to the hypothesis. Therefore \(f\) must vanish at some point of \(B^{n+1}\).