Theorem: If $f : S^n \rightarrow S^n$ is of degree zero, then there are points $x,y$ with $f(x) = x$ and $f(y) = -y$.

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Proof: If $f$ does not have a fixed point, then it is of degree $\pm 1 \neq 0$. Conversely, if $f$ is of degree zero it must have a fixed point.

Since $f$ is of degree zero, we have $\deg (-f) = 0$ as well since degree is multiplicative with respect to composition. Then, by the same argument as before, we conclude that $-f$ must have a fixed point $y$, so that $-f(y) = y$ and thus $f(y) = -y$.

Corollary: A continuous non-vanishing vector field $F$ on $D^n \subset \mathbb{R}^n$ has a point on the boundary where $F$ points radially inward, and another where it points radially outward.

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Proof: Since $F$ is non-vanishing, we can normalize it, letting $\what{F} = F / \norm{F}$. Then, $\im \what{F} \subseteq S^{n-1}$. Now, we have a map $G: S^{n-1} \xrightarrow{i} D^n \xrightarrow{\what{F}} S^{n-1}$. Since the map factors through a contractible space, $G_* = 0$, and $G$ is of degree zero. Then, applying the theorem we have $x$ and $y$ with $G(x) = x$ and $G(y) = -y$. Then, since the first map in $G$ is just the inclusion, we have $\what{F} = x$ and $\what{F}(y) = -y$. Then, for the original map $F$, we conclude $F(x)$ is parallel to $x$ and $F(y)$ is opposite in direction to $y$. Since $x$ and $y$ point in the radial direction, this means that $F(x)$ "points outward" and $F(y)$ "points inward".