How can I conclude that F a continuous vector field on unit ball gives a map of degree zero? It is a problem for a Hatcher's book, and it is my homework problem.
It is a section 2.2 problem 3, stating:
Let $f:S^n\to S^n$ be a map of degree zero. Show that there exist points $x,y \in S^n$ with $f(x)=x$ and $f(y)=-y$. Use this to show that if $F$ is a continuous vector filed defined on the unit ball $D^n$ in $\mathbb{R}^n$ such that $F(x) \neq 0$ for all $x$, then there exists a point on boundary of $D^n $ where $F$ points radially outward and another point on the boundary of $D^n $ where $F$ points radially inward.
I could get the first statement by the property of a degree. However, in order to use this fact to conclude that this fact applies to the second statement, I should know that $F$ restricted to $S^{n-1}$ and being normalized so that $\bar F:S^{n-1} \to S^{n-1}$ is of degree zero. If I can conclude that $\bar F$ is not surjective, then it's all done. However, I am not sure to show why $\bar F$ is of degree zero. 
Any comment about this would be grateful!  
 A: First, as you mentioned, normalize $F$ to be $\overline{F}:D^n\rightarrow S^{n-1}$.
Next, define $\overline{F}_t: S^{n-1}\rightarrow S^{n-1}$ to be $\overline{F}_t(x)=\overline{F}(tx)$, then $\overline{F}_0$ is a constant map and $\overline{F}_t$ construct a homotopy between $\overline{F_0}$ and $\overline{F_1}$. Since degree is defined by homology hence invariant under homotopy, we have $\deg \overline{F}_1=\deg \overline{F}_0=0$, and you can apply your first statement to $\overline{F}_1$ and then obtain the second statement.
A: Observe that the diagram
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
H_n(D^{n+1}) & = & H_n(D^{n+1}) \\
\ua{inc^*} & & \da{(F/|F|)^*}  \\
H_n(S^n) & \ra{\overline{F}^*} & H_n(S^n)\\
\end{array}
$$
commutes. Hence, as the group $H_n(D^{n+1})$ vanishes, the map $\overline{F}^{*} = 0$ as desired.  
