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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!

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up vote 2 down vote accepted

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.

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Thank you! I see that lots of approach can be possible to solve a problem! – Emily Apr 23 '12 at 16:31

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.

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Thank you! This helps a lot! – Emily Apr 23 '12 at 16:30
The idea seems marveling! – Emily Apr 23 '12 at 16:37

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