Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

2 Answers 2

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.

share|improve this answer
    
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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.