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.

I'm studying for an algebraic topology exam, and the following question has me stumped.

Problem. For $n \geq 1$, prove there does not exist a continuous map $f : S^n \rightarrow S^{n-1}$ such that $f(-x) = -f(x)$ for all $x \in S^n$.

The case $n = 1$ is fairly obvious since $S^0$ is disconnected, but I don't expect this to help for $n \geq 2$.

I'm interested in any answer, but one based in algebraic topology would be especially helpful.

share|improve this question
1  
I imagine you can get something out of the fact that $x\mapsto-x$ is orientation preserving on $S^n$ for odd $n$, but not for even $n$. –  Harald Hanche-Olsen May 1 '12 at 16:30

2 Answers 2

up vote 5 down vote accepted

This is an application of the Borsuk-Ulam theorem (can be found for example in Hatcher's book).

Thm: A map $g:S^{n-1}\rightarrow S^{n-1}$ satisfying $g(x)=-g(-x)$ has odd degree (i.e. the induced map on the $(n-1)^{st}$ homology is multiplcation by an odd number).

If a map $f:S^n\rightarrow S^{n-1}$ with $f(x)=-f(-x)$ would exist, then the restriction to the equator $g:S^{n-1}\rightarrow S^{n-1}$ would fulfill the requirements of the theorem and would therefore be odd. On the other hand $g$ is null-homotopic, via the restriction of $f$ to the upper hemisphere. This is a contradiction as $g$ now induces both multiplication by $0$ and by an odd number on homology.

share|improve this answer
    
It's a shame we haven't covered Borsuk-Ulam, but this definitely answers the question. Thanks! –  Hans Parshall May 1 '12 at 16:56

Such a map will induce a continuous map $g: \mathbb RP^n \to \mathbb RP^{n-1}$.

Consider what happens when you apply $\pi_1$ to $g$: you must obtain the zero map, otherwise you will have an isomorphism on $H_1$ and therefore an isomorphism on $H^1$, and then the contradiction that $x^n = 0$ in $H^n(\mathbb RP^n, \mathbb Z/2) \subset \mathbb Z/2[x]/(x^{n+1})$.

It now follows by covering space theory that $g$ factors through the cover $S^{n-1}\to \mathbb RP^{n-1}$. By uniqueness of lifts, the composition $S^n \to \mathbb RP^n \to S^{n-1}$ must agree with either $x\mapsto f(x)$ or $x\mapsto f(-x)$, since at a single point $t$ the possible preimages of $g(t)$ are $f(t)$ and $f(-t)$.

But then, we see that the map $f$ must take antipodal points to the same point, and that is a contradiction.

share|improve this answer

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.