# Continuous Map $f : S^n \rightarrow S^{n-1}$ with $f(-x) = -f(x)$

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.

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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

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.

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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.

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