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Let $f:S^n\rightarrow S^n$ be of odd degree, i.e. $f^*(1)$ is odd where $f^*:H_n(S^n)\rightarrow H_n(S^n)$ is the induced map on homology. Prove that there exists an $x\in S^n$ with $f(-x)=-f(x)$.

I tried to imitate the proof of Borsuk-Ulam theorem, but with no achievements.

Even in the case of $S^1$, I can not see how this happens, mainly because I don't know how to turn the condition on homology to some more intuitive ones. Should I use alternative definition of degrees in this case?

Anonymous Coward
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1 Answers1

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If $f(x)$ and $f(-x)$ are never opposite points, then we can deform $f$ to a new map $g$ whose value at both $x$ and $-x$ is the midpoint of the unique shortest arc joining $f(x)$ to $f(-x)$. But the map $g$ by construction factors through $RP^n$. Therefore its degree is necessarily even (for example by counting inverse images of a generic point).

John Gowers
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Mikhail Katz
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  • I should have thought of this. The proof that any $f:S^n\rightarrow S^n$ with no fixed point is homotopic to the antipodal is very similar to this. In both cases, you the segment linking the two functions $f,g$ ($f(x),f(-x)$ and $f(x),-x$ explicitly) does not pass through the origin. Then we can form $ft+g(1-t)$ and projective it to the sphere to get the homotopy between $f,g$ (and any intermediate maps for a definite $t$). In this case we stop at the midpoint to form an even function, to render an even degree. – Anonymous Coward Jul 08 '13 at 16:57
  • Nice answer +1. – Elchanan Solomon Jul 08 '13 at 22:24
  • @MikhailKatz the answer is incomplete since $\mathbb{RP}^n$ is orientable only when $n$ is odd. To complete the proof one needs to consider the case in which $n$ is even, then $f$ would be homotopic to its precomposition with the antipodal map, yielding a contradiction. – Aaron Maroja Feb 19 '17 at 20:31