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?
Best Answer
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).