Borsuk–Ulam theorem

Remarks

An equivalent form says that there is no continuous odd map $g \colon S^n \to S^{n-1}$, that is, no map with $g(-x) = -g(x)$ for all $x$. Indeed, if $f(x) \neq f(-x)$ for all $x$, then $g(x) = (f(x) - f(-x)) / \norm{f(x) - f(-x)}$ is such a map. One proof uses the fact that an odd map $S^{n-1} \to S^{n-1}$ has odd degree. The restriction of an odd map $S^n \to S^{n-1}$ to the equator is an odd map that extends over a hemisphere. So it is null-homotopic and has degree $0$, a contradiction.

Example: for $n = 1$, let $h(x) = f(x) - f(-x)$ on the circle. Then $h(-x) = -h(x)$, so $h$ takes values of both signs or vanishes, and by the intermediate value theorem it has a zero.

Example of use: at any moment there are two antipodal points on the Earth with the same temperature and the same air pressure, if both are continuous functions on $S^2$. The theorem fails for maps $S^n \to \R^{n+1}$: the inclusion $S^n \subset \R^{n+1}$ takes different values at $x$ and $-x$.

Going deeper

For a student

The theorem is the starting point of the use of topology in combinatorics and geometry. It implies the ham sandwich theorem, the Lusternik–Schnirelmann covering theorem, and Lovász's computation of the chromatic number of Kneser graphs. Its combinatorial counterpart is Tucker's lemma, from which it can also be proved.