= Solution
Suppose an odd continuous map $f:S^2\to S^1$ existed. Restrict to the equator, parametrized by $e^{i\theta}$. Choose a continuous real lift $F$ with $f(e^{i\theta})=e^{iF(\theta)}$. Oddness gives $F(\theta+\pi)-F(\theta)=\pi+2\pi k$ for a fixed integer $k$, since the integer-valued difference is continuous. Thus $F(\theta+2\pi)-F(\theta)=2\pi(1+2k)$, and the equatorial map has odd <degree of a continuous mapping>, hence nonzero degree. But it extends over a hemisphere, which is a disk, so the equatorial loop is null-homotopic and has degree zero. This contradiction proves the dimension-two <Borsuk-Ulam theorem> in the stated form.
If a continuous injection $h:S^2\to\mathbb R^2$ existed, then $h(x)-h(-x)$ would never vanish. Normalization would define
$$
f(x)=\frac{h(x)-h(-x)}{|h(x)-h(-x)|}\in S^1,
$$
a continuous odd map, already ruled out. In particular \b[the sphere cannot be homeomorphic to any planar subset].
Back to article page
