 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 definea continuous odd map, already ruled out. In particular the sphere cannot be homeomorphic to any planar subset.