Let $g: S^2 \to S^2$ be continuous and $g(x) \neq g(-x)\ \forall x$. Prove that $g$ is surjective.

Question

Let $g: S^2 \to S^2$ be continuous and $g(x) \neq g(-x)\ \forall x$. Prove that $g$ is surjective.

The hint that if $p \in S^2$, then $S^2 - \{p\}$ is homeomorphic to $\mathbb{R^2}$.

It is pretty obvious that one can use the Borsuk-Ulam Theorem, to prove that if we remove a point in the co-domain, this is homeomorphic to $\mathbb{R}^2$, so there must be an $x$ such that $g(x) = g(-x)$ which is obviously not possible with our assumption.

The main question I ask myself is, why does it fail for $g(X) = A \subset S^2$, where $A$ can be a set in the form of $S^2 - \bigcup_\alpha \{p_\alpha\}$. This is not necessarily homeomorphic to $\mathbb{R^2}$ or is it?

Kees

Answer 1

If $g(S^2) = A \subset S^2$, then $A \subseteq S^2\setminus\{p\}$ for some $p \in S^2$. If $\varphi : S^2\setminus\{p\} \to \mathbb{R}^2$ denotes stereographic projection, then $\varphi\circ g : S^2 \to \mathbb{R}^2$. By the Borsuk-Ulam Theorem, there is $x \in S^2$ such that $(\varphi\circ g)(x) = (\varphi\circ g)(-x)$, and therefore $g(x) = g(-x)$.

If $S^2\setminus A$ is more than a single point, then $\varphi\circ g$ is not a surjective map $S^2 \to \mathbb{R}^2$, but the Borsuk-Ulam Theorem applies to any continuous map $S^2 \to \mathbb{R}^2$, not only surjective ones.