Show that if \(g: S^2 \to S^2\) is continuous and \(g(x) \neq g(-x)\) for all \(x\), then \(g\) is surjective. [Hint: If \(p \in S^2\), then \(S^2 - \ts{p}\) is homeomorphic to \(\mathbb{R}^2\).]
Solution
Suppose, toward a contradiction, that \(g\) is not surjective. Choose \[ p\in S^2-g(S^2). \] Since \(S^2-\{p\}\) is homeomorphic to \(\mathbb R^2\), choose a homeomorphism \[ \phi:S^2-\{p\}\longrightarrow\mathbb R^2. \] Then \[ F=\phi\circ g:S^2\to\mathbb R^2 \] is continuous. By the Borsuk–Ulam theorem there exists \(x\in S^2\) with \[ F(x)=F(-x). \] Since \(\phi\) is injective, this implies \[ g(x)=g(-x), \] contrary to the hypothesis. Therefore \(g\) must be surjective.