Let \(S^1\subseteq\RR^2\) denote the unit sphere (with the subspace topology), and \(\RR\) the real line with the usual topology. Show that for every continuous map \(f:S^1\to\RR\) there is an \(x\in S^1\) with \(f(x)=f(-x)\).
Solution
Define \(g\colon S^1\to\RR\) by \(g(x)=f(x)-f(-x)\); it is continuous because \(f\) and \(x\mapsto-x\) are.
1. \(g(-x)=-g(x)\) for every \(x\in S^1\).
\(g(-x)=f(-x)-f(x)=-g(x)\).
2. Q.E.D.
Fix \(x_0=(\cos\theta_0,\sin\theta_0)\in S^1\). If \(g(x_0)=0\), then \(x=x_0\) works. Otherwise, by step 1, \(g(x_0)\) and \(g(-x_0)\) are nonzero with opposite signs. Let \(\gamma(t)=(\cos(\theta_0+\pi t),\sin(\theta_0+\pi t))\), a path in \(S^1\) from \(x_0\) to \(-x_0\). By the intermediate value theorem applied to \(g\circ\gamma\) on \([0,1]\), there is \(t^*\) with \(g(\gamma(t^*))=0\), and \(x=\gamma(t^*)\) satisfies \(f(x)=f(-x)\).