Let \(h: S^1 \to S^1\) be continuous and antipode-preserving with \(h(b_0) = b_0\). Show that \(h_*\) carries a generator of \(\pi_1(S^1, b_0)\) to an odd power of itself. [Hint: If \(k\) is the map constructed in the proof of Theorem 57.1, show that \(k_*\) does the same.]
Solution
Identify \(S^1\) with \(\mathbb R/\mathbb Z\) by \(t\mapsto e^{2\pi it}\), and take \(b_0=1\). Since \(h(b_0)=b_0\), lift \(h\) to a continuous map \[ H:\mathbb R\to\mathbb R \] with \(H(0)=0\) and \[ e^{2\pi iH(t)}=h(e^{2\pi it}). \]
Antipode preservation gives \[ h(e^{2\pi i(t+1/2)})=-h(e^{2\pi it}), \] so \[ H(t+1/2)-H(t)\in \frac12+\mathbb Z. \] The left side is continuous in \(t\), while \(\frac12+\mathbb Z\) is discrete; hence it is constant. Thus for some \(m\in\mathbb Z\), \[ H(t+1/2)=H(t)+m+\frac12 \] for all \(t\). Applying this twice, \[ H(t+1)=H(t)+2m+1. \] Therefore the degree of \(h\) is the odd integer \(2m+1\). If \(a\) denotes the standard generator of \(\pi_1(S^1,b_0)\cong\mathbb Z\), then \[ h_*(a)=a^{\,2m+1}, \] an odd power of the generator, as required.