Show that if \(n > 1\), every continuous map \(f: S^n \to S^1\) is nulhomotopic. [Hint: Use the lifting lemma.]
Solution
Let \(n>1\) and \(f:S^n\to S^1\). Choose a basepoint \(x_0\in S^n\), and let \[ p:\mathbb R\longrightarrow S^1,\qquad p(t)=e^{2\pi i t} \] be the universal covering map. Since \(S^n\) is simply connected for \(n>1\), \[ f_*\pi_1(S^n,x_0)=0\subset p_*\pi_1(\mathbb R,\tilde b_0)=0. \] The lifting criterion therefore gives a lift \[ \tilde f:S^n\to\mathbb R,\qquad p\tilde f=f. \] Because \(\mathbb R\) is contractible, \(\tilde f\) is homotopic to a constant map. Composing this homotopy with \(p\) gives a homotopy of \(f\) to a constant map in \(S^1\). Hence every map \(S^n\to S^1\) is nullhomotopic.