Show that if \(X\) is a path-connected, locally path-connected topological space with finite fundamental group \(\pi_1(X, x_0)\), then every continuous map \(f : X \to S^1\) is homotopic to a constant map (nullhomotopic).
Solution
Let \(p\colon\RR\to S^1\), \(p(t)=e^{2\pi it}\), be the universal covering map, and choose \(t_0\in\RR\) with \(p(t_0)=f(x_0)\).
1. The induced homomorphism \(f_*\colon\pi_1(X,x_0)\to\pi_1(S^1,f(x_0))\cong\ZZ\) is trivial.
The group \(\pi_1(X,x_0)\) is finite by hypothesis, so its homomorphic image \(f_*(\pi_1(X,x_0))\) is a finite subgroup of \(\ZZ\). Every nontrivial subgroup of \(\ZZ\) is infinite, so \(f_*(\pi_1(X,x_0))=\{0\}\).
2. The map \(f\) lifts to a continuous map \(\widetilde f\colon X\to\RR\) with \(\widetilde f(x_0)=t_0\) and \(p\circ\widetilde f=f\).
Because \(\RR\) is simply connected, \(p_*(\pi_1(\RR,t_0))\) is the trivial subgroup of \(\pi_1(S^1,f(x_0))\). By step 1, \(f_*(\pi_1(X,x_0))\subseteq p_*(\pi_1(\RR,t_0))\). Since \(X\) is path-connected and locally path-connected, the lifting criterion for covering spaces, applied with the point \(t_0\in p^{-1}(f(x_0))\), gives the lift.
3. The lift \(\widetilde f\) is homotopic to the constant map with value \(t_0\).
Define \(H\colon X\times[0,1]\to\RR\) by \(H(x,s)=(1-s)\widetilde f(x)+s t_0\). This map is continuous, \(H(x,0)=\widetilde f(x)\), and \(H(x,1)=t_0\).
4. Q.E.D.
3.By step 2, \(F(x,0)=p(\widetilde f(x))=f(x)\), and \(F(x,1)=p(t_0)=f(x_0)\) for every \(x\in X\).Hence \(F\) is a homotopy from \(f\) to the constant map with value \(f(x_0)\), and \(f\) is nullhomotopic.