Suppose that $\pi_1(X)$ is a finite group. Show that any map $f:X \to S^1$ is nullhomotopic.
My attempt:
Since $\pi_1(X)$ is finite and $\pi_1(S^1)=\mathbb{Z}$ torsion-free, then the induced homomorphism $f_*: \pi_1(X) \to \pi_1(S^1)$ has to be trivial. Therefore it is homotopic to a constant map and hence by definition nullhomotopic.
Is my reasoning correct? I have seen the solution to this problem using the covering spaces, lifting $f$ to $\mathbb{R}$ and then using the fact that $\mathbb{R}$ is contractible, but is this additional machinery really needed?
Best Answer
Fancy: $\pi_1(X)$ finite implies $H_1(X)$ finite implies $0 = Hom(H_1(X), \mathbb Z) = H^1(X, \mathbb Z) = [X, K(\mathbb Z, 1)] = [X, S^1]$.