I have this question of the Hatcher book:
Show that if a path-connected, locally path-connected space $X$ has $\pi_1(X)$ finite, then every map $X\rightarrow S^1$ is nullhomotopic. [Use the covering space $\mathbb{R} \rightarrow S^1$]
I thought it could be solved considering that if I have a map $f\colon X\rightarrow S^1$, then it generates an homomorphism $f_{\ast}\colon \pi_1(X)\rightarrow \pi_1(S^1)=\mathbb{Z}$, but since $\pi_1(X)$ is finite, all the elements of $\pi_1(X)$ are of finite order, then they only can be sent to $0\in \mathbb{Z}$. So $f_{\ast}([\gamma])=0$ for any loop class, so where I use all the other information?
Could anyone tell me where I'm wrong?
Best Answer
Showing that $f_\ast$ is the zero map is an important step, but it isn't enough.
In general, knowing that $f_\ast$ is the zero map does not imply that $f$ is nullhomotopic. Consider the identity map $id:S^2\rightarrow S^2$. Since $\pi_1(S^2)=0$, $id_\ast=0$. However, if $id$ were nullhomotopic, then $S^2$ would be contractible, which it isn't.
In your case, you'll need to use the universal property of covering spaces. Observe that since $f_\ast(\pi_1(X))=0$, it is a subset of $p_\ast(\pi_1(\mathbb{R}))$ where $p:\mathbb{R}\rightarrow S^1$ is the covering space map. Therefore, the map $f:X\rightarrow S^1$ lifts to a map $\tilde{f}:X\rightarrow\mathbb{R}$.
Notice that $\mathbb{R}$ is contractible. Can you take it from here?