I'm trying to show that if $n>1$ then any continuous map $f: S^n\to S^1$ is nullhomotopic.
I can only prove this when f is not surjective. I guess this has something to do with the fact that $\pi_1(S^n)$ is trival for $n\geqslant2$ but I can't see how to proceed.
Thanks in advance!
Here is the answer by Lord Shark the Unknown in more detail.
Leaving basepoints implicit, the lifting criterion for covering maps gives you that for any covering space $p:\widetilde{X}\to X$ and a map $f:Y\to X$, with $Y$ path connected and locally path connected, then a lift $\tilde{f}:Y\to \widetilde{X}$ exists if and only if $f_*(\pi_1(Y))\subset p_*(\pi_1(\widetilde{X}))$.
For the simply connected cover, $\pi_1(\widetilde{X})\cong 0$, and hence a lift exists provided $f_*(\pi_1(Y))\cong 0$, which is trivially satisfied if $Y$ is simply connected. Taking $Y=S^n$, $X=S^1$ and $\widetilde{X}=2\pi i\mathbb{R}\overset{\rm exp}{\longrightarrow}S^1$ the universal cover of $S^1$ gives you a lift $\widetilde{f}:S^n\to 2\pi i \mathbb{R}$. Since $\mathbb{R}$ is contractible, you can choose a nullhomotopy for $\tilde{f}$, which (after post-composing with ${\rm exp}$) will induce a nullhomotopy for $f$.