Show that for $n \geq 2$, every continuous function $f: S^{n} \to S^{1}$ is null-homotopic
This question has been already asked here but I don't understand why it is necessairy to use the universal covering:
Since $S^n$ is contractile for $n \geq 2$ we have that for any given topological space $Z$, and for every continuous function $f: S^n \to Z$, $f \simeq ct_z$ for some $z \in Z$ where $ct_z$ is the constant function $ct_z :S^n \to Z$ such that $ct_Z(x)=z$ for every $x$. Why doesn't this work for this case to show that $f$ is null homotopic?
Best Answer
$S^n$ is not contractible! loops in $S^n$ are contractible, which is to say: $S^n$ is simply connected for $n \geq 2$.
One can check this by observing that $H_n(S^n)$ or $\pi_n(S^n)$ is nontrivial, and noting that it does not have the homotopy type of a point.
Hint: use the fact that $\pi_1(S^n)=0$ for $n \geq 2$ (the easiest way I know to prove this is via cellular approximation) to note that for any map $f:S^n \to S^1$, we have that $f_*: \pi_1(S^n) \to \pi_1(S^1)$ is trivial, and hence a subgroup, so we can apply the lifting lemma to $\mathbb R$ and the covering map $p:\mathbb R \to S^1$.