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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?

Joaquin Liniado
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1 Answers1

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$S^n$ 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 $p:\mathbb R \to S^1$.

Andres Mejia
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