If $n\in \Bbb{N}, n\geq 2$ and $f : \Bbb{S}^n \hookrightarrow \Bbb{S}^1$ is continuous, then $f$ is null-homotopic.
This is a fact that appears as trivial on my class-notes, but I don't see why it is. I guess that it is related to the fact that $\Bbb{S}^n$ is simply connected and so $f_* : \pi_1 (\Bbb{S}^n) \hookrightarrow \pi_1 (\Bbb{S}^1)$ is the trivial homomorphism, which implies that for every path $\alpha : [0,1] \hookrightarrow \Bbb{S}^n$ we have that $f \circ \alpha$ is homotopic to a constant path on $\Bbb{S}^1$
I have tried using this to find a homotopy between $f$ and a constant map, but I am not reaching any point, so any help would be appreciated.
Let $\tilde{f}:\mathbb{S}^n\rightarrow \mathbb{R}^1$ be the induced map. It is clear that $\tilde{f}$ is null-homotopic. So $f$ is null homotopic.