Prove that there does not exist an surjective and continuous map from $S^1= \{(x,y) \in \mathbb{R}^2:x^2+y^2=1 \}$ (provided with the metric induced by the usual $ \mathbb{R}^2$) in $ \mathbb{R}$.
How can I extend my proof below to prove that there does not exist an surjective and continuous map from $S^1$ to $\mathbb{R}$?
Let us take $K=[0,2π]\subset \mathbb{R}\rightarrow S^1$, $L=S1=\{(x,y)\in \mathbb{R}^2:x^2+y^2=1\}$ and $Φ:[0,2π]\rightarrow \mathbb{R}^2$, given by
$$
Φ(t)=(\cos t,\sin t)
$$
Proof: We have that $[0,2π]$ and $S^1$ are compact and $Φ:[0,2π]\rightarrow S^1$ is continuous and surjective. Let $g:[0,2π]\rightarrow \mathbb{R}^n$ be continuous such that $g(0)=g(2π)$. Let’s define $f:S^1\rightarrow \mathbb{R}^n$ from $g$ as: $f(\cos t,\sin t)=g(t)$, as $g(0)=g(2π)$, $f$ is well defined. We note that $g=f(g(Φ))$ is continuous.
Thus, to define a continuous map in the circle $S^1$, it is sufficient to define it in $[0,2π]$ so that it is continuous in this interval and assumes equal values in the extremes $0$ and $2π$.
Best Answer
Perhaps the method that you started could be finished, but here's a quick way. It will use a common result, that the image of a compact space under a continuous map is compact. I suspect one of the above comments was hinting at this.
Suppose there were a continuous surjective map $f:S^1\rightarrow\mathbb{R}$. We know $S^1$ is compact, so by the above stated result, $f(S^1)=\mathbb{R}$ is compact. This is a contradiction, however, since $\mathbb{R}$ is not compact.