There does not exist a continuous surjective function from $S^1$ onto $\mathbb R$.

How can we go about proving that there does not exist any continuous surjective function from $S^1$ onto $\mathbb R$. Here $S^1$ is the unit circle in $\mathbb R^2$.

If it were bijective function, we could have just invoked the result that continuous image of a connected set is connected. But as it is only surjective, removing one point from the domain could still leave the image connected.

• Do you know about compactness? – Cheerful Parsnip Dec 3 '16 at 16:46
• $S_1$ is compact. – zhw. Dec 3 '16 at 16:47
• 1. As a interesting question you can also try to show that there is no injective continuous map from $S^1 \to \mathbb R^1$.(Infact $1$ can be replaced with any natural number but the proof for general $n$ is bit involved). 2. Also continuous image of any connected space is connected.(Don't require any bijectivity) but this result does not tell you anything as $\mathbb R$ is connected. – Arpit Kansal Dec 3 '16 at 17:08
• @ArpitKansal I was going to invoke connected space argument by removing one point from $S^1$ and restricting the function to the new subset. Removing one point from $\mathbb R$ would leave it disconnected. I will try to prove the first part. – Prince Kumar Dec 4 '16 at 5:26

$S^1$ is compact and $\mathbb R$ is not.