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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.

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    $\begingroup$ Do you know about compactness? $\endgroup$ Commented Dec 3, 2016 at 16:46
  • $\begingroup$ $S_1$ is compact. $\endgroup$
    – zhw.
    Commented Dec 3, 2016 at 16:47
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    $\begingroup$ 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. $\endgroup$ Commented Dec 3, 2016 at 17:08
  • $\begingroup$ @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. $\endgroup$ Commented Dec 4, 2016 at 5:26

1 Answer 1

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

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    $\begingroup$ (and the continuous image of a compact set is compact) $\endgroup$ Commented Dec 3, 2016 at 16:52
  • $\begingroup$ It's been very long since I have done a Topology course. Now that I look at it, I realise that it was a very basic question. Thank you! $\endgroup$ Commented Dec 3, 2016 at 16:57
  • $\begingroup$ @Prince No problem. $\endgroup$ Commented Dec 3, 2016 at 16:59

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