Why cannot a smooth (or piecewise linear) map $S^1 \to S^2$ be surjective? There are space-filling curves, but the usual examples have very "twisty" definitions.

UPD A bit of background for this problem. It's part of he proof that all the normal vector fields on $S^1 \subset \mathbb{R}^4$ are homotopic. Once it's proved that every map $S^1 \to S^2$ is homotopic to piecewise linear or smooth that is what is left to prove the statement.

UPD One more thing. This is a first semester set of tasks. Sard's theorem is not exactly what gives intuition besides this problem. Given answer for PL is what I was looking for. ;) It would be great to find a reasoning like this for the smooth case.

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    $\begingroup$ The proposals to close this as off topic are absurd. ${}\qquad{}$ $\endgroup$ Commented Jul 15, 2015 at 20:39
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    $\begingroup$ @MichaelHardy Why? No effort is shown. $\endgroup$
    – user223391
    Commented Jul 15, 2015 at 20:46
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    $\begingroup$ @Michael You are perfectly right and unfortunately I have noticed that analogous proposals occur with unpleasant frequency. Especially for advanced topics in which most closers probably have no clue. I wrote about that in Meta and my post got closed and deleted :-) $\endgroup$ Commented Jul 15, 2015 at 20:48
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    $\begingroup$ @avid19 There is no obvious way to show an effort in exactly this problem. If it was not clear it's all about statement that all normal vector fields on $S^1$ in $\mathbb{R}^4$ (and higher) are homotopic. $\endgroup$
    – Stan
    Commented Jul 15, 2015 at 20:55
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    $\begingroup$ @StanO. for future reference, you would show effort/provide background by either telling us any thoughts you've had about the problem, fruitful or not, or giving us some sense of how much topology and differential topology you know. $\endgroup$ Commented Jul 15, 2015 at 21:39

1 Answer 1


It may be easier for a PL map than a smooth one: the image of a PL curve lies in finitely many great circles. But no topological space is the union of finitely many nowhere dense sets (or for an unnecessarily big hammer use the Baire category theorem.)


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