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Given a smooth map $f:S^1\to S^1$, there is a smooth map $g:\Bbb R\to\Bbb R$ such that $f(\cos t,\sin t)=(\cos g(t),\sin g(t))$ and $g(2\pi)=g(0)+2\pi q$ for some $q\in\Bbb Z$.

I'm fairly sure one can prove this without using algebraic topology/path lifting. The idea is that $g$ measures the "signed distance along $S^1$" that the image of $f$ travels (at least on $[0,2\pi]$). One then extends $g$ to $\Bbb R$ by setting $g(t+2\pi)=g(t)+2\pi q$. I don't know how to measure that in a smooth manner (I can't even get it to be continuous).

Is there an elementary proof of this?

Ryan Unger
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    @Huy The linked question involves path lifting, which is not considered elementary in this context. – Ryan Unger Jul 19 '16 at 21:30
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    To the close-voters: neither of the answers to the suggested duplicate answers this question. – Rob Arthan Jul 19 '16 at 21:39
  • By the way, even if one doesn't use a theorem that says lifts of paths exist, the exercise is in fact asking you to prove a lift of a certain path exists. – anon Jul 19 '16 at 21:49

2 Answers2

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Pick a diffeomorphism $S^1 \to \Bbb R/\Bbb Z$ and work in the latter space instead for convenience of notation. Now what you want to show is that if $f: \Bbb R/\Bbb Z \to \Bbb R/\Bbb Z$, there exists a lift $\tilde f: \Bbb R \to \Bbb R$ such that $[\tilde f(x)] = f([x])$. Your idea that you should use distance traveled is a perfectly good one; define, for $x \in \Bbb R, \tilde f(x) = \int_0^x f'(t)dt$. Note that, while $f(x)$ is not a well-defined real number, the derivative $f'(x)$ is!

(In fancy language, I'm using a trivialization of $T(\Bbb R/\Bbb Z)$ to get a well-defined real number $f'(x) \in \Bbb R$ instead of a tangent vector.)

You should verify that this formula satisfies the requisite properties.

  • I also note that this is literally an instantiation of the path-lifting lemma, which is in any case not difficult to prove in any level of generality; though I don't know a way off the top of my head to use this particular proof. –  Jul 19 '16 at 21:47
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    Reason for the downvote, please? – Rob Arthan Jul 19 '16 at 21:48
  • This is a neat trick - one is using a trivialization of $TS^1$ to identify the fibers canonically with a copy of $\Bbb R$, which then enables one to write down $\tilde{f}(x)$. Thanks for this answer. – Balarka Sen Jul 19 '16 at 22:08
  • That's cute and answers the OP's question, but the result holds for arbitrary continuous $f$ and isn't difficult to prove in the more general setting. – Rob Arthan Jul 19 '16 at 22:41
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    @RobArthan I agree with you. –  Jul 19 '16 at 22:42
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Hint: you can't really avoid a theorem about path lifting, but the particular case you need isn't too hard to prove from first principles and requires no assumptions other than continuity. Have a look at Chapter 1 of May's Concise Introduction to Algebraic Topology.

Rob Arthan
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