TITLE: Smooth path lifting on $S^1$
QUESTION [3 upvotes]: 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?

REPLY [7 votes]: 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.
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