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Let $f\colon S^1\to S^1$ be any smooth map. Prove that there exists a smooth map $g\colon\mathbb{R}\to\mathbb{R}$ such that $f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))$ and satisfying $g(2π)=g(0)+2\pi q$.

The book told me to show that $g(2\pi)=g(0)+2\pi q$, then extend $g(t+2\pi) =g(t)+2\pi q$.

But I don't understand what they mean. How can I get function $g$ from $f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))$, and what is $g$?

ChoF
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Diane Vanderwaif
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    It seems to me, knowing a little bit about such problems, that $q$ must be an integer, any integer, so that the statement $g(t + 2\pi) = g(t) + 2 \pi q$ means that $e^{ig(t + 2\pi)} = e^{i(g(t) + 2\pi q)} = e^{ig(t)}$; $g(t + 2\pi)$ and $g(t) + 2\pi q$ map to the same point of $S^1$. As for establishing the existence of such $g(t)$, I have to think more about it. Cheers. – Robert Lewis Oct 19 '14 at 19:15

2 Answers2

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Take a particular value of $t$; compute $(\cos(t), \sin(t))$. Evaluate $f$ at that point, to get a new point $(\cos u, \sin u)$, where $u$ is defined only up to multiples of $2 \pi$. Now tentatively define $g(t) = u$. This gives a definition, but it's not necessarily continuous, right? (Or smooth). But that's the main idea: $g(t)$ is the angle for $f(\cos t, \sin t)$. And $q$ is an integer that you might have to add to make things work out smoothly at the endpoints of an interval.

To really solve the problem properly, you need to the path lifting theorem. I can write out detail if you like, but since you seemed to be confused by the basics, I thought I'd start here.

One small suggestion: write '\cos' rather than 'cos' in math to make formatting look nicer.

Details: Let $I$ denote $[0, 2\pi]$.

Let $$ h: I \to S^1: t \mapsto f(\cos t, \sin t). $$

Since $h: I \to S^1$ is continuous, there's a map $H:I \to \mathbb R$, the universal cover of $S^1$ such that $h = p \circ H$, where $p : \mathbb R \to S^1: x \to (\cos x, \sin x)$ is the universal covering map. (This is because of the path lifting theorem.)

Now compose $H$ with the map $u \mapsto u \bmod 2\pi$ to get a map $$ K: I \to \mathbb R : u \mapsto H(u \bmod 2\pi) $$ which is evidently continuous as well. $K$ is the map $g$ that you're looking for.

You now need to show that everything works out nicely, but that's fairly straightforward. The number $2\pi q$ that you're looking for is just $H(2 \pi) - H(0)$, which is an integer multiple of $2 \pi$ because of the conclusions of the path-lifting theorem, one of which is that $p(H(0)) = p(H(2\pi))$.

For smoothness, you've got several maps that are compositions of each other, and two out of every 3 are known to be smooth; it's not too hard to conclude that the third is as well.

John Hughes
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  • Can you state the path lifting theorem for me please? I'm not sure I'm familiar with the name. – Diane Vanderwaif Oct 19 '14 at 19:57
  • Look at http://en.wikipedia.org/wiki/Homotopy_lifting_property; in the special case where $X$ is an interval, this is called the "Path Lifting Property". That definition just says what the Homotopy Lifting Property is; the Path Lifting Theorem says that covering spaces have the Path lifting Property. (They also have the homotopy lifting property; in fact, they're sort of the archetype for fibrations.)

    [1]:

     – John Hughes Oct 19 '14 at 22:11
  • I have a question: how can one show that $f(\cos t, \sin t)= (\cos g(t), \sin g(t))$ or $g(2\pi)-g(0)=2\pi q$ if there is no explicit formula for $H$ and hence for $g$? – user557 Mar 18 '18 at 00:16
  • To clarify my comment above: the first equality I stated probably can be verified without the knowledge of an explicit formula for $g$. Namely, it follows from the commutativity (of the corresp. diagram) that $f(\cos t, \sin t)=(\cos H(t), \sin H(t))$, and then one writes $\mod 2$ to the right of $t$ everywhere to get the desired formula (if I understand correctly). But to compute $g(2\pi)-g(0)$ shouldn't one know a formula for $g$? How does $g(2\pi)-g(0)=2\pi q$ follow? Also, a little remark: I think your $K$ has domain $[0, 2\pi]$ (since $H$ does), and $g$ is a suitable extension of $K$. – user557 Mar 18 '18 at 01:00
  • Thanks for the tip about the domain of $K$; I'll fix that. As for $g(2\pi) - g(0)$, what's missing in the question is any condition on $q$, but the point is that $g(2\pi) - g(0)$ is $2 \pi$ times some integer $q$. This follows from the path lifting theorem (the last sentence of my second-to-last paragraph). Without constructing $H$, you can't compute $q$, but you know it's an integer by the mere existence of an $H$ with the required properties, an existence guaranteed by the Path Lifting Theorem. – John Hughes Mar 18 '18 at 12:51
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$$f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))=(\cos(g(t)+2\pi q),\sin(g(t)+2\pi q))$$ On the other hand $$f(\cos(t),\sin(t))=f(\cos(t+2\pi q),\sin(t+2\pi q))=(\cos(g(t+2\pi q)),\sin(g(t+2\pi q)))$$ From these two results you get $$g(t)+2\pi q=g(t+2\pi q)$$ where $q\in\mathbb{Z}$. $f$ can be thought as a map taking points from the unit circle and mapping them to another unit circle through the rule $$f(\cos(t))=\cos(g(t))$$ and $$f(\sin(t))=\sin(g(t))$$ For example of $g(t)=t$ is such a function and $f$ is the identity map.

Arian
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  • if you defined $g(t)=t$ then $g(2\pi)= 2\pi= 0 +2\pi *(1)=g(0) +2\pi q$ for $q=1$. Do I understand It correctly? I'm known for misunderstand things easily. – Diane Vanderwaif Oct 19 '14 at 19:34
  • @DianeVanderwaif: Yes because $2\pi=g(2\pi)=g(2\pi+0)=g(0)+2\pi=0+2\pi=2\pi$. – Arian Oct 19 '14 at 19:37