For every map $\gamma:[0,1]\to S^1$, show that there is a map $\hat\gamma:[0,1]\to\mathbb{R}$ with $\gamma(t)=P(\hat\gamma(t)),$ where $P(s)=(\cos 2\pi s,\sin 2\pi s)\in S^1$.

I want to prove this proposition using Lebesgue covering lemma, which states that for a compact metric space with open cover $\cup V_\lambda$, there is an $\epsilon>0$ such that for every $x$, its $\epsilon$-ball is completely contained in some sets in the cover.

In order to use this theorem, I suppose we first need to find an open cover for $[0,1]$. So I considered a finite cover of $S^1$ (which is possible, since $S^1$ is compact0, then the preimage of these sets under $\gamma^{-1}$ form an open cover for $[0,1]$. But then I don't know how to proceed.

Any thought would be helpful. Thank you very much.

  • $\begingroup$ Why do you want to use Lebesgue Covering Lemma? Isn't $P$ a homeomorphism already? $\endgroup$ – Alp Uzman Jun 5 '15 at 2:03
  • $\begingroup$ @Uzman Because that's what the question is. $\endgroup$ – Toujou Jun 5 '15 at 2:05
  • $\begingroup$ @Uzman $P : \mathbb{R} \to S^1$ is certainly not injective, so not a homeomorphism. It is a local homeomorphism, though. $\endgroup$ – André 3000 Jun 5 '15 at 2:27
  • $\begingroup$ p. 32 of these notes goes through the proof. $\endgroup$ – André 3000 Jun 5 '15 at 2:33
  • $\begingroup$ Right! I automatically restricted $P$ to some unit interval.. $\endgroup$ – Alp Uzman Jun 5 '15 at 3:39

You need to start with a particular cover of $S^1$. Compactness says that any such cover has a finite subcover, not just that $S^1$ has some finite cover! (Every space admits finite covers.)

The trick is to cover $S^1$ with sets $U_1,U_2$ such that the restriction $p:p^{-1}U_i\to U_i$ is a local homeomorphism. A simple choice is $U_1=\{e^{2\pi i\theta},\theta\in(0,1)\}$ and $U_2=\{e^{2\pi i\theta},\theta\in (-1/2,1/2)\}$. The inverse image $p^{-1}U_1$, for instance, is $\mathbb{R}\setminus \mathbb{Z}$, and on each interval $(m,m+1), p$ is a homeomorphism onto $U_1$.

Now via Lebesgue covering you get small intervals $[k/N,(k+1)/N]$ such that $\gamma([k/N,(k+1)/N])\subset U_i$ for one of the $i$s. Then there are infinitely many liftings of $\gamma$ restricted to $[k/N,(k+1)/N]$ to $\mathbb{R}$: just invert any of the homeomorphisms $p:(m,m+1)\to U_1$ or $p(m-1/2,m+1/2)\to U_2$, as appropriate. Which interval you lift into is determined by $\tilde\gamma(k/N)$. So to lift all of $\gamma$ you just pick any lift you want when $k=0$. That determines the lift when $k=1$, because $\tilde\gamma(1/N)$ is already chosen; that lift determines the lift of $[2/N,3/N]$, because $\tilde\gamma(2/N)$ is already chosen, and so on: the entire lift is uniquely determined by $\tilde\gamma(0)$.


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