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How do I prove that $f:\mathbb R\to \mathbb S^1$, $f(x)=(\cos2\pi x,\sin2\pi x)$ is an open map? I'm thinking about a simple solution.

Maybe it's a hard question

I need help here

Thanks

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    Open subsets of $\mathbb R$ are unions of open intervals, and $f(\cup I_i)=\cup f(I_i)$, so it suffices to show that $f(I)$ is open in $S^1$ when $I$ is an open interval.2012-11-22
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    @JonasMeyer why $f(\cup I_i)=\cup f(I_i)$?2012-11-22
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    Rafael: That is a property that always holds for images and unions, regardless of the function. You can prove it.2012-11-22
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    @JonasMeyer ha, yes of course, sorry2012-11-22

2 Answers 2

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As Jonas Meyer pointed out in the comments, it suffices to show that $f(I)$ is open in $S^1$ for an open interval. Moreover, it should not be hard to convince yourself that we may assume $I \subset [0,1)$ by $2\pi$-periodicity and the fact that if $I$ contains an interval of the form $[a,b)$ with $b-a = 1$, then its image is all of $S^1$. Let $I = (a,b)$. Then the image $f(I)$ is precisely those points on $S^1$ with angle between $2\pi a$ and $2\pi b$. Consider the open set $O$ in $\mathbb{R}^2$ given in polar coordinates by $O = \{(r,\theta): r > 0, 2\pi a < \theta < 2 \pi b\}$. Then $f(I) = S^1 \cap O$, which is open in the relative topology on $S^1$.

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    when you said "Let $I=(a,b)$. Then the image $f(I)$ is precisely those points on $S^1$ with angle between $a$ and $b$." I think it's not write. Thank you for your answer.2012-11-22
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    another problem is how we can prove that $f(I)=S^1\cap O$?2012-11-22
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    Recall that I assumed (without loss of generality) that $I$ was contained in $[0,1)$. Note that $f$ wraps $[0,1)$ around the circle in such a way that $f(\theta)$ is precisely the point on $S^1$ with angle $\theta$. Then $f$ maps $(a,b)$ to the set of points with angles between $a$ and $b$. Of course, you can think of this as the set of all nonzero points in $\mathbb{R}^2$ with angle between $a$ and $b$ (this was the set $O$) restricted to $S^1$. Is that any clearer?2012-11-22
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    take for example $\theta = 1/2$ the angle of $f(1/2)$ is $1/2$?2012-11-22
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    Sorry, I edited it, the angle of $f(\theta)$ should be $2\pi \theta$. So if $\theta = 1/2$, we should have an angle of $\pi$, but $f(1/2) = (\cos \pi, \sin \pi) = (-1,0)$, which indeed has angle $\pi$.2012-11-22
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    thank you for the edit, it's correct now.2012-11-22
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    it remains one problem, how can we prove that $f(I)=S^1\cap O$? I think I've already tried to prove it, I thought difficult.2012-11-22
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    $f(I)$ is the set of points on the circle with angles between $2 \pi a$ and $2 \pi b$. $O$ contains all the points in $\mathbb{R}^2$ that satisfy that inequality (besides for the origin). If you restrict that to the circle, you get the set of points on the circle between those angles. It should follow immediately from the definitions.2012-11-22
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    yes, I understand your answer, but in order to be really formal, we have to use geometry or another tool, because how can you know when you restrict to the circle, you get the set of points on the circle between those angles? I know it's intuitive, but it's not a formal proof.2012-11-23
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    If a point is on the circle between those two angles, it is in the restriction by definition. Conversely, if a point is in the restriction, it is on the circle and between those angles. Thus the two sets are the same. I assure you that this constitutes a formal proof.2012-11-23
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    what is the definition of arc are you using?2012-11-23
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    I do not think I made any mention of an arc.2012-11-23
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We identify $\mathbb{S}^1$ with $\mathbb{T} = \{z \in \mathbb{C}|\ |z| = 1\}$. Then $f$ becomes $f(x) = exp(i x)$. Hence $f$ is a continuous homomorphism of topological groups. $f$ factors to $\mathbb{R} \rightarrow \mathbb{R}/2\pi \mathbb{Z} \rightarrow \mathbb{T}$. Since $\mathbb{R}/2\pi \mathbb{Z}$ is compact, $\mathbb{R}/2\pi \mathbb{Z} \rightarrow \mathbb{T}$ is an isomorphism. Since $\mathbb{R} \rightarrow \mathbb{R}/2\pi \mathbb{Z}$ is open, $f$ is open.