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I'd like help for the problem:

Let the additive group $2\pi \mathbb{Z}$ act on $\mathbb{R}$ on the right by $x · 2\pi n = x+2\pi n$, where $n$ is an integer. Show that the orbit space $\frac{\mathbb{R}}{2\pi n\mathbb{Z}} $ is a smooth manifold.

I proved that $\frac{\mathbb{R}}{2\pi n\mathbb{Z}} $ is hausdorff and second countable, but I don't know how to find an atlas, I was thinking about $ \psi([x])=e^{ix}$ but I don't know how to show that this is a homeomorphism.

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    Keep in mind the geometric interpretation that $\mathbb R/2\pi\mathbb Z$ is a circle.2012-02-26
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    A continuous bijection between compact Hausdorff spaces is an homeo.2012-02-26
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    @MarianoSuárez-Alvarez I'm under the impression that that's a rather hefty theorem though.2012-02-26
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    @AlexBecker the idea is to conclude that,so suppose I don't know this.2012-02-26
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    @MarianoSuárez-Alvarez How does one know that $\frac{\mathbb{R}}{2\pi \mathbb{Z}} is compact?2012-02-26
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    It is the image of $[0,2\pi]$ under the projection map.2012-02-26
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    Alex: not really. Since the function $f$ we start with is bijective, it has an inverse, so we want to show it is continuous: it is more or less obvious that it is enough to show that the image under $f$ of a closed set is closed. Closed sets in the domain are compact, the continuous image of a compact set is compact, and in the codomain compacts are closed. In any case, I'd expect this little piece of general topology to be known by anyone who knows the definition of a smooth manifold... His life *will* be hard if not!2012-02-26
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    @AlexBecker I think the that the "theorem" is a consequence of the definition of homeomorphism2012-02-27
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    @MarianoSuárez-Alvarez Sure! Thanks for the comments.2012-02-27

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