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Can we provide the set $\{(x,y,z)\in\mathbb{R^3}|x^2+y^2=1\}$ with a 2-dimensional manifold structure involving only 1 chart? I can see it with 2 charts with cylindrical coordinates, but not with only one...

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    Well, Hans's answer now handles it. If you restrict to open balls, then you need at least 2 charts. If you restrict to general open subsets of $\mathbb{R}^2$ an annulus (or a unit disc with origin removed) works.2012-11-23

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You can't do it if you want a chart that is homeomorphic to (e.g.) the unit disk in $\mathbb{R}^2$. The latter is homotopy equivalent to a point, but your manifold is not. But you can do it with the unit disk with the origin removed.

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    No, I'm taking points from the punctured disk. In Cartesian coordinates, the map is $(u,v) \mapsto (\frac{u}{\sqrt{u^2 + v^2}}, \frac{v}{\sqrt{u^2 + v^2}}, \tan \left( \pi \sqrt{u^2+v^2} - \pi/2 \right))$.2012-11-23