$S^{1}\times (0, 1)$ admits a single smooth chart?

Question

I would like to know if this assertion is true and, in case it is, how to proceed. I already proved that $S^{1}\times [0,1]$ can't be covered by a single chart. But I get stuck in this case.

Edit: as my idea was wrong, I deleted it, so people that look for it won't be confused.

Thanks, for any help.

Answer 1

The map \begin{align}\phi:\Bbb R^2\setminus\{0\}&\to S^1\times(0,1)\\ x&\mapsto \left(\frac{x}{\lVert x\rVert},\frac1\pi\operatorname{arccot}(-\ln \lVert x\rVert)\right)\end{align} is a diffeomorphism. Its inverse is $(\omega,t)\mapsto e^{-\cot(\pi t)}\omega$.