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I might misunderstand something. But it seems that if we choose an atlas ($\mathbb{R}, f:x\mapsto x^{1/3}$), there is nothing to check for compatibility, because there is only one in there, hence even though $f$ is not smooth, it still qualifies as an atlas. Also, $f$ is homeomorphism between a real number to a real number. Am I thinking right?

able20
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There is a priori no notion of smoothness on the domain. While this is not smooth with respect to the standard manifold structure on $\mathbb R$, it doesn't matter for the purpose of defining an atlas. You're right that we don't have a compatibility condition to check in this case (perhaps besides the trivial transition function from the given chart to itself), but we still need to make sure that the chart is a homeomorphism onto its image (which is not difficult to see in this case).

Tipping Octopus
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  • Thanks! Just to make sure if I understand.. you mean that the above one is technically atlas according to the definition, right? – able20 Feb 07 '21 at 01:40
  • This atlas defines a differentiable structure for $\mathbb R$ different from the usual one. – GEdgar Feb 07 '21 at 01:45
  • It's a different structure than the usual one, but still diffeomorphic to it (can you guess what will the diffeomorphism be?) – Ivo Terek Feb 07 '21 at 02:27
  • The only chart in this atlas is $f$, so the only transition map is I guess Identity? – able20 Feb 07 '21 at 05:31