Local diffeomorphism is diffeomorphism provided one-to-one.

Isn't this false, though? For instance, consider the map of $(0,1)\to S$ where $S\subset\mathbb{R}^2$ is the set indicated by the black line in the picture

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Follow up: Is this not a counter example for the reason that $S$ itself is not a manifold, and hence the exercise does not apply to the map $(0,1)\to S$?

  • 1
    $\begingroup$ Yes, that's right; $S$ isn't a manifold so the exercise doesn't apply. $\endgroup$ – Qiaochu Yuan May 10 '18 at 18:18

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