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
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 $(0,1)\to S$?
