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Let’s say I have an immersion $f:\Omega \to \mathbb{R}^n$, with $\Omega$ an open set of $\mathbb{R}^n$. It is a local diffeomorphism (because its differential is injective and thus bijective), so it is an open map. If I restrict the image and consider it to be not $\mathbb{R}^n$, but $f(\Omega)$, it would also be surjective. The only way it could not be a diffeomorphism is to not be injective, so I was wondering: are there counterexamples? Maybe that work on all dimensions (I don’t see how to give a counterexample in dimension 1, but it might be a special case where there are actually no counterexamples, while in greater dimensions there are)

tommy1996q
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  • Could one not just take an open tubular neighbourhood of a self-intersecting smooth curve in $\mathbb{R}^n$? The only case this wouldn't work in is dimension 1, as you say. – Dan Rust Jul 11 '19 at 21:14
  • I haven’t studied open tubular neighborhoods yet (I will very soon, though!) – tommy1996q Jul 11 '19 at 21:27
  • By the way, I originally wrote the post confusing the word embedding with immersion, and I have discovered there is no “embedding” tag! Isn’t it a very important topic in differential geometry? – tommy1996q Jul 11 '19 at 21:29
  • All I mean is, take an open cylinder around the $x$-axis. Map the $x$-axis to a self-intersecting smooth curve, and then extend that map to the entire cylinder in a smooth way (if you really wanted, you could write a formula, but it's unnecessary). – Dan Rust Jul 11 '19 at 21:30
  • I should also mention of course that Rolle's theorem ensures that any smooth non-injective function $\mathbb{R}\supset U \to \mathbb{R}$ for connected open interval $U$ must have a point with zero derivative, so it can't be an immersion. – Dan Rust Jul 11 '19 at 21:40
  • Yeah, the thing about Rolle’s theorem is exactly what I thought – tommy1996q Jul 11 '19 at 22:04
  • So you take advantage of the fact of being in dimension at least 2 to do this self-intersection without having to have derivative 0. Not sure how to extend things smoothly in the intersection, though. Shouldn’t it give some trouble? – tommy1996q Jul 11 '19 at 22:06
  • You could even have a surjective local diffeo from $R^n$ to itself which is not a diffeomorphism. – Moishe Kohan Jul 11 '19 at 23:41

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The fundamental counterexample which shows up repeatedly as you proceed is the exponential mapping $f\colon \Bbb C\to\Bbb C-\{0\}$, $f(z)=e^z$. (In real coordinates, this is, of course, $f(x,y) = (e^x\cos y,e^x\sin y)$.) Each point has infinitely many preimages, but the map is a local diffeomorphism at every point of the domain.

Ted Shifrin
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