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This is Problem 3 in Guillemin & Pallock's Differential Topology on Page 18. So that means I just started and am struggling with the beginning. So I would be expecting a less involved proof:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a local diffeomorphism. Prove that the image of $f$ is an open interval and maps $\mathbb{R}$ diffeomorphically onto this interval.

I am rather confused with this question. So just identity works as $f$ right?

  • The derivative of identity is still identity, it is non-singular at any point. So it is a local diffeomorphism.

  • $I$ maps $\mathbb{R} \rightarrow \mathbb{R}$, and the latter is open.

  • $I$ is smooth and bijective, its inverse $I$ is also smooth. Hence it maps diffeomorphically.

Thanks for @Zev Chonoles's comment. Now I realized what I am asked to prove, though still at lost on how.

WishingFish
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2 Answers2

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If $f$ is a local differeomorphism then the image must be connected, try to classify the connected subsets of $\mathbb{R}$ into four categories. Since $f$ is an open map, this gives you only one option left. I do not know if this is the proof the author has in mind.

Bombyx mori
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  • Hello, thanks for your help. I am not familiar with the classification. Would you point some reference? Thanks! – WishingFish Jul 01 '13 at 01:49
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    This is intended for your own exercise to fill in. You can skip the classification to find a better proof yourself as well. – Bombyx mori Jul 01 '13 at 01:50
  • let me think about it, and thanks, zuchongzhi.. – WishingFish Jul 01 '13 at 01:56
  • Oh, do you mean closed, open, half closed half open, and half open and half closed...? – WishingFish Jul 01 '13 at 02:04
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    Yes. Should not be difficult to prove. – Bombyx mori Jul 01 '13 at 03:05
  • Thank you. You sound like a geometer. – WishingFish Jul 01 '13 at 03:40
  • Sorry I'm still a bit uncertain here: so $f$ is open map because it is a diffeomorphism from $\mathbb{R}$? – WishingFish Jul 01 '13 at 03:47
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    $f$ is an open map because locally it is a diffeomorphism, and thus it maps open intervals to open intervals. You can prove this rigorously by bounded the interval in a compact set and consider $f'$'s value on this set. Now, we know $f(A\cup B)=f(A)\cup f(B)$. So use the fact open sets in $\mathbb{R}$ is a disjoint union of open intervals the image must be a union of (not necessarily disjoint) open intervals. $f$ is not necessarily a diffeomorphism if you consider $f=\arctan[x]$, for example. I am not sure if this is the slickest proof possible. – Bombyx mori Jul 01 '13 at 04:16
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    Sorry for providing a wrong example, which is indeed a global differeomorphism. But I think such counter-example should not be difficult to find. – Bombyx mori Jul 01 '13 at 04:28
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    A better way is to prove $f^{-1}$ is continuous. This follows automatically from the fact $f$ is a local homeomorphism. – Bombyx mori Jul 01 '13 at 13:12
  • oh, thank you! But this relies on local diffeomorphism is equivalent to local homeomorphism. Is this trivial? – WishingFish Jul 01 '13 at 17:26
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    No, a local diffeomorphism is not equivalent to a local homeomorphism. Consider a curve $y=x$ when $x\in [0,1]$, $y=2-x$ when $x=[1,2]$. Then $f(x)=y$ is a local homeomorphism but not a diffeomorphism. – Bombyx mori Jul 01 '13 at 18:44
  • oh, ok, thanks. so local diffeomorphism $\Rightarrow$ local homeomeorphism, but not otherwise..? – WishingFish Jul 01 '13 at 19:35
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    A diffeomorphism by definition is equivalent to a map $\mathbb{R}^{n}\rightarrow \mathbb{R}^{n}$ which is a homeomorphism, and $\det(f)\not=0$ at any point. Local diffeomorphism just means it is a diffeomorphism in a local chart, but may not be a global one. Consider the map $\mathbb{R}^{1}\rightarrow \mathbb{S}^{1}$ by $x\rightarrow e^{2\pi ix}$, for example. – Bombyx mori Jul 01 '13 at 19:44
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Hint: Since $f$ is a local diffeomorphism, $f$ is a local homeomorphism. Use this fact to prove that $f$ is an open map. By continuity, the image of $f$ is connected. What is the only type of open, connected subset of $\mathbb{R}$?

Here's the proof that $f$ is an open map. Let $U \subseteq \mathbb{R}$ be an open set. Given $y \in f(U)$, there exists $x \in U$ such that $f(x) = y$. Take an open set $U_x$ around $x$ for which $f_{U_x}$ is a homeomorphism onto its image. Since $U \cap U_x$ is an open subset of $U_x$, $f(U \cap U_x)$ is open in $f(U_x)$. However, $f(U_x) \subseteq \mathbb{R}$ is open, and therefore, $f(U \cap U_x) \subseteq \mathbb{R}$ is open. We have

$$y \in f(U \cap U_x) \subseteq f(U)$$

and so $f(U)$ is open.

For the second part, it suffices to show that the local diffeomorphism $f: \mathbb{R} \to (a, b)$ is a bijection. We know it's surjective, so assume it's not $1-1$. Then by Rolle's theorem (or MVT), there is some point $x \in \mathbb{R}$ such that $f'(x) = 0$. However, the pushforward $df_x$ at $x$ must be a linear isomorphism. This is a contradiction.

Ink
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  • Thanks so much Ink. But I have a really really silly question: Isn't it given that $f: \mathbb{R} \rightarrow \mathbb{R}$, which the latter is an open interval?? – WishingFish Jul 01 '13 at 04:54
  • Or it can be any subspaces and subsets open or closed? – WishingFish Jul 01 '13 at 05:06
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    @user83036 We're given that $f: \mathbb{R} \to \mathbb{R}$ is a local diffeomorphism. It's not clear a priori what the image of $f$ is. Indeed, it is an open interval, but you need to prove this by using the fact that $f$ is an open map. Is that helpful? – Ink Jul 01 '13 at 05:34
  • Yes, thank you very much Ink – WishingFish Jul 01 '13 at 05:57
  • Actually, after a second thought, I don't quite get your $U_x$ idea - why we need to introduce it? Thank you. – WishingFish Jul 01 '13 at 20:10
  • Since I found this stands as a question itself, I posted here http://math.stackexchange.com/questions/433958/the-general-idea-of-prove-openness – WishingFish Jul 01 '13 at 20:30
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    @user83036 We need to introduce it because we're using the fact that $f$ is a local homeomorphism to prove that $f$ is open. – Ink Jul 01 '13 at 21:00
  • For the second part, local diffeomorphism implies $f$ and $f^{-1}$ is globally smooth, right? – WishingFish Jul 01 '13 at 22:08
  • And also, why $f: \mathbb{R} \to (a, b)$ is a bijection? – WishingFish Jul 01 '13 at 22:14
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    @user83036 For any local diffeomorphsim, $f$ and $f^{-1}$ are both smooth provided that $f^{-1}$ is defined, since any bijective local diffeomorphism is a diffeomorphism. – Ink Jul 01 '13 at 22:18
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    @user83036 I've shown that the image of $f$ is an open interval which I am calling $(a, b)$. Therefore, $f: \mathbb{R} \to (a, b)$ is surjective, and I have proven that $f$ is also injective. – Ink Jul 01 '13 at 22:22
  • Oh got it. I was under the (wrong) impression that $f$ maps every element in $\mathbb{R}$ to every element in $\mathbb{R}$. – WishingFish Jul 01 '13 at 22:23
  • So for this question, $f^{-1}$ is defined locally since it is a diffeomorpohism locally. But how about globally? Is it defined? – WishingFish Jul 01 '13 at 22:24
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    @user83036 The given local diffeomorphism $f: \mathbb{R} \to \mathbb{R}$ isn't necessarily surjective, so $f^{-1}$ may not be defined. However, it is defined for $f: \mathbb{R} \to (a, b)$. – Ink Jul 01 '13 at 22:27
  • Great. Can I rephrase it as: \underline{$f$ is a diffeomorphism:} Since local diffeomorphism implies $f$ and $f^{-1}$ are globally smooth, it suffices to show that the local diffeomorphism $f: \mathbb{R} \to$ image$f$ is a bijection. We know it's surjective since we restricted to its image. – WishingFish Jul 01 '13 at 22:28
  • But I restricted $f$ to its image, I thought it as your $(a,b)$ – WishingFish Jul 01 '13 at 22:44
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    @user83036 I misread that last comment. Yeah, you can phrase it like that. – Ink Jul 01 '13 at 23:07
  • something interesting (hopefully): the next question (GP 1.3.5 on page 18): Prove that a local diffeomorphism $f: X \rightarrow Y$ is actually a diffeomorphism of $X$ onto an open subset of $Y$, provided that $f$ is one-to-one. - confused: isn't it exactly what we just did here? – WishingFish Jul 01 '13 at 23:10
  • @user83036 I used that property, but I didn't exactly prove it. As for that exercise, the jist of it is that differentiability is a local property. It should be straightforward to prove that $f^{-1}$ is smooth directly from the definition. – Ink Jul 01 '13 at 23:25
  • let us continue this discussion in chat – WishingFish Jul 01 '13 at 23:34
  • You proved by Rolle's theorem, right...? – WishingFish Jul 01 '13 at 23:35
  • The second part of this question proved question 1.3.5: Since local diffeomorphism implies $f$ and $f^{-1}$ are globally smooth, it suffices to show that the local diffeomorphism $f: \mathbb{R} \to$ image$f$ is a bijection. We know it's surjective since we restricted to its image. But if it is not surjective, by mean value theorem, there is some point $x \in \mathbb{R}$ such that $f'(x) = 0$. However, the pushforward $df_{U_x}$ at $x$ must be a linear isomorphism. This is a contradiction. – WishingFish Jul 01 '13 at 23:36
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    @user83036 The second part doesn't prove GP 1.3.5 since the questions are different. And I'm assuming $X, Y$ are smooth manifolds not necessarily $\mathbb{R}$. If you have a question about proving 1.3.5, I suggest you ask it as a new question, and I'll be happy to answer it. – Ink Jul 02 '13 at 00:03
  • Oh, thank you so much. Let me try my attempt first. Thank you Ink. – WishingFish Jul 02 '13 at 00:04
  • btw, the reason I was confused is that I find the proof of 1.3.3 is more or less independent of the fact that the manifold happen to be $\mathbb{R}$. – WishingFish Jul 02 '13 at 00:06
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    @user83036 The manifold was $\mathbb{R}$, so I was able to apply Rolle's theorem. – Ink Jul 02 '13 at 00:12
  • /_\ but injectivity is given at 1.3.5! :=D – WishingFish Jul 02 '13 at 00:15
  • Aha, here is my attempt. please see if it is right? http://math.stackexchange.com/questions/434090/local-diffeomorphism-on-mathbbr-and-on-manifolds – WishingFish Jul 02 '13 at 00:17
  • Yeah, so in a way, I cheated because I used the result of problem 1.3.5 (bijective local diffeo $\implies$ diffeo) for this problem. But hey, I don't have Pollack, so I couldn't have known. – Ink Jul 02 '13 at 00:19
  • Thanks Ink. The point is I think you proved 1.3.5, at least in the $\mathbb{R}$ case. And I also found the general manifold case is almost the same. – WishingFish Jul 02 '13 at 00:21