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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.

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    $\begingroup$ Demonstrating an example is very different from proving it works for any $f$ satisfying those properties. $\endgroup$ – Zev Chonoles Jul 1 '13 at 1:25
  • $\begingroup$ Oops. I guess not I get what the problem is asking for - though still at lost for the problem... $\endgroup$ – WishingFish Jul 1 '13 at 1:40
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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.

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    $\begingroup$ @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? $\endgroup$ – Ink Jul 1 '13 at 5:34
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    $\begingroup$ @user83036 We need to introduce it because we're using the fact that $f$ is a local homeomorphism to prove that $f$ is open. $\endgroup$ – Ink Jul 1 '13 at 21:00
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    $\begingroup$ @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. $\endgroup$ – Ink Jul 1 '13 at 22:18
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    $\begingroup$ @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. $\endgroup$ – Ink Jul 1 '13 at 22:22
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    $\begingroup$ @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 $f: \mathbb{R} \to (a, b)$. $\endgroup$ – Ink Jul 1 '13 at 22:27
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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.

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    $\begingroup$ This is intended for your own exercise to fill in. You can skip the classification to find a better proof yourself as well. $\endgroup$ – Bombyx mori Jul 1 '13 at 1:50
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    $\begingroup$ Yes. Should not be difficult to prove. $\endgroup$ – Bombyx mori Jul 1 '13 at 3:05
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    $\begingroup$ $f$ 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. $\endgroup$ – Bombyx mori Jul 1 '13 at 4:16
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    $\begingroup$ Sorry for providing a wrong example, which is indeed a global differeomorphism. But I think such counter-example should not be difficult to find. $\endgroup$ – Bombyx mori Jul 1 '13 at 4:28
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    $\begingroup$ A better way is to prove $f^{-1}$ is continuous. This follows automatically from the fact $f$ is a local homeomorphism. $\endgroup$ – Bombyx mori Jul 1 '13 at 13:12

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