local diffeomorphism on $\mathbb{R}$ and on manifolds.

I find the proof of diffoemorphism in Guillemin & Pallock's Differential Topology 1.3.3 is more or less independent of the fact that the manifold happen to be $\mathbb{R}$, and therefore are the same.

Then I am asking if my two proofs (primarily the latter one) are correct.

GP 1.3.3: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a local diffeomorphism. Prove that the image of $f$ is an open interval and that, in fact $f$ maps $\mathbb{R}$ diffeomorphically onto this interval.

Proof: 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.

And right after:

GP 1.3.5: 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.

Proof: Since local diffeomorphism implies $f$ are globally smooth and $f^{-1}$ are globally smooth within the image of $f$, it suffices to show that the local diffeomorphism $f: X \to$ image$f \subseteq Y$ is a bijection. We know it's surjective since we restricted to its image. Given $f$ is one-to-one, it is bijective. Hence a diffeomorphism onto an open subset of $Y$.

• What is GP? Here and elsewhere you seem to be quoting from somewhere, but never bothered to tell us where form, afaics? – Mariano Suárez-Álvarez Jul 2 '13 at 0:22
• Notice that here you did not ask anything... – Mariano Suárez-Álvarez Jul 2 '13 at 0:23
• Like I said, I did not know that the fact I used in 1.3.3 was stated as a problem in 1.3.5, but your proof of 1.3.5 is problematic. – Ink Jul 2 '13 at 0:25
• Please add all the relevant information and the actual question to the body of the question itself. In particular, I do not see where you aked a question. – Mariano Suárez-Álvarez Jul 2 '13 at 0:26
• I think the idea is that the $\mathbb{R} \to \mathbb{R}$ case is easily visualisable, so you first get an easy one to warm up. Then you get the general one, to see what parts you actually needed, and which special properties of $\mathbb{R}$ that you may have used for the first were actually irrelevant. – Daniel Fischer Jul 2 '13 at 0:46

The problem with your first proof is this:

1. Why is $f$ an open map?
2. Why is the image an interval?
3. Also tautologically $f$ is surjective onto its image so what's going on with the MVT in 1.3.3?

Well the first actually this follows immediately from $f$ being a local diffeomorphism! For any $f(y) \in f(\Bbb{R})$, $f$ a local diffeomorphism implies there is $U$ open about $y$ such that $f(U)$ is open in $\Bbb{R}$ about $f(y)$. Necessarily $f(U) \subseteq f(\Bbb{R})$ and so immediately this means $f(\Bbb{R})$ is open. For 2) the image is an interval because $f(\Bbb{R})$ is an open connected set which is an interval from basic real analysis. To complete the problem, you just need to tell us why $f$ is injective: If $x,y$ with $x\neq y$ are such that $f(x) = f(y)$, the MVT implies there is $d$ between $x$ and $y$ so that $f'(d) = 0$. Now how does this contradict what we have?

Exercise: Complete the proof above for why $f$ has to be injective with the following hints:

1. Let $M$ be a smooth manifold. For any $U \subseteq M$ open and $p\in M$, $T_pM \cong T_pU$.

2. A functor always takes isomorphisms to isomorphisms. If $f'(d) = 0$ then the induced map on tangent spaces at $d$ is the (.....) map.

• I think, there is a mistype in the problem: The sentence starting by "But if it is not surjective" should rather be "But if it is not bijective".. – Berci Jul 2 '13 at 0:57
• 2. A functor always takes isomorphisms to isomorphisms. If $f'(d) = 0$ then the induced map on tangent spaces at $d$ is the (identity) map? – WishingFish Jul 2 '13 at 1:43
• However, for 1, I don't really know how to prove. Can you give me some hint? – WishingFish Jul 2 '13 at 1:52
• @WishingFish I just proved 1 for you. If $f'(d) = 0$, then this means that $f'$ is the zero map! – user38268 Jul 2 '13 at 2:00
• OH, I mean 1 in the exercise. – WishingFish Jul 2 '13 at 2:03

We know local diffeomorphisms are open maps. Let $N = f(X)$. By assumption we have a bijective local diffeomorphism $f: X \to N$. To prove that $f$ is smooth let $x \in X$. There exists an open set $U \subseteq X$ around $x$ such that $f_U : U \to f(U)$ is a diffeomorphism. Hence, there exists charts $(U_x, \phi)$ $(U_{f(x)}, \psi)$ of $x$ and $f(x)$ such that $f(U_x) \subseteq f(U_{f(x)})$ and

$$\psi \circ f \circ \phi^{-1} : \phi(U_x) \to \psi(U_{f(x)})$$

is smooth as a map between Euclidean space. Hence, $f$ is smooth. For any $f(x) \in N$, the map $f_U^{-1} : f(U) \to U$ is a diffeomorphism, so the exact same argument shows that $f^{-1}$ is smooth. I think this proves 1.3.5.

• okay, now I see why these two questions are different. :-P – WishingFish Jul 2 '13 at 1:54
• How about surjectivity? Is that for free when we restrict the image of $f$ to $f(N)$? – WishingFish Jul 2 '13 at 2:07
• Hello Ink - and why $$\psi \circ f \circ \phi^{-1} : \phi(U_x) \to \psi(U_{f(x)})$$ is smooth as a map between Euclidean space, then $f$ is smooth...? Thank you~~~ -) – WishingFish Jul 2 '13 at 3:15
• Is that because locally, $\psi, f, \phi$ are all smooth, and the composition carry any point in $\mathbb{R}^m$ smoothly to $\mathbb{R}^n$, so $f$ is globally smooth? Feel I lost something here.. – WishingFish Jul 2 '13 at 3:18
• A new question is posted here demonstrates the recent progress - thanks in advance! math.stackexchange.com/questions/434187/… – WishingFish Jul 2 '13 at 3:48

Yes, we can get 1.3.3, by applying the more general 1.3.5, $\$ and $\$ the additional fact for $\Bbb R$ that each local homeomorphism $f:\Bbb R\to\Bbb R$ is injective (that you showed by the mean value theorem).