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

Question

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 $Y$.

Answer 1

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:

<ol>
<li><p>Let $M$ be a smooth manifold. For any $U \subseteq M$ open and $p\in M$, $T_pM \cong T_pU$.</p></li>
<li><p>A functor always takes isomorphisms to isomorphisms. If $f'(d) = 0$ then the induced map on tangent spaces at $d$ is the (.....) map.</p></li>
</ol>

Answer 2

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.

Answer 3

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).