Differential Topology – Local Diffeomorphism on $\mathbb{R}$ and Manifolds

differential-topology

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 $Y$.

Best Answer

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.

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