Diffeomorphism $\Leftrightarrow$ injective

Question

My question is really simple, I would like to prove a local diffeomorphism $f:U\to \mathbb R^m$ is a global diffeomorphism over its image $V=f(U)$ if and only if it's an injective function.

The $\Rightarrow$ part is easy, I've already proved a local diffeomorphism is an open function, but I don't know how to use this fact to prove the converse.

It's bijective so it has an inverse. The inverse is smooth locally near any $p$ by the inverse function theorem, and thus globally. — juan arroyo, Jan 10 '17 at 04:51
@juanarroyo could you give me what your version of the inverse function theorem is? the version I know is http://www.math.ucsd.edu/~nwallach/inverse[1].pdf — user42912, Jan 10 '17 at 05:24
That's the one. Using that you can define a smooth inverse in a neighborhood of each point in $f(U)$. A map that is smooth in a nbhd of each point is smooth. — juan arroyo, Jan 10 '17 at 13:29
Related? Is open (topological) smooth embedding equivalent to injective local (homeomorphism) diffeomorphism? — BCLC, Apr 21 '21 at 10:44

score 2 · Accepted Answer · answered Jan 10 '17 at 04:40

2

Given $x\in V$, let $W\subseteq U$ be an open neighborhood of $f^{-1}(x)$ on which $f$ is a diffeomorphism onto its image. That is, $f$ restricts to a diffeomorphism $f|_W:W\to f(W)$. Then the restriction of $f^{-1}$ to $f(W)$ is smooth, since it is the inverse of the diffeomorphism $f|_W$. But $f(W)$ is an open neighborhood of $x$ and smoothness is a local property, so $f^{-1}$ is smooth at $x$. Since $x\in V$ was arbitrary, this proves $f^{-1}$ is smooth.

answered Jan 10 '17 at 04:40

Eric Wofsey

342,377

Thank you for your answer. Did you use the fact $f$ is an open function? – user42912 Jan 10 '17 at 04:51
I used the fact that $f(W)$ is open...this follows from the fact that $f$ is open, but also is probably just directly part of your definition of "local diffeomorphism". And you need to know that $V$ is open for it to even make sense to talk about $f:U\to V$ being a diffeomorphism. – Eric Wofsey Jan 10 '17 at 04:57
@EricWofsey, how you use the hypothesis the $f$ is injective? I imagine that you use the hypothesis to ensure that $|f^{-1}(x)| = 1$ and obtain the existence of this $W$, but I think $W$ exists because $f$ is a local diffeomorphism. I can't see how the injectiveness of $f$ was used. – George Jul 13 '19 at 16:44
$f^{-1}$ does not even exist as a function unless $f$ is injective. – Eric Wofsey Jul 13 '19 at 16:45
Related? Is open (topological) smooth embedding equivalent to injective local (homeomorphism) diffeomorphism? – BCLC Apr 21 '21 at 10:44

Diffeomorphism $\Leftrightarrow$ injective

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