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A differentiable transformation of R^n at each point has an invertible derivative. Does it imply that the transformation is a global diffeomorphism?

flag	asked Dec 4 2011 at 17:12 Sz_Z 112●1●5

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Look on the exponential map on the real line... – Petya Dec 4 2011 at 17:37

I think you mean the exponential function on the complex numbers (considered as ${\mathbb R}^2$). The statement would be true for $C^1$ maps of $\mathbb R$ to itself. – Robert Israel Dec 4 2011 at 18:21

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Robert -- I think Petya meant that the exponential is not a diffeomorphism $\mathbb{R}\to\mathbb{R}$. – algori Dec 4 2011 at 18:36

stephanos · Accepted Answer · 2011-12-04 18:20:11Z UTC

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Let $f:\mathbb{R}^n\rightarrow M\subseteq\mathbb{R}^n$ be the differentiable transformation, with $M=f(\mathbb{R}^n)$. If $M\neq\mathbb{R}^n$, then obviously $f$ isn't an global diffeomorphism of $\mathbb{R}^n$. But it is global diffeomorphism between $\mathbb{R}^n$ and $M$.

*This was intended to be a comment not an answer, but I cannot comment, sorry. *

answered Dec 4 2011 at 18:20

stephanos
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stephanos -- re "it is a global diffeomorphism ...": it isn't in general: take $f$ to be the complex exponential, then $M=\mathbb{R}^2$ minus the origin, which is not diffeomorphic to the plane. Even if the image of $f$ is the whole $\mathbb{R}^n$, then $f$ is not necessarily a diffeomorphism. To see this set $g$ to be a degree 3 complex 1-variable polynomial whose derivative has no double roots. The universal cover of $U=\mathbb{C}\setminus$ the roots of $f'$ is $\mathbb{R}^2$; set $h$ to be the universal covering map $\mathbb{R}^2\to U$. Set $f=g\circ h$.... – algori Dec 4 2011 at 21:08

... this is a local diffeomorphism with image $\mathbb{C}=\mathbb{R}^2$ but not a global diffeomorphism as it takes each value infinitely many times. – algori Dec 4 2011 at 21:10

Global invertibility

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