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Could someone give me an example of a local diffeomorphism from $\mathbb{R}^p$ to $\mathbb{R}^p$ (function of class say $C^k$ with an invertible differential map in each point) that is not a diffeomorphism..

in the real line (1 dim case) that would mean a function with a continuous non null derivative on an open $V$ of $\mathbb{R}$ that is not bijective which does not make sense thus any local diffeomorphism on the real line is a diffeo..

Could one give me a counterexample in a higher dimension?

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    $\begingroup$ Take the complex exponential $e^z$ as a function from $R^2$ to $R^2$. For every horizontal strip [iy+ i(y+2Pi) ) (i.e., including iy, but not i(y+2Pi)) there is an inverse--a logz --, but there is no global inverse. $\endgroup$ – gary May 15 '11 at 1:37
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    $\begingroup$ there is no inverse around zero $\endgroup$ – yoyo May 15 '11 at 2:27
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    $\begingroup$ Correction: that should be the strip [iy, i(y+2Pi)) $\endgroup$ – gary May 15 '11 at 2:38
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    $\begingroup$ @yoyo: I assume you mean the target zero , right? Otherwise, the inverse function guarantees the existence of a local diffeomorphism at each point, including 0 in the domain, since d/dz($e^z$)=1 at z=0; same for all other points' or using the $R^2$ version of the inverse function theorem. $\endgroup$ – gary May 15 '11 at 22:36
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    $\begingroup$ You are wrong about the one dimensional case: the exponential function $\mathbb R\to \mathbb R: x\mapsto e^x$ is a non surjective local diffeomorphism. $\endgroup$ – Georges Elencwajg Sep 1 '14 at 20:09
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Consider $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ defined by $f(x,y)=(e^x \cos y,e^x \sin y)$

$Df(x,y)$ is always invertible because $\det Df(x,y)=e^{2x}$ but clearly $f$ is not one to one. It is periodic with period $2\pi$.

In higher dimensions, one can use $f(x)=(e^{x_1}\cos x_2, e^{x_1}\sin x_2, x_3, \dots, x_n)$.

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  • $\begingroup$ To draw the link to complex analysis, treating x and y as the usual coordinates on the complex plane, f = e^z, z = x + iy. The inverse, ln(f(z)) is multi-valued, and hence a branch cut must be taken, e.g. 0 < arg(f(z)) = y < 2pi. $\endgroup$ – Rascalniikov May 13 '17 at 6:28

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