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Let $f:\mathbb{R}^{n}\to\mathbb{R}^{n}$ be a continuously differentiable mapping. We assume that the set $$\{x\in\mathbb{R}^{n};j(f)(x)=0\}$$ is a hypersurface of $\mathbb{R}^{n}$, where $j(f)(x)$ denotes the jacobian determinant of $f$ at $x$.

I wonder if it is possible to apply a weak variant of the Inverse theorem function to $f$ at any point of $\mathbb{R}^{n}$ which states as follow:

For a given $a\in\mathbb{R}^{n}$, there is some neighborhood $V$ of $a$ in $\mathbb{R}^{n}$ such that $f(V)$ is a neighborhood of $f(a)$.

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The problem is fold singularities. Take $\mathbb{R}^n$, fold it in half and project. The function $x^2$ as a function from the reals to the reals has a fold singularity at 0. You need some additional hypotheses about the derivatives of your map along the singular set to get the conclusion you want. If you have a graduate level understanding of mathematics, there is a difficult book by Gromov, called "Partial differential relations" (1986) that has a lot of information about questions like these. A better starting place if you are committed to understanding such questions is Guillemin and Golubitsky's "Stable Mappings and their Singularities" (1973), though there are probably more modern books that you might find more accessible.

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$f(x,y)=(xy,y)$ is a counterexample.

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  • $\begingroup$ I am sorry, please see the edit. $\endgroup$
    – driss-alamilouati
    Commented Aug 23, 2013 at 11:59
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    $\begingroup$ $f(x)=x^2$ and $a=0$ is a counterexample. $\endgroup$
    – Michael Bächtold
    Commented Aug 23, 2013 at 12:04

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