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Let $f:R^N \to R^N$ be a differentiable mapping, and $J_f$ its Jacobian determinant. Suppose that $\exists a,b \in R^N : J_f(a)<0,J_f(b)>0$. Is it right that on every continous curve connecting $a$ and $b$ there exists a point $c$ such that $J_f(c)=0?$

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    $\begingroup$ What am I missing? $J_f$ is a continuous function on $\mathbb{R}^N$, so it is continuous when restricted to any continuous curve connecting $a$ to $b$. That curve is just the image of some continuous $\gamma:[0,1]\to\mathbb{R}^N$. So $F=J_f\circ\gamma:[0,1]\to\mathbb{R}$ is continuous and satisfies $F(0)<0$ and $F(1)>0$, so by the intermediate value theorem it must vanish for some value in $(0,1)$. $\endgroup$
    – Joe Silverman
    Commented Mar 27, 2014 at 20:43
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    $\begingroup$ @JoeSilverman I guess some people by "differentiable" mean just differentiable, not necessarily $C^1$. Personally, I also disapprove such maps :) $\endgroup$
    – Alex Degtyarev
    Commented Mar 27, 2014 at 21:07
  • $\begingroup$ @AlexDegtyarev Ah, well, if $J_f$ is allowed to be (mildly) discontinuous, then can't one construct a counterexample by taking a standard example of a map $f$ that is differentiable, but not $C^1$. $\endgroup$
    – Joe Silverman
    Commented Mar 28, 2014 at 1:27
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    $\begingroup$ @JoeSilverman This I don't know, as for ordinary derivatives $f'(x)$ there is an intermediate value theorem, even without the assumption that it's continuous. (Don't remember the exact name.) So, the question about the Jacobian may be not as simple as it seems. $\endgroup$
    – Alex Degtyarev
    Commented Mar 28, 2014 at 8:52
  • $\begingroup$ @JoeSilverman The only hope that I see is that the restriction of the Jacobian to a curve is some kind of ordinary derivative. But standard proofs of such things usually assume class $C^1$, so one should go all the way over the proofs to see of they work in the more general setting. Boring :) $\endgroup$
    – Alex Degtyarev
    Commented Mar 28, 2014 at 8:55

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The answer is negative. To construct a counterexample I used Jan Maly's paper THE DARBOUX PROPERTY FOR GRADIENTS, 1996. First let's consider $$ \varphi(x,y):= \begin{cases} \displaystyle \frac {2xy^4}{x^2+y^4} -x, (x,y) \ne(0,0); \\0,(x,y)=(0,0). \end{cases} $$ We observe that $\varphi(x,y)$ is everywhere differentiable, $\varphi^\prime_x(0,0)=-1$, $\varphi^\prime_x(0,y)=1 $ $\forall y\ne0$ and $\varphi^\prime_y(0,y)=0$. Now let's consider $f(x,y):=(\varphi(x,y),x+y)$. It's clear that $f:R^2\to R^2$ is everywhere differentiable and $$ J_f(0,y) = \begin{cases} 1, y\ne0; \\-1, y=0. \end{cases} $$ Thus intermediate value (Darboux) property for Jacobian determinant doesn't hold even on a straight line $x=0$.

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