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Let $f:R^N \to R^N$ be a differentiable mapping, and $J_f$ its Jacobian. Suppose that $\exists a,b \in R^N : J_f(a)<0,J_f(b)>0$. I want to prove two things that seem intuitively right: 1) $f$ is not injective, 2) $\exists c \in R^N : J_f(c)=0$. I thought that such statements (or their disproof) must be well-known, but haven't found any useful information by now. If somebody has any thoughts on this issue, I'll be glad!

[Edit] Thanks for participation, further generalizations can be discussed here.

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  • $\begingroup$ Possibly related question: mathoverflow.net/questions/152605/… $\endgroup$
    – Vít Tuček
    Commented Mar 25, 2014 at 17:43
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    $\begingroup$ For the 2nd question apply the intermediate value theorem to $J_f$-restricted to the line joining a and b. $\endgroup$
    – pinaki
    Commented Mar 25, 2014 at 19:41
  • $\begingroup$ @auniket this shows only that there is c s.t. $J_f(c)[b-a]=0$ $\endgroup$
    – user126154
    Commented Mar 25, 2014 at 19:44
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    $\begingroup$ @auniket but $J_f$ is what? if it is the differential of $f$ it does not take value in $\mathbb R$, if it is the Jacobian determinant, then yes, your are right. $\endgroup$
    – user126154
    Commented Mar 25, 2014 at 19:54
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    $\begingroup$ @user126154: you are right. I saw in the question $J_f(a) < 0$, and immediately interpreted $J_f$ as the Jacobian determinant. $\endgroup$
    – pinaki
    Commented Mar 25, 2014 at 20:01

2 Answers 2

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As you have already seen from the comments, the real difficulty is to show that $f$ is not injective. This can be proved by using the so-called topological degree theory. Namely, if $f$ is injective, then by the invariance of domain, $f(\mathbb{R}^N)$ will be a domain. Then the topological degree is constant on $f(\mathbb{R}^N)$. On the other hand, since $f$ is differentiable everywhere, at points where the Jacobian is non-zero, one can easily prove that the local degree equals to the sign of the Jacobian, and hence the Jacobian of a differentiable homeomorphism is either non-negaitive or non-positive. So in your situation, $f$ cannot be injective.

If you want to know more on degree theory, you can read any book on topological degree theory to figure out the detailed proof of my indication.

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In these assumptions, the topological degree $\operatorname{deg}\big(f,B(0,r),p\big)$ is well defined whenever $p\in f\big(B(0,r)\big)$; by invariance, it is the same for all such $p$ and $r$, and it's possible values are either $1$ or $-1$. In the case that $f(x)=p$ and $J_f(x)\neq0$, by the Jacobian formula for the degree, this degree it's simply $\operatorname{sgn} J_f(x)$. This explains why either $J_f(x)\ge0$ for all $x$ or $J_f(x)\le0$ for all $x$.

[edit] As to your further question, a smooth map $f:\mathbb{R}^N\to\mathbb{R}^N$ may well be injective on a path connecting two points with Jacobian of opposite sign. Think e.g. to $f(x,y)=(x,y^2)$ : it's restriction to the diagonal is injective, while the Jacobian has the sign of $y$.

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