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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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Alexander Kuleshov
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  • Possibly related question: http://mathoverflow.net/questions/152605/generalization-of-darbouxs-theorem – Vít Tuček Mar 25 '14 at 17:43
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    For the 2nd question apply the intermediate value theorem to $J_f$-restricted to the line joining a and b. – pinaki Mar 25 '14 at 19:41
  • @auniket this shows only that there is c s.t. $J_f(c)[b-a]=0$ – user126154 Mar 25 '14 at 19:44
  • @user126154: I meant the function $\phi: \mathbb{R} \to \mathbb{R}$ defined by $\phi(t) := J_f(a + t(b-a))$. – pinaki Mar 25 '14 at 19:51
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    @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. – user126154 Mar 25 '14 at 19:54
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    @user126154: you are right. I saw in the question $J_f(a) < 0$, and immediately interpreted $J_f$ as the Jacobian determinant. – pinaki Mar 25 '14 at 20:01
  • Yes, right, thanks! 2) is a generalization of Darboux theorem, and this fact follows if it's proved, that f is not injective. Moreover, i think that such a point c exists on every curve from a to b! So the main point is to prove 1). Though it looks quite natural, but it doesn't seem simple to prove... – Alexander Kuleshov Mar 25 '14 at 20:01
  • Thanks for your answers! I didn't think that this problem requires such instruments... And can this degree approach be used to prove, that f is not injective on every curve connecting a and b?? – Alexander Kuleshov Mar 25 '14 at 22:50

2 Answers2

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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.

Changyu Guo
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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 $y$.

Pietro Majer
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