Jacobian of an injective mapping

Question

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.

Answer 1

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.

Answer 2

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 $y$.