Given that $F:R^n\rightarrow R^n$ is $C^1$ on $R^n$ s.t. the jacobian matrix $Df(p)$ is invertible for all $p \in R^n$, prove that f must be injective.

To show this in $R\rightarrow R$, I applied mean value theorem to show why $f$ it must be injective. However, I cannot figure out how I should approach this in $R^n$ in general.

One method I have in mind is to apply mean value theorem in $R^n$ (though I am not familiar with this theorem on $R^n$, and another is to invoke implicit function theorem. Both method seems to make sense to me, only that I don't know how to begin and articulate the proof.

What are ways to tackle this proof in $R^n$?


This is already false in $\mathbb R^2$. Consider, for instance,$$\begin{array}{rccc}f\colon&\mathbb R^2&\longrightarrow&\mathbb R^2\\&(x,y)&\mapsto&e^x(\cos y,\sin y).\end{array}$$The Jacobian matrix is invertible at every point, but $f(0,0)=f(0,2\pi)$.

| cite | improve this answer | |
  • $\begingroup$ Hello, could you also explain why inverse function theorem, which state that in this case I can always find some set $V$ around $x_0$, $W$ around $f(x_0)$ for all points in $R^n$ such that the function is bijective (and thus injective), fail to prove this statement for $R^n$? Thank you very much. $\endgroup$ – LHC2012 Jan 13 at 0:56
  • $\begingroup$ As my example shows, the fact that, near each point of $\mathbb R^2$, $f$ is injective, is not enough to deduce that $f$ is injective globally. $\endgroup$ – José Carlos Santos Jan 13 at 6:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.