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Let $f : \mathbb{R}^n \to \mathbb{R}^n$ have Jacobian $J\,f : \mathbb{R}^n\to\mathsf{M}_n(\mathbb{R})$ which has positive eigenvalues everywhere. Is $f$ (globally) invertible on it's range/injective?

If $J\,f$ was also guaranteed to be symmetric, this would be true by this question. We also know $f$ is locally invertible by the inverse function theorem.

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    $\begingroup$ Take $f=\arctan$. Then $f' >0$ but it is not surjective. $\endgroup$ – copper.hat Aug 7 '18 at 0:34
  • $\begingroup$ @copper.hat I guess I should specify invertible on it's range i.e. injective $\endgroup$ – cdipaolo Aug 7 '18 at 0:35
  • $\begingroup$ There was nothing about the linked proof which required symmetry. All you need is that $\langle (Jf)x,x\rangle >0$ for all $x\neq 0$, which follows since the smallest eigenvalue is $\min_{\|x\|=1} \langle (Jf)x,x\rangle$. $\endgroup$ – Mike Earnest Aug 7 '18 at 0:59
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    $\begingroup$ @MikeEarnest Forgive me if I'm missing something, but I'm pretty sure that only works for self-adjoint $Jf$. $A = \begin{bmatrix}1&-3\\0&1\end{bmatrix}$ is a counterexamples with $x=[1,1]^*$. $\endgroup$ – cdipaolo Aug 7 '18 at 1:03
  • $\begingroup$ You are right, apologies for my hasty comment. $\endgroup$ – Mike Earnest Aug 7 '18 at 1:09

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