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Suppose $f : \mathbb{R}^n \to \mathbb{R}^n$ has Jacobian $Jf : \mathbb{R}^n\to\mathsf{M}_n(\mathbb{R})$ with positive eigenvalues everywhere. Is $f$ (globally) injective (invertible on its range)?

If $Jf$ 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
    Commented Aug 7, 2018 at 0:34
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    $\begingroup$ @copper.hat I guess I should specify invertible on it's range i.e. injective $\endgroup$
    – cdipaolo
    Commented Aug 7, 2018 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 $\min_{\|x\|=1} \langle (Jf)x,x\rangle$. $\endgroup$ Commented Aug 7, 2018 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
    Commented Aug 7, 2018 at 1:03
  • $\begingroup$ You are right, apologies for my hasty comment. $\endgroup$ Commented Aug 7, 2018 at 1:09

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