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Consider a surjective mapping $F \colon \mathbb R^n \to \mathbb R^n$, $F\in C^1$, $\dfrac{\partial F_i}{\partial x_j} > 0$, and $\det \left(\!\left( \dfrac{\partial F_i}{\partial x_j} \right)\!\right) \neq 0$.

Does the inverse mapping $x = F(y) \colon \mathbb R^n \to \mathbb R^n$ exist? Are there general results concerning this issue?

For instance, if $n = 2$, and we suppose that two roots $a$ and $b$ of $F(x) = y$ exist then $\det J(a) > 0$ and $\det J(b) < 0$ because the vectors $\nabla F_i$ are oriented with respect to each other in different ways at the roots, so $\det J(x)$ takes zero which leads to contradiction (one may depict the isolines of zero level of functions $F_1$ and $F_2$).

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  • $\begingroup$ In the case $n = 3$, consider the domain of constant signs of $F_i(x)-y$ for given $y$. Look from one root directly to another root locating at the same domain of prescribed signs (on its boundary). Then, if from one root, the surface $F_1(x)=0$ is "to the left", then, from the other one, it is "to the right". So, the direction of rotation from one gradient to another changes to the opposite. So, $\det J$ changes its sign from one root to another. $\endgroup$
    – jokersobak
    Commented Aug 25, 2022 at 6:07
  • $\begingroup$ If $F_i > 0$, $F_j < 0$ then $(\nabla F_i \times \nabla F_j, e_k)$ changes sign from one root to another. If we state the additional requirement that all diagonal minors of $J$ are not zero, then we obtain $\det(\frac{\partial F_i}{\partial x_j}) = 0, i, j = 1, 2$ in some point, and this is contradiction, so the root is unique. $\endgroup$
    – jokersobak
    Commented Aug 25, 2022 at 9:46
  • $\begingroup$ Let $f(x) = F(x)-y$ and consider the projection to the plane $x_i,x_j$ at which $f_i$ and $f_j$ have different signs in the domain of constant signs whose end points are two roots $a$ and $b$. Tho these roots are located at the opposite ends of the projection of $D$ onto $x_i,x_j$, so $\det (\frac{\partial F_i}{\partial x_i}, \frac{\partial F_i}{\partial x_j}, \frac{\partial F_j}{\partial x_i},\frac{\partial F_j}{\partial x_j})$ changes sign from $a$ to $b$, so it is 0 in an intermediate point - contradiction if, e.g., $J$ is Gram matrix. $\endgroup$
    – jokersobak
    Commented Aug 25, 2022 at 21:03
  • $\begingroup$ fig.if.usp.br/~marchett/fluidos/… $\endgroup$
    – jokersobak
    Commented Dec 23, 2022 at 10:25

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