Injectiveness of a monotonic surjective mapping $\mathbb R^n \to \mathbb R^n$ with $\det J \neq 0$

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 $$F_2$$).

• 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. Commented Aug 25, 2022 at 6:07
• 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. Commented Aug 25, 2022 at 9:46
• 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. Commented Aug 25, 2022 at 21:03
• fig.if.usp.br/~marchett/fluidos/… Commented Dec 23, 2022 at 10:25