Is a strictly monotone continuous function defined on $\mathbb{R}^m$ has a continuous inverse?

Question

Let $F: \mathbb{R}^m \to \mathbb{R}^m$ be a strictly monotone, surjective, continuous function. Here, the strict monotonicity means that $(x - y)^T (F(x) - F(y)) > 0$ for all $x,y \in \mathbb{R}^m$ as shown at

Is every monotone map the gradient of a convex function?

Then, by strict monotonicity of $F$, it is also injective, and hence bijective. So, there is its inverse function $F^{-1}: \mathbb{R}^m \to \mathbb{R}^m$.

My question is "Is the inverse $F^{-1}$ also continuous?"

When $m = 1$, it seems true as shown at

Prove that if f is a continuous strictly monotone function defined on an interval, then its inverse is also a continuous function.

Many thanks in advance for your answer and discussion.

Answer 1

Any continuous bijection $F:\mathbb{R}^m\to\mathbb{R}^m$ has continuous inverse. This is (a special case of) a famous theorem known as invariance of domain.

Answer 2

Maybe it is helpful the local inversion theorem, that is a consequence of Dini's Theorem for systems. It states that if $f:A \subset \mathbb{R}^n \rightarrow \mathbb{R}^n$, $A$ open set, $f \in C^1(A)$ and $x_0 \in A$. If $\det J_f(x_0) \neq 0$ then there exixts $\sigma, \delta>0$ and a unique function $\varphi$ such that $$ \varphi: B_\sigma(y_0) \rightarrow B_\delta (x_0) $$ is $C^1$ in $B_\sigma(y_0)$, $$ \varphi(f(x))=x,\quad f(\varphi(y))=y,\quad \forall x \in B_\delta (x_0), \quad\forall y \in B_\sigma(y_0) $$ where $y_0=f(x_0)$. Here the inverse is $\varphi$ that belongs to $C^1(B_\sigma(y_0))$, so in particular $\varphi$ is continuous over $B_\sigma(y_0)$. Your function $F$ is invertible then satisfies the hypothesis $\det J_f(x_0) \neq 0$.