I'm reading inverse function theorem (IVT) at page 306 of Amann's Analysis I, i.e.,
Amann's IVT: Let $X$ be an open subset of $\mathbb{K} \in \{\mathbb{C}, \mathbb{R}\}, f: X \rightarrow \mathbb{K}$, and $Y:=f(X)$. Let $f$ be injective and consider the inverse $f^{-1}: Y \rightarrow X$ of $f$. Suppose that $f$ is differentiable at $a \in X$, and $f^{-1}$ is continuous at $b:=f(a) \in Y$. Then $f^{-1}$ is differentiable at $b$ if and only if $f^{\prime}(a) \neq 0$. In this case, $\left(f^{-1}\right)^{\prime}(b)=1 / f^{\prime}(a)$.
This version does not require $f'$ to be continuous, which distinguishes itself from the standard version. Moreover, a continuous bijection $f:\mathbb C \to Y \subset\mathbb C$ is a homeomorphism by invariance of domain theorem. So Amann's IVT is indeed powerful.
I would like to ask if Amann's version can be extended to multivariate setting, i.e.,
Let $X$ be an open subset of $\mathbb{K}^n$ where $\mathbb{K} \in \{\mathbb{C}, \mathbb{R}\}, f: X \rightarrow \mathbb{K}^n$, and $Y:=f(X)$. Let $f$ be injective and consider the inverse $f^{-1}: Y \rightarrow X$ of $f$. Suppose that $f$ is differentiable at $a \in X$, and $f^{-1}$ is continuous at $b:=f(a) \in Y$. Then $f^{-1}$ is differentiable at $b$ if and only if $f^{\prime}(a)$ is invertible. In this case, $\left(f^{-1}\right)^{\prime}(b)=[f^{\prime}(a)]^{-1}$.
Thank you so much for your help!
