How can I say a function $F:\mathbb{R}^n \longrightarrow \mathbb{R}^m$ $nMy doubts are about an application of the Implicit Function Theorem: I do not understand how a function $F: \mathbb{R}^n \longrightarrow \mathbb{R}^m$, with $n<m$ (!!!) can locally be an homeomorphism between a neighbourhood of $x$ and a neighbourhood of $F(x)$, for all $x$ such that $rnk(DF(x))=n$.
I just cannot understand how I can apply the Inverse Function Theorem (which I know for Banach spaces), given that $DF(x)$ cannot be invertible! I'm asking this because I found this result written in many texts, for example the solution of exercise 2.2.5 of Berkley-Problems in Mathematics.
To be precise, the exercise was:
$F: \mathbb{R}^n \longrightarrow \mathbb{R}$, $F\in C^2(\mathbb{R})$. Let0s say that a point $x$ is a nondegenerate critical point if $DF(x)=0$ and $rnk(D^2F(x))=n$. Show that every nondegenerate crytical point is isolated.
The solution provided from the book stated:
Define $G: \mathbb{R}^n \longrightarrow \mathbb{R}, G(y):= |DF(y)|^2 $, where $|.|$ is the Euclidean norm.Then, $G$ is $C^1$, $G'(x) != 0 $ and $G(x)=0$, so by the Inverse Function Theorem G is locally diffeomorphism to a neighbourhood of 0, hence injective.
I don't understand this solution!Thanks in advance to everybody who'll answer me! 

The proposed solution is nonsense, you should complain to the people who wrote it. (In the 3rd edition of ``Berkeley Problems in Mathematics" that I have, this is a "solution" of Problem 2.2.10.) A correct solution is to consider the gradient function $G(x)=\nabla F(x), x\in {\mathbb R}^n$. This function is $C^1$ and has zero at some $x_0$. The assumption that the Hessian of $F$ has rank $n$ at $x_0$ amounts to the assumption that the derivative $DG(x)$ has (maximal) rank $n$ at $x_0$. Regarding $G$ as a map ${\mathbb R}^n\to {\mathbb R}^n$ we then can apply the IFT to $G$ at $x_0$ and conclude that $G$ is a local diffeomorphism at $x_0$, implying that $x_0$ is an isolated zero of $G$. In other words, $x_0$ is an isolated critical point of $F$. 
