Explain why this set is not a differentiable manifold I want to figure out why the set of zeros of the function $g:\mathbb{R}^{2} \to \mathbb{R}$ defined as $g(x,y) = x^2 - y^2$ is not a differentiable manifold. So what I want to use is the following result: 
Let $A \subset \mathbb{R} $ open, $p<n$, $g:A \to \mathbb{R}^p$ of class $C^1$ such that $g'(x)$ has rank $p$ for all $x \in A$ such that $g(x)=0$ therefore $g^{-1}(\{0\})$ is a differentiable manifold in $\mathbb{R}^n$ of dimension $n-p$.   
So I computed the Jacobian matrix 
$$(2x \;\; -2y )$$
Then in the origin this does not have rank 1, is has rank $0$ and we note that this is the only point where this happens therefore it can't be a differentiable manifold.
The other thing I was trying to do is to give an example of a system of coordinates (commonly charts) where some condition fails, and I thought in $F:\mathbb{R}^2 \to \mathbb{R}^3$ as $F(x,y)=(x,y,x^2-y^2)$ and its Jacobian in zero has rank 3, but I don't know how to proceed to find the contradiction to my definition of differentiable manifold :
A subset $M \subset \mathbb{R}^n$ is a differentiable manifold of dimension $k$ if, for each $x \in M$, exists open sets $U \subset \mathbb{R}^k$ and $V \subset \mathbb{R}^n$ and a class $C^1$ function $f:U \to V$ such that:
1) $x \in V$ 
2) $f(U)=V \cap M$, $f$ is an homeomorfism
3) for each $y \in U$ the jacobian matrix has rank k 
Can you help to proceed in this or tell if my approach are right? thanks a lot :) 
 A: Take any neighborhood of the origin, we want to show it isn't diffeomorphic to the open ball $B^n$.  The standard arguments from topology are:


*

*If we delete the origin, we disconnect the neighborhood.  No point in $B^n$ has this property for $n \geq 2$.

*If we delete the origin, there are at least 4 connected components.  On the other hand deleting a point in the interval gives only two components.


Thus, the neighborhood is not even homoeomorphic to any open ball.
A: Here is a "non-topological" proof.
The set $M$ is simply the union of two lines $M = L_1 \cup L_2$ where $L_1$ is the line $y=x$ and $L_2$ is the line $y=-x$. 
Arguing by contradiction, let's assume that $M$ is a differentiable manifold. It's dimension must equal $1$, because $M-\{(0,0)\}$ clearly is a differentiable manifold of dimension $1$, and removing a single point from a differentiable manifold of dimension $k$ produces a differentiable manifold of dimension $k$. 
Let $L_3$ be the tangent line of $M$ at the point $(0,0)$. For every $C^1$ curve 
$$\gamma : (-1,+1) \to M
$$
such that $\gamma(0)=(0,0)$, the vector $\vec \gamma'(0)$ is tangent to the line $L_3$.
Since $L_1 \ne L_2$, the line $L_3$ is unequal to one of $L_1$ or $L_2$ (possibly both).
If $L_3 \ne L_1$ then we obtain a contradiction by using the curve $\gamma(t) = (t,t)$, whose tangent vector $\vec \gamma'(0)$ is tangent to $L_1$ but not to $L_3$. 
If $L_3 \ne L_2$ then we obtain a similar contradiction using $\gamma(t) = (t,-t)$.
