Let $f(x,y)=(x^2-y^2,2xy)$
Prove it is not restricted the domain of $f$ then $f$ don't have inverse.
I proved this: "Prove for all point $(x_0,y_0)\not =(0,0)$ the restriction of $f$ to neighborhood $(x_0,y_0)$ have inverse."
I don't sure of how attack this exercise. But i think this:
Suppose $(x_0,y_0)=(0,0)$, then
$$f'(x_0,y_0)=\begin{pmatrix} \nabla f_1(x_0,y_o) \\ \nabla f_2(x_0,y_0) \end{pmatrix}=\begin{pmatrix} 2x_0 &-2y_0\\ 2y_0 & 2x_0 \end{pmatrix}\implies \det(f'(x_0,y_0))=4x_0^2+4y_0^2=0$$
Then $f'(x_0,y_0)$ is not invertible.
Here i'm stuck. can someone help me?
If the domain is the whole ${\bf{R}}^{2}$, then $f$ has no inverse because it is not one-to-one: For any pair $(x,y)\ne(0,0)$, then $(-x,-y)\ne(x,y)$ but $f(-x,-y)=f(x,y)$.