Function $F:\Bbb R^4\to\Bbb R^2$ is given by the formula $F(x,y,u,v)=(x^2+y^2+u^2+v^2-2,x^2-y^2+u^2-v^2).$ Find all points $(x,y,u,v)\in\Bbb R^4$ in which the equation $F(x,y,u,v)=0$ defines a functional dependence of $(u,v)$ on $(x,y).$
My thoughts:
Let $S$ denote the set of all the points we're looking for. The Implicit Function Theorem gives us the sufficient condition for $(x,y,u,v)\in\Bbb R^4$ to be in $S$, so if $S_0=\{(x,y,u,v)\in\Bbb R^4: F(x,y,u,v)=0,\frac{\partial F}{\partial(u,v)}(x,y,u,v)\text{ is regular}\},$ then $S_0\subseteq S.$
$$\det\left[\frac{\partial F}{\partial (u,v)}\right]=\begin{vmatrix}2u&2v\\2u&-2v\end{vmatrix}=-8uv\ne 0\iff u\ne 0\text{ and }v\ne 0, $$ so $$S_0=\{(x,y,u,v)\in\Bbb R^4\mid uv\ne 0\}\cap F^{-1}(\{(0,0)\}).$$
I'm not sure about the other points.
Suppose $(x,y,u,v)\in S.$ Then we should be able to solve the following system $$\begin{cases}u^2+v^2=2-x^2-y^2\\u^2-v^2=-x^2+y^2.\end{cases}$$
I obtained $$\begin{cases}u^2=1-x^2\\v^2=1-y^2\end{cases}.$$
Let $U,V\subseteq\Bbb R^2$ be some open neighbourhoods of $(0,0),$ so that $U\times V$ is an open neighbourhood of $(0,0,0,0)\in\Bbb R^4.$
I think, just like with the circle $x^2+y^2=1$ in $\Bbb R^2,$ in this case $u$ and $v$ must be non-zero because every open neigbourhood $V$ of $(u,v)=(0,0)$ contains four different points to which we might map $(x,y)\in U$ so we cannot define $(u,v)$ as a function of $(x,y)$ in any open neighbourhood $U\times V.$
Taking everything above into consideration, I believe $S=S_0.$
Is this valid?