**Set-up:** Consider the action of $\mathbb{C}^*$ on $\mathbb{C}^4$ defined as follows: $(t,(x,y,z,w))=(tx,ty,t^{-1}z,t^{-1}w)$. I know that the affine GIT quotient is equal to $$\phi: \mathbb{C}^4 \to \mathbb{C}^4//\mathbb{C}^*=Z(XW-YZ),$$ $$(p_1,p_2,p_3,p_4)\to (p_1p_3,p_1p_4,p_2p_3,p_2p_4).$$

**Question:** Consider now the open subset $U=\{p\in \mathbb{C}^4\mid \lim_{t\to 0}tp \text{ doesn't exists} \}$. I would like to prove that the restriction morphism $\phi:U\to \phi(U)\subset \mathbb{C}^4//\mathbb{C}^*$ is a geometric quotient.

**My trial:** In order to visualize it, I did the smaller case $\mathbb{C}^*$ acting on $\mathbb{C}^2$ by $(tx,t^{-1}y)$, and there I had no problems as $U=\{(x,y)\mid y\neq 0 \}$, therefore the orbit $\mathbb{C}^*\cdot (0,y)$ is mapped to $0$, the orbits $\mathbb{C}^*\cdot (x,y)$ (with $x,y\neq 0$) are mapped to a point $c\neq 0$ (the product of the coordinates), therefore the preimage of $\phi$ at every point was a single orbit and I was able to conclude.

The problem in this new example is that $U=\{p\in \mathbb{C}^4\mid p_3\neq 0 \vee p_4\neq 0\}$, and now when I consider the morphism $\phi$ the following orbits are still mapped to the origin:

- $\mathbb{C}^*\cdot (0,0,z,0)$
- $\mathbb{C}^*\cdot (0,0,0,w)$
- $\mathbb{C}^*\cdot (0,0,z,w)$

therefore $\phi^{-1}(0,0,0,0)$ is not a single orbit, hence it is not a geometric quotient.

Am I doing something wrong? I intuitively thought that, even $U$ does not coincide with the stable points (which would give me a geometric quotient), still throwing away only some bad points would have helped me