I'm reading now D. Mumford's Geometric Invariant Theorem and have faced an argument in the proof of proposition 0.1 on page 4 that I don't understand:
main setting: let $X/S$ a $S$-scheme and a group and $S$ scheme $G/S$ acts at $X$ with $\sigma: G \times_S X \to X$ (c.f. page 2).
Proposition 0.1 Let $\sigma$ be an action of $G/S$ on $X/S$ and suppose $(Y, \phi)$ is a geometric quotient (see definition 0.6 below) of $X$ by $G$. Then $(Y, \phi)$ is a categorical quotient of $X$ by $G$, hence is is unique up to isomorphism. ...
Proof. Suppose $\psi:X \to Z$ is any $S$-morphism such that $\psi \circ \sigma = \psi \circ p_2$, as morphisms from $G \times_S X$ to $Z$.
$p_2$ is the canonical projection $G \times_S X$ to $X$.
To construct a morphism $\chi: Y \to Z$, let $\{V_i\}$ be any affine open covering of $Z$. Then for each $i$, $\psi^{-1}(V_i)$ is an invariant open subset of $X$, hence by condition (ii) of definition 0.6 $\psi^{-1}(V_i)= \phi^{-1}(U_i)$ for some subset $U_i$ of Y. ...
the Definition 0.6 refered few times:
Q: why Definition 0.6 (ii) imply that such $U_i \subset Y$ with $\psi^{-1}(V_i)= \phi^{-1}(U_i)$ exist. (ii) tells me that $\phi$ is surjective and the image of canonical map $\Psi=(\sigma, p_2): G \times_S X \to X \times_S X$ is $X \times_Y X$. why this imply the existence of the $U_i$ with described property? what I need is that $\psi^{-1}(V_i)$ has property of a saturated set w.r.t. $\phi$, i.e. that $\phi^{-1}(\phi(\psi^{-1}(V_i)))=\psi^{-1}(V_i)$. then we could define $U_i :=\phi(\psi^{-1}(V_i))$. on the other hand Definition (ii) gives no reason that $\psi^{-1}(V_i)$ have this property.