I'm reading Geometric Invariant Theory by Mumford-Fogarty, but I can't understand some details in the proof that any geometric quotient is categorical.
Let $\sigma$ be an action of $G/S$ on $X/S$ and $(Y, \phi)$ a geometric quotient of $X$ by $G$. Then $(Y, \phi)$ is a categorical quotient of $X$ by $G$.
The proof goes like this. Let $\psi \colon X \to Z$ be such that $$\begin{matrix} G \times X&\stackrel{\sigma}{\longrightarrow}&X\\ \downarrow{p_2}&&\downarrow{\psi}\\ X&\stackrel{\psi}{\longrightarrow}&Z \end{matrix} $$ commutes. The goal is to find a unique $\chi \colon Y \to Z$ with $\psi = \chi \circ \phi$.
Let $\{V_i\}$ be an affine open cover of $Z$. Then $\psi^{-1}(V_i)$ is an invariant open subset of $X$ and hence $\psi^{-1}(V_i) = \phi^{-1}(U_i)$ for some subset $U_i$ of $Y$. Now since $\phi$ is subversive, $U_i$ is open. Also, since $\phi$ is surjective, $\{U_i\}$ cover $X$. Now, such $\chi$ has to satisfy $\chi(U_i) \subset V_i$ and therefore it must be defined by a set of homomorphisms $h_i$ such that
$$\begin{matrix} \Gamma(V_i, \mathscr{O}_Z)&\stackrel{h_i}{\longrightarrow}&\Gamma(U_i, \mathscr{O}_Y)\\ \downarrow{\psi^*}&&\downarrow{\phi^*}\\ \Gamma(\psi^{-1}(V_i), \mathscr{O}_X)&\stackrel{=}{\longrightarrow}&\Gamma(\phi^{-1}(U_i), \mathscr{O}_X) \end{matrix} $$ commutes. Since $\phi^*$ is injective, the $h_i$'s are uniquely determined if they exist. But $\psi^*(g)$ is an invariant element of $\Gamma(\phi^{-1}(U_i), \mathscr{O}_X)$ for $g \in \Gamma(V_i, \mathscr{O}_Z)$, and hence it is in the sub-ring $\phi^*[\Gamma(U_i, \mathscr{O}_Y)]$, which defines $h_i$ and hence $\chi$.
There are some steps that I don't quite understand and I'd be grateful if someone could add some more detail to them:
- How exactly do the ring homomorphisms $h_i$ determine the morphism $\chi \colon Y \to Z$?
- Why is $\psi^*(g)$ an invariant element of $\Gamma(\phi^{-1}(U_i), \mathscr{O}_Z)$?