I am trying to prove the following lemma -
Lemma - Let $X$ be a variety and let $G$ be an algebraic group acting algebraically on $X$. Let $\pi:X\rightarrow X//G$ be a good categorical quotient. Then the following are equivalent:
$X$ has a $G$ - invariant Zariski dense open subset $U_0$ such that $G\cdot x$ is closed in $X$ for all $x\in U_0$
$X//G$ has a Zariski dense open subset $U$ such that $\pi|_{\pi^{-1}(U)}:\pi^{-1}(U)\rightarrow U$ is a geometric quotient.
I have managed to prove $1\Rightarrow 2$. For $2\Rightarrow 1$ here is my attempt -
Let $U_0=\pi^{-1}(U)$. Since $\pi|_{U_0}:U_0\rightarrow U$ is a geometric quotient we have that $U_0$ is $G$ - invariant and $G\cdot x$ is closed in $U_0$ for all $x\in U_0$. From here on I'm stuck.
(a) Why is $G\cdot x$ closed in $X$?
(b) Why is $U_0$ dense? I can show it is dense if $\pi$ is open but How would I show $\pi$ is open?
Thank you.
For (a): You have that $U_0=\pi^{-1}(U)$. Let $x\in U_0$ and $y\in \overline{Gx}$ (the closure in $X$). Then, by definition of the quotient, $\pi(y)=\pi(x)$ and therefore you have $y\in\pi^{-1}(\pi(x))\subseteq \pi^{-1}(U)=U_0$. Since $y$ was arbitrary, you have $\overline{Gx}\subseteq U_0$ and since $Gx$ was already closed in $U$, we have $\overline{Gx}=Gx$.
For (b): Note that since $U_0$ is $G$-invariant, so is $Z:=\overline{U_0}$. We want to show that $Z=X$. Let $I$ be the ideal sheaf of $Z$ in $X$. We want to show that it is the zero sheaf. If it were not, then since $Z$ is $G$-invariant, so is $I$ and we may pick an open, $G$-invariant subset $V\subseteq X$ such that $I(V)$ contains a nonzero element $f$. Hence, $f\in\mathcal O_X(U)^G$ and $f(U_0)=0$. Since $\pi$ is a good quotient, we have $f=\pi^\sharp(\tilde f)$ and $\tilde f(U)=\tilde f(\pi(U_0))=f(U_0)=0$. Since $U$ is dense in $X/\!\!/G$, this means that $\tilde f=0$. Consequently, $f=0$.