TITLE: Affine GIT is an open map?
QUESTION [8 upvotes]: Let $k$ be a field, $X= \text{Spec}\,A$ be an affine scheme, with $A$ a finitely generated $k$-algebra. $G=\text{Spec}\,R$ is a linearly reductive group acting rationally on A, i.e. every element of $A$ is contained in a finite-dimensional $G$-invariant linear subspace of $A$. By Nagata's theorem, $A^G$ is a finitely generated $k$-algebra. We have the affine GIT quotient $X \rightarrow X/G := \text{Spec}\,A^G$ induced by the inclusion $A^G \rightarrow A$ of $k$-algebras.
Question. Is the affine GIT quotient, viewed as a map of the underlying topological spaces, necessarily an open map? It does not need to be an open immersion of schemes. If not, any counterexample?

REPLY [9 votes]: Categorical quotients are in general very far away from being open. In fact, it is a theorem of Chevalley (I think) that a morphism $\pi:X\to Y$ onto a normal variety $Y$ is open if and only it is equidimensional.
Unless $G$ is finite, this is a very rare condition for quotient morphisms.
The idea behind this is the following: Assume that the fiber $X_y$ has bigger dimension than the generic fiber dimension $\dim X-\dim Y$. Then there is a curve $C\subseteq Y$ passing through $y$ such that the fiber dimension jumps over $C$. Let $Z$ be the closure of $\pi^{-1}(C)\setminus X_y$ in $X$. Then for dimension reasons $X_y\not\subseteq Z$. Consider the open subset $U:=X\setminus Z$ of $Z$. Then $\pi(Z)=(Y\setminus C)\cup\{y\}$ is only constrictible but not open.
More formally, the openness of a morphism is expressed by the so-called going-down property of Cohen-Seidenberg.
Let's try this out for the quotient of $G=\mathbf G_m$ acting on $\mathbf A^3$ by $(tx,ty,t^{-1}z)$. Then
$$
\pi:\mathbf A^3\to\mathbf A^2:(x,y,z)\mapsto (u,v):=(xz,yz)
$$
is the quotient morphism mapping the plane $\{z=0\}$ to $(0,0)$. Take $C=\{u=0\}$. Then $\pi^{-1}(C)=\{x=0\}\cup\{z=0\}$ and the image of $U=\mathbf A^3\setminus\{x=0\}$ is $\mathbf A^2\setminus\{u=0\}\cup\{(0,0)\}$ which is not open.
PS: Quotient maps have some properties not shared by other morphism:

Images of closed $G$-stable subsets are closed.
A subset of $Y$ is open iff its preimage in $X$ is open.
The morphism is surjective.

These properties hold universally, i.e., even after base change.