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?



Affine GIT is an open map?
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.


