# $G$-variety maps send closed orbits to closed orbits if $G$ is linear reductive

Is it true that if $$G$$ is a linear reductive group and $$X$$ and $$Y$$ are two $$G$$-varieties with a surjective $$G$$-map $$f:X\rightarrow Y$$ then $$f$$ sends a closed orbit to a closed orbit? If not, is there an easy condition to guarantee this property?

I tried to use the fact that we have a commutative diagram $$\newcommand{\ra}{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} X & \ra{f} &Y\\ \da{q_1} & & \da{q_2}\\ Z & \ra{g} & T \\ \end{array}$$ where $$Z,T$$ are the quotients $$X/G,Y/G$$ respectively but I could not figure out anything out of it.

No. Take $$G = \mathbb{C}^{\ast}$$, $$X = \mathbb{C}^2$$ and $$Y = \mathbb{C}$$, with $$t \in G$$ acting by $$(x,y) \mapsto (tx, t^{-1} y)$$ and $$z \mapsto tz$$. Let the map $$X \to Y$$ be projection on the first coordinate. Then the closed orbit $$xy=1$$ is sent to the non-closed (in fact, open) orbit $$z \neq 0$$.
As a more sophisticated example, to show that replacing "reductive" with "simple" doesn't help, let $$G = SL_2$$, let $$X$$ be $$2 \times 2$$ matrices and let $$Y = \mathbb{C}^2$$. We take $$G$$ to act by left multiplication in each case, and map $$X \to Y$$ by sending a matrix to its first column. Then the closed orbit $$\det \left[ \begin{smallmatrix} u& v \\ w & x \end{smallmatrix} \right]= 1$$ is sent to the non-closed (in fact open) orbit $$\left[ \begin{smallmatrix} u \\ w \end{smallmatrix} \right] \neq \left[ \begin{smallmatrix} 0 \\ 0 \end{smallmatrix} \right]$$.
• Thank you for your answer and for your very simple examples. Are aware of a condition to guarantee that this is true? In my case, $G=SL_3$. – Levent Jan 20 at 0:42