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 $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} X & \ra{f} &Y\\ \da{q_1} & & \da{q_2}\\ Z & \ra{g} & T \\ \end{array} $$ where $Z,T$ are the quotients $X/G,Y/G$ respectively but I could not figure out anything out of it.
Thanks in advance.
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 $\left[ \begin{smallmatrix} u \\ w \end{smallmatrix} \right] \neq \left[ \begin{smallmatrix} 0 \\ 0 \end{smallmatrix} \right]$.