Let $(X,G)$ be a group action, must the restriction of the orbit map be an open mapping?
In other words, $\phi: G \times X \to X$ is a group action, for any $x\in X$, $\phi_x:G \to Gx$ is an open mapping, where $Gx=\{gx:g \in G \}$?
I have known that for any $x\in X$, the orbit map $\phi_x:G \to X$ not be necessarily an open mapping.
Thanks a lot.
The answer depends on which topology you put on the orbit $Gx$. Namely, there are two possibilities:
(1) Since the orbit $G.x$ is the coset space $G/G_x$, where $G_x$ is the isotropy subgroup at $x$ you can put on the orbit $Gx$ the quotient topology. Then, by definition, the projection map $\pi : G \to G/G_x$ is open hence the map $\phi_x : G \to Gx$ is indeed open.
(2) Since the orbit $Gx$ is a subset of $X$ you put on $G.x$ the induced topology. In this case the map $\phi_x : G \to Gx$ is not necessarily open. Here it is the standard example: Let $X$ to be the torus $T^2 := \mathbb{R}^2/\mathbb{Z}^2$. I am going to use the notation $[(x,y)] \in T^2$ to denote the equivalence class of $(x,y) \in \mathbb{R}^{2}$. The group $G = \mathbb{R}$ acts on $X$ as follows $$\phi: \mathbb{R} \times X \to X$$ given by $$\phi(t,[(x,y)]) = [(t + x , y + t \sqrt{2})]$$ you can check that this is indeed an smooth action. Finally, since $\sqrt{2}$ is an irrational number each orbit $G x$ is dense in $T^2$. So the induced topology on $G x$ is different from the topology of $G / G_x = \mathbb{R}$ explained in (1). Then the map $\phi_x : G \to Gx$ is not open.
A typical hypothesis on the action of $G$ that implies both topologies on $Gx$ are the same is the so called proper action. You can look in internet for such a definition. So if the $G$ action is proper then the map $\phi_x : G \to Gx$ is indeed an open map.