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Suppose $G$ is a topological group that acts on a connected topological space $X$. Show that if this action is transitive (and continuous), then so is the action of the identity component of the group.

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Well $(g,x)\mapsto g x$ is an open map simply because for fixed $g$ the map $x\mapsto gx$ is open. But what i really need is, that for fixed $x$ the map $g\mapsto g x$ is open. This would suffice to prove the upper statement, but I think it is not true. – fk44 May 3 '12 at 15:29
Are you sure there aren't any other hypothesis on $X$? I think I have a stupid counter example. Take $H$ to be your favorite connected topological group and let $G = H\times \mathbb{Z}/2\mathbb{Z}$. So the identity component of $G$ is $H$. Let $X=G$ as a set and a group, but with the indiscrete topology. Then $X$ is connected, $G$ acts transitively on $X$ by the usual group multiplication, but $H$ does not. – Jason DeVito May 3 '12 at 18:02
I also have the feeling that it is just not true. Your counterexample seems to be right. This method should work for any group with non-trivial indentity component acting on a copy of itself with the indiscrete topology. But this is the first example in the first chapter of Dave Witte Morris' "Introduction to arithmetic groups". And he applys it alot in the first chapter. Especially he says that for the isometry group of a Riemannian manifold, if it acts transitively, i.e. for homogeneous spaces. – fk44 May 4 '12 at 10:31
Dear fk44, Isometry groups of Riemannian manifolds are pretty far from the counterexamples that you and @Jason are considering. In the context of Dave Witte Morris's book, all the actions $G\times X \to X$ will be very well-behaved, and in particular, the openness property that you want will be true. Regards, – Matt E May 4 '12 at 11:11
It's entirely possible that the author has, in a preface or something, a statement along the lines of "Every topological space considered will be Hausdorff...". But barring that, I'm in agreement that the counterexample works. Then again, I'm biased and am often wrong... ;-). – Jason DeVito May 4 '12 at 13:47

Once you assume the space $X$ is a connected homogeneous space with the group acting on it as $G$, the quotient map $\pi:G\to G/G_x$, where $G_x$ is a stabilizer, is an open map and $G/G_x\to X$ is a homeomorphism. This tells us that $g\mapsto gx$ is an open map. Now it follows that (in particular) $G_0$, the identity component, acts transitively on $X$ (since X is connected). This is what Morris assumed in (the exercise) his book I think.

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For your conclusion, i.e. $G_0$ acts transitively because $X$ is connected and $g\mapsto gx$ is open, don't you need some assumption on $G_0$? Specifically, $G_0$ need not be open. Considering the context it's probably okay to assume $G$ is a Lie group. – Donjim Feb 5 '15 at 19:02
I just thought of one more thing, it's not clear why $g\mapsto gx$ is closed, which is necessary in order to make $G_0$ clopen. – Donjim Feb 5 '15 at 19:33

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