Homeomorphism from the coset space $G/G_x$ to the orbit $xG$

compactnessgeneral-topologygroup-actionsgroup-theorytopological-groups

I'm currently working through a set of notes on fiber bundles and I'm struggling to prove something mentioned in my notes.

Question

Let $G$ be a topological group and $X$ be a Hausdorff, paracompact topological space with right $G-$action. That is, there exists a continuous map, $\mu: X \times G \to X : (x, g) \mapsto xg$ that satisfies the group action axioms.

For $x \in X$, let $xG = \{xg : g \in G \}$ and $G_x = \{ g \in G : xg = x\}$. Finally, consider the set $ G \big / G_x = \{G_xg : g \in G \}$, with quotient topology defined by the projection $p: G \to G \big / G_x : g \mapsto G_xg$. I want to prove the claim that $G \big / G_x $ is homeomorphic to $xG \subset X$.

My attempts

Consider the map $ \phi: G \big / G_x \to xG : G_xg \mapsto xg$. I was able to prove that this map is well-defined, bijective and continuous. So in order for it to be a homeomorphism, it remains to prove that it is an open map.

Now, if $G \big / G_x $ were compact, then the claim would be proved because continuous bijective functions from a compact space to a Hausdorff space are homeomorphisms. However, I do not know if $G \big / G_x $ is in fact compact and I could not think of a way to prove/disprove this fact. So, I tried a different method.

Suppose $U \subset G \big / G_x $ is an open set. Then, $p^{-1}(U) = \{ g : G_xg \in U \}$ is open in $G$ and $\phi(U) = \{ xg : G_xg \in U \} = \{xg : g \in p^{-1}(U) \} $. If the map $g \to xg$ is open, then $\phi(U)$ is open and the theorem has been proved. However, I also do not know how to prove this.

Summary

More specifically, I am asking the following questions.

Let $G$ be a topological group and $X$ be a Hausdorff, paracompact topological space with right $G-$action that is both continuous and proper.

  1. Is $G \big / G_x $ necessarily compact? If so, how could I go about proving this?

  2. Is the map $g \to xg$ is open? If so, how can I prove this?

  3. If 1 and 2 are false, how do I prove $G \big / G_x \simeq xG$?

Note: Thanks to Moishe Kohan's comment, I included the condition that the G-action be proper.

Best Answer

Instead of the paracompactness of $X$ I will assume that $X$ is 1st countable (e.g. metrizable). I will keep the Hausdorff assumption for $X$. I will also assume that $G$ is metrizable (equivalently, is Hausdorff and 1st countable). I am quite sure, all these assumptions will suffice for the purpose of discussion of fiber bundles in the notes that you are reading.

Lemma. Suppose that $G\times X\to X$ is a proper action. Then for every $x\in X$, the orbit map $o_x: G\to Gx$ descends to a homeomorphism $\phi: G/G_x\to Gx$ (where the orbit $Gx$ is equipped with the subspace topology induced from $X$).

Proof. You already know that $\phi$ is bijective (clear) and continuous. Hence, we just need to verify that its inverse $\psi: Gx\to G/G_x$ is continuous. Since $X$ is assumed to be 1st countable, it suffices to prove that for every sequence $g_n\in G$, if $g_nx$ converges to $y=gx\in Gx$, then the sequence $[g_n]\in G/G_x$ converges to $[g]\in G/G_x$. First, observe that the subset $$ K=\{y\}\cup \{g_nx: n\in {\mathbb N}\} $$ is compact. Hence, its preimage in $G$ (under $o_x$) is a compact $C$ (by properness of the action). Clearly, $g_n\in C$ for each $n$. After extraction, we can assume that $g_n$ converges to some $h\in C$ (here I am using the assumption that $G$ is metrizable). By the continuity of the orbit map, $o_x(h)=y$, hence, $[h]=[g]$ in $G/G_x$. This proves continuity of $\psi$. qed

See also my answer here for a proof which assumes only that $Gx$ is closed (instead of properness of the action) but makes further assumptions: $G$ is locally compact and $X$ is completely metrizable.

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