I am trying to prove that every connected topological manifold $M$ (assume Hausdorff and second countable) is homogeneous in the sense defined by Bourbaki's General Topology Ch.3 Sec.2.5 pg 232.
The requirement on $M$ for topological homogeneity is that it admits a group action $\theta:G\times M\to M$ which is continuous and transitive and also has for every $x\in M$ that the orbit map $\theta_x:G\to M$ defined by $\theta_x(g):=\theta(g,x)$ is an open map.
The requirement that all of these orbit maps are open is here included so that the quotient of $G$ with its stabilizer subgroup $G/G_x$ ends up being homeomorphic to $M$. Essentially what I am asking is: Is every connected topological manifold $M$ homeomorphic to the quotient of some two topological groups?
Firstly, I know that $M$ being connected implies that $\text{Homeo}(M)$ acts transitively on it, see here. Moreover, I know from Arens (1946) that $\text{Homeo}(M)$ is a topological group in this case (edit: with the compact-open topology). In fact, I know from here that this is the coarsest topology for $\text{Homeo}(M)$ which makes it a topological group and which makes its natural action on M continuous in both variables simultaneously.
Thus, all that remains is to show that the orbit maps $\theta_x:\text{Homeo}(M)\to M$ defined by $\theta_x(h)=h(x)$ are open maps. Let $U$ be a generic open set in $\text{Homeo}(M)$. Our goal is to show that the image of this map, $\theta_x(U)=U(x)$, is an open set in $M$.
I think I could prove this if I restrict my attention to topological manifolds which can be given a smooth structure. Then I can talk about $\text{Diff}(M)$ instead and get the open map condition by thinking about the Lie group exponential. I'd prefer to keep the result as general as possible, however. Any thoughts on how to proceed?
Best Answer
I now have a full solution which I will sketch below. This proof relies on what I will call the ``Continuous Bump Lemma'' which guarantees the following. Let $x\in\mathcal{M}$ and $N\subset \mathcal{M}$ an open neighborhood thereof, $x\in N$. There exists an open neighborhood $T\subset N$ with $x\in T$ such that for every $y\in T$ there exists a homeomorphism $\phi:\mathcal{M}\to \mathcal{M}$ which maps $T$ into itself while acting trivially outside of $T$ and while also mapping $x$ to $y$ as $\phi(x)=y$.
Proof using this lemma: Recall that our goal is to show that $U(x)\subset \mathcal{M}$ is an open set whenever $U\subset\text{Homeo}(\mathcal{M})$ is an open set according to the compact-open topology. Recall also that the following sets form a sub-basis for the compact-open topology, $V(K,W):=\{h\in \text{Homeo}(\mathcal{M})\vert h(K)\subset W\}$, where $K\subset \mathcal{M}$ is any compact set and $W\subset \mathcal{M}$ is any open set. This set contains all of the homeomorphisms which map $K$ into $W$. A generic open set $U\subset \text{Homeo}(\mathcal{M})$ is an arbitrary union of finite intersections of such sets, $U=\cup\cap U_\alpha = \cup\cap V(K_\alpha,W_\alpha)$ for some index $\alpha$. In order to prove that this generic $U$ maps to an open set $U(x)\subset \mathcal{M}$ it is sufficient to prove that every $V(K_\alpha,W_\alpha)$ does so individually.
For any $U=V(K,W)$, its image under $\theta_x$ is \begin{align} U(x)=\{y\in \mathcal{M} \ \vert \ \exists h\in \text{Homeo}(\mathcal{M}) \text{ with } y=h(x) \text{ and } h(K)\subset W\}. \end{align} Namely, $U(x)$ is a collection of all of the places, $y\in \mathcal{M}$, where $\text{Homeo}(\mathcal{M})$ might send $x$ while also mapping $K$ into $W$. Our goal is to show that this set $U(x)$ is open for every compact set $K$ and open set $W$. I will now prove this by showing that for any $y\in U(x)$ there is an open neighborhood $T_y\ni y$ with $T_y\subset U(x)$. Taking a union over all of these $T_y$ reveals that $U(x)=\cup_y T_y$ is a union of open sets and hence is itself an open set.
Let us first consider the case where $y\not\in h(K)$. Since topological manifolds are regular Hausdorff spaces we can find an open neighborhood $N\ni y$ which is disjoint from $h(K)$. Applying the Continuous Bump Lemma we know that there exists an open set $T\subset N$ with $y\in T$ such that for every $z\in T$ there exists a homeomorphism $\phi:\mathcal{M}\to\mathcal{M}$ which maps $y$ to $z$ as $\phi(y)=z$ while being the identity outside of $T\subset N$.
Recall that we have assumed that $y\in U(x)$. Hence, there exists some homeomorphism, $h\in \text{Homeo}(\mathcal{M})$, with $y=h(x)$ and with $h(K)\subset W$. Now consider the homeomorphism, $\phi\circ h$, and note that it too has $\phi\circ h(K)\subset W$ since $\phi$ acts trivially outside of $N$. (Recall that $h(K)$ and $N$ are disjoint). Hence, we have that $z=\phi\circ h(x)$ is in $U(x)$. Note that we can here let $z$ range over $T$. This confirms as desired that for any $y\in U(x)$ with $y\not\in h(K)$ we have an open set $T\subset U(x)$ with $y\in T$.
Let us next consider the case of $y\in U(x)$ with $y\in h(K)\subset W$. The proof in this second case is nearly identical to the first case, except that we take the neighborhood $N\ni y$ to be contained within $W$ as $N\subset W$. The construction of $T$ and the $\phi$ map then proceeds exactly as before. Note that this time we have $T\subset N\subset W$. In this case, we still have that $\phi\circ h(K)\subset W$ by the following two observations. Firstly, it follows from $\phi$ acting trivially outside of $T$ that the parts of $h(K)$ which are outside of $T$ stay within $N\subset W$. Secondly, it follows from the fact that $\phi$ maps $T$ into itself that the parts of $h(K)$ which are within $T$ will stay within $N\subset W$. Thus (just as in the first case) we have proved the desired claim. For any $y\in U(x)$ with $y\in h(K)$ there exists an open set $T\subset U(x)$ with $y\in T$.