Does a continuous group action induce an open map? I am currently reading the book "The Topology of Fibre Bundles" written by Norman Steenrod. The book says a sentence without a proof and I was trying to give a rigorous proof of the sentence.
The settings for the sentence are:
Let $B$ be a topological group which acts on a topological space $X$ continuously, that is, the map $B\times X \to X$ is continuous.
Let $e\in B$ be the identity element of $B$.
Fix an element $x_0\in X$ and define a map $p:B\to X$ by $p(b)=b\cdot x_0$ for $b\in B$.
The statement I wish to prove is:
Suppose that $p$ maps an open neighborhood of $e$ onto an open neighborhood of $x_0$.
Then $p$ is an open map.
It seems that the context of the book assumes the Hausdorff condition on the space $X$ but I guess that condition is irrelevant to this problem.
 A: Without further assumptions, the claim is simply false. As an example, consider the group $G={\mathbb Q}$ equipped with discrete topology, $X=G$ equipped with the standard order topology. Assume that the action $G\times X\to X$ is the one given by the left multiplication. I will leave it to you to verify that the action is continuous. Take $x_0=0$. The corresponding orbit map $p: G\to X$ maps $G$ onto $X$. Thus, we found an open subset of $G$ (i.e. $G$ itself) whose image under $p$ is open. However, the map $p$ clearly is not open (just take the image of any singleton under the orbit map). So, what Tsemo proved for you is false: His mistake was to assume that every neighborhood of $e$ has open image under $p$.
However, the statement you are trying to prove holds under the following mild extra assumptions which hold in most natural applications of the fiber bundle  theory:

*

*Assume that your group $G$  (I do not like using the letter $B$ for groups)
is locally compact and 2nd countable. (For instance, every Lie group is.)


*Assume that $X$ is completely metrizable (i.e. admits a complete metric metrizing its topology). - Note that ${\mathbb Q}$ (with the order topology) in my example is not completely metrizable.
Then indeed the orbit map is open provided that one of the two equivalent conditions holds:
a. There exists an (open, as customary in the US literature on general topology) neighborhood $U$ of $e$ such that $p(U)$  is open in $X$.
b. $p(G)$ is an open subset of $X$.
A proof of openness of $p$ under these extra assumptions is an application of the Baire Category theorem (arguing as in the proof of Arens Lemma in the theory of transformation groups, see here).
A: Let $V$ be a non empty open subset of $B$, and $v\in V$, $W=v^{-1}V=\{v^{-1}w, w\in V\}$ is an open subset which contains $e$, we deduce that $p(W)$ is open. Since $e\in W$, $p(V)=v.p(W)$ is open since the map defined on $X$ by $f:y\rightarrow v.y$ is invertible, we deduce that $p(V)=f(p(W))$ is open.
