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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.

9sven6
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S.Y.Park
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2 Answers2

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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:

  1. 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.)

  2. 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).

Moishe Kohan
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  • This is very helpful. Can you provide a link to some discussion of the proof assuming that G is locally compact and 2nd countable? The link you have about Aren's Lemma seems to use the completely metrizable assumption. – Daniel Grimmer Jul 12 '23 at 21:00
  • @DanielGrimmer: See the reference I gave here. – Moishe Kohan Jul 12 '23 at 21:33
  • Thanks! One more quick question, I see in Theorem 2.13 of "Topological Transformation Groups'' the following condition on $G$. Let $G^\prime$ be an open subgroup of $G$ such that $G/G^\prime$ is countable and $G^\prime/G_0$ is compact. Here $G_0$ is the identity component. From what you have said above, I take it that the existence of such a $G^\prime$ is equivalent to $G$ being second countable. Is that correct? Neither $G^\prime$ nor $G_0$ are used in the proof. – Daniel Grimmer Jul 12 '23 at 23:09
  • Assuming the group is 2nd countable, it has countably many components. Thus, take $G'=G_0$. – Moishe Kohan Jul 12 '23 at 23:42
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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.

Tsemo Aristide
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  • The first sentence ends oddly. But for the rest it is a clear answer – 9sven6 Aug 17 '20 at 13:17
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    It's not clear why $W$ is open though. The assumption is that $p$ maps an open neighborhood $U$ of $e$ to an open set, this $U$ might has nothing to do with $W$. – Arctic Char Aug 17 '20 at 17:15
  • Indeed, your argument is incomplete: The assumption is that there exists an open neighborhood $U$ of $e$ such that $p(U)$ is open. You still have to prove that for every open nbd $W$ of $e$ the subset $p(W)$ is open. – Moishe Kohan Aug 18 '20 at 23:53
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    Dear Tsemo: I wonder if you realize that the result you "proved" is simply false. – Moishe Kohan Aug 19 '20 at 22:03