# group action on a fiber bundle

I am in the following situation,

Let $F \rightarrow E \xrightarrow{p} B$ be a fiber bundle and suppose that we have a group $G$ acting freely on $E$ and $B$, satisfying $p(g\cdot x) = g \cdot p(x)$; thus we have a well defined map

$\overline{p}: E/G \rightarrow B/G$, my question is: is that map a fiber bundle with fiber $F$?

I have seen this argument, for instance, proving that the borel construction $EG \times_G X \rightarrow BG$ is a fiber bundle with fiber $X$, by applying the above construction to the projection map $EG \times X \rightarrow EG$.

So, is that a general result or there are some conditions that I am missing

• I think your question has been addressed (at least in part) in mathoverflow.net/questions/52224/…. I'll also point to a statement I found in tom Dieck's "Transformation Groups" as Exercise 3.21 on page 121. – Tyrone Oct 26 '17 at 7:56
• @Tyrone: I don't currently have access to that book - would you mind writing out the statement? – Jason DeVito Aug 22 '18 at 20:10
• @JasonDeVito, it's too long for me to reasonably type, but the relevant parts of the book are available on google books. – Tyrone Aug 22 '18 at 21:19
• Tyrone: I looked on google books and got the "these pages are not part of the preview" issue. Anyway, I will try to get my hands on a copy. I feel that the OPs question must be true in general, but haven't been able to figure out a proof... – Jason DeVito Aug 22 '18 at 21:29

I am going to assume that all the relevant spaces $(F,E,B,G)$ are compact, smooth manifolds and that the $G$ action is smooth.
Now, the map $E\rightarrow B$ is a submersion, as is $B\rightarrow B/G$, so the composition $E\rightarrow B/G$ is a submersion. On the other hand, this is equal to the composition $E\rightarrow E/G\rightarrow B/G$. It follows from the chain rule that the map $E/G\rightarrow B/G$ is a submersion.
Since $E$ is compact, so is $E/G$, so the map $E/G\rightarrow B/G$ is proper. By Ehressman's theorem, the map $E/G\rightarrow B/G$ is a fiber bundle.
It remains to determine the fiber. But the fact that $E\rightarrow B\rightarrow B/G$ is the same map as $E\rightarrow E/G\rightarrow B/G$ exactly says the the bundle $E\rightarrow B$ is the pull back of $E/G\rightarrow B/G$. Since $E\rightarrow B$ has fiber $F$, that must be the fiber for $E/G\rightarrow B/G$ as well.