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