I would like to understand why, given a principal $G$-bundle $(P,M,G)$ and an associated $G$ bundle $P_F=P \times F / \ \sim $, the fiber of $P_F$ is isomorphic to $F$, if this is true.
I was thinking something on these lines. We want to have a map from the fiber $\pi_{P_F}^{-1}(m)$ to $F$. To construct this, let $x \in \pi_{P_F}^{-1}(m)$. We can fix an element $p \in \pi_P^{-1}(m)$ on the principal bundle $P$ and write $x=[p,f(x)]$ for some $f$. Note that $p$ does not depend on $x$, only $f$ does according to the choice made.
Fixing $p$, we have built a map $u : \pi_{P_F}^{-1}(m) \rightarrow F$ such that $x \rightarrow f(x)$.
Does it make sense this approach ? Can it be completed to a proof ?
Further, if this statement is true, can we say rigorously that $P_F$ is a $G-$bundle with fiber $F$ ? I am a bit confused here because if this was the case $F$ would be isomorphic to $G$ (because in any principal G-bundle the fibers should be isomorphic the the structure group G), which actually would make the previous statement trivial... Can somebody help me putting together the pieces of the puzzle and correct me if I am making wrong statements ?
This follows from the fact that the action of $G$ is free, let $p:P\rightarrow M$ the projection of the principal bundle. Denote by $p_F:P_F\rightarrow M$ the projection of the associated bundle, that is the quotient of $P\times F$. Let $m\in M$, and $y\in p^{-1}(m)$. Let $y,y'\in F$, we denote by $[x,y], [x,y']$ the image of $[x,y]$ and $[x,y']$ in $P\times F/\simeq$; $p_F([x,y])=p_F([x,y'])$ if and only if there exists $g\in G$ such that $g.(x,y)=(g.x,g.y)=(g.x,g.y)=(x,y')$. This implies that $g=Id_G$ and $y=y'$. Thus the map $F\rightarrow p_F^{-1}(m)$ defined by $y\rightarrow [x,y]$ is injective. It is also surjective since $G$ acts transitively on $p^{-1}(x)$. For every $x'\in p^{-1}(m), x'=g.x, [x',y]=[x,g^{-1}.y]$.