Vector bundle and principal bundle

Question

Why fiber of $(P\times V)/G\rightarrow P/G$ isomorphic to $V$ ?

I think the fiber should be $V/G$, but it is not isomorphic to $V$

Picture below is from the 66 page of Jost's Riemannian Geometry and Geometric Analysis

Answer 1

Fix a point $x$ in the base. Then the fiber over $x$ consists of the orbits of all pairs $(p,v)\in P\times V$ such that $p$ lies in the fiber over $x$. Now fix a point $p_0$ in that fiber and consider the map from $V$ to the fiber of $P\times_GV$ over $x$ which sends $v$ to the orbit of $(p_0,v)$. This is injective since $p_0\cdot g=p_0$ implies $g=e$, so $(p_0,v)$ and $(p_0,w)$ lying in the same orbit implies $v=w$. On the other hand, for each $p$ in the fiber, there is an element $g\in G$ such that $p=p_0\cdot g$. Hence the orbit of $(p,v)$ contains $(p_0,g\cdot v)$ which implies surjectivity.