The first trivial example of a fibre bundle $E$ is a product bundle $E=F \times B$, with fibre $F$ and base space $B$. Of course in this trivial example, one can exchange base space and fibre and think of the product bundle $E$ as a fibre bundle with fibre $B$ and base space $F$.
Under which necessary conditions can one exchange fibre and base space, i.e. regard $E$ as an $F$-bundle over $B$ as a $B$-bundle over $F$?
It seems that when one has a flat connection, one "knows" how to move from fibre to fibre (that is, from $F_x$ to $F_y$) and makes sure that one ends up in the same place after a small loop. When global loops also end up in the same spot and not on the other side of the fibre (i.e. the holonomy is not just discrete, but trivial), then it looks like given any point in the fibre, one can trace out the base space $B$, i.e. one could interpret $E$ as a $B$-bundle over $F$.
So, is a flat $F$-bundle over $B$ with trivial holonomy also a $B$-bundle over $F$? If not, could someone provide a counter-example?