I learnt that if I have a principal $G$-bundle, say $(X,p,B)$, and if $F$ is a left $G$-space, I can build a fibre bundle $((X\times F)/G,p',B)$ with the action $(x,f)g=(xg,g^{-1}f)$ and the fibres would be homeomorphic to $F$. But if I start with a given fibre bundle $(E,q,B')$ with the structure group $G$, can I always find a principal $G$-bundle that gives me this fibre bundle following the above way?
I am thinking if there is always a representation $\rho:\pi_1(B)\to\text{Homeo}(F)$, then I can build a bundle with the total space $(\tilde{B}\times F)/\pi_1(B)$ where the action is given by $(\tilde{b},f)\gamma=(\tilde{b}\gamma,\rho(\gamma)^{-1}f)$, but the structure group considered here then would be $\pi_1(B)$ instead of $G$, isn't it? This is my another confusion related to the monodromy representation that I read from Farb and Margalit's Primer on MCG... And it seems it's not always true that we can find such a representation.
Thanks in advance for any idea or references...
Yes. Explicitly, what it means for the fiber bundle to have structure group $G$ is that it can be described using transition functions taking values in $G$, and you use those transition functions to describe the corresponding principal $G$-bundle.
Principal $G$-bundles are classified by maps $B \to BG$, and of these, the ones that correspond to maps $\pi_1(B) \to G$ are "flat."