Nice to meet you. I have a question about associated fiber bundle.
Let $P(M,G)$ be a principal $G$-bundle. The action of $G$ on a manifold $F$ is defined from left and the action of $G$ on $P\times F$ is defined by $$(u,f) \to (ug,g^{-1}f)$$
The associated fiber bundle is defined as a quotient of $P\times F$ by the relation $\sim$: $$E = P\times F / \sim$$
where $$\quad(u,f) \sim (ug,g^{-1}f)$$
Let $\pi_E$ be the projection of $E$ onto $M$.
Every point $x\in M$ has a neighborhood $U$ such that $\pi^{-1}(U)$ is isomorphic to $U\times G$.
Many books don't have a detailed explanation of "It follows that the isomorphism $\pi^{-1}(U)\approx U\times G$ induces an isomorphism $\pi^{-1}_{E}(U)\approx U\times F$". Could you teach me how to prove this? Why $\pi^{-1}_{E}(U)\approx U\times F$?
Here's how you construct the local trivializations. I'll write an arbitrary element of $E=(P\times F)/G$ as $[u,f]$. Let $$\{(U_{\alpha},\phi_{\alpha})\ |\ \alpha\in J\}$$ be the atlas for the original bundle $\pi:P\to M$, where the local trivializations have the form $\phi_{\alpha}:U_{\alpha}\times G\to\pi^{-1}(U_{\alpha})$. Define the new trivializations by: $$\Phi_{\alpha}:U_{\alpha}\times F\to\pi_F^{-1}(U_{\alpha}),\quad \Phi_{\alpha}(p,f) = [\phi_{\alpha}(b,1),f].$$ That is, $\Phi_{\alpha}$ is defined as the composition $$U_{\alpha}\times F\xrightarrow{\ \phi_{\alpha}(-,1)\times id_F \ }\pi^{-1}(U_{\alpha})\times F\xrightarrow{\ \pi_G \ }\pi_F^{-1}(U_{\alpha})$$ and therefore must be continuous. To construct the inverse, let $\rho:G\times F\to F$ be the action of $G$ on the manifold $F$. then, you can show that the composition $$\pi^{-1}(U_{\alpha})\times F\xrightarrow{\ \phi_{\alpha}^{-1}\times id_F \ }U_{\alpha}\times G\times F\xrightarrow{\ id\times\rho \ }U_{\alpha}\times F$$ is $G$-equivariant, and therefore induces a map $$\pi_F^{-1}(U_{\alpha})\to U_{\alpha}\times F,\quad [u,f]\mapsto(id\times\rho\circ\phi_{\alpha}^{-1}\times id_F)(u,f)$$ by the universal property of quotients. This induced map is exactly $\Phi_{\alpha}^{-1}$.