Triviality of a bundle given that the corresponding bundle of covering spaces is trivial Let $F,\ E,\ B$ be connected manifolds and  $$F\hookrightarrow E\to B$$ be a fiber bundle. Suppose that $F$ is simply-connected and that $\pi_1(E)=\pi_1(B)= \mathbb Z_2$. 
Now consider the universal covering manifolds $\tilde E$ and  $\tilde B$ of $E$ and $B$ respectively. Suppose that we have the following fiber bundle 
$$F\hookrightarrow \tilde E\to \tilde B$$
and given that this bundle is trivial i.e. $\tilde E=F\times \tilde B$. Does this imply that also $E=F\times B$?
 A: No.  For instance, let $B=\mathbb{R}P^2$ and let $E$ be the $S^2$-bundle on $B$ classified by the map $B\to \mathbb{R}P^\infty=B\mathbb{Z}_2\to BO(3)$ where the first map is the inclusion and the second map is induced by the map $\mathbb{Z}/2\to O(3)$ sending a generator to $-I$.  This $S^2$-bundle is nontrivial since it is not orientable: going around a loop that generates $\pi_1(\mathbb{R}P^2)$, the orientation on the fiber gets reversed since $-I$ is orientation-reversing.  Since $F=S^2$ is simply connected, the long exact sequence on homotopy groups immediately implies $\pi_1(E)\cong\pi_1(B)\cong \mathbb{Z}_2$.
On the other hand, the universal cover bundle $\tilde{E}$ on $\tilde{B}$ is the bundle classified by the composition $\tilde{B}\to B\to B\mathbb{Z}_2\to BO(3)$.  But since $\tilde{B}$ is simply connected, the composition $\tilde{B}\to B\to B\mathbb{Z}_2$ is already nullhomotopic, and thus the classifying map of the bundle $\tilde{E}$ is nullhomotopic so it is a trivial bundle.
