Fibration: if map $i^*: H^*(X, G) \to H^*(F, G)$ is surjective, then action of $\pi_1(B)$ on $H^*(F, G)$ trivial?

Question

For a fibration $F \overset{i}{\to} X \overset{p}{\to} B$ with $B$ path-connected, if the map $i^*: H^*(X, G) \to H^*(F, G)$ is surjective, then does it necessarily follow that the action of $\pi_1(B)$ on $H^*(F, G)$ is trivial?

Answer 1

Yes. If you look at how the action is defined, the maps given by inclusion $F\to X$ and the homotopy equivalence on the fiber followed by inclusion $F\to F \to X$ are homotopic. This just follows from the fact that the map $F\to F$ is constructed by lifting $F\times I \to I \to B$. Thus, if we can extend a cohomology class $F\to K(G, n)$ to $X\to K(G,n)$, it will follow that the action on this class is trivial. If we can extend all the classes, then the action is trivial.