Fibration: if map $i^*: H^*(X, G) \to H^*(F, G)$ is surjective, then action of $\pi_1(B)$ on $H^*(F, G)$ trivial? 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?
 A: 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. 
A: The edge map for the serre spectral sequence of this fibration is $H^n(E) \to E_\infty^{0,n} \subset E_2^{0,n}=H^{0}(B,\mathscr{H}^n(F)) \subset E_1^{0,n}=H^n(F)$.  Since the edge map is the induced inclusion map, and this map is surjective by hypothesis, $H^n(F)^{\pi_1(B)}:=H^{0}(B,\mathscr{H}^n(F))=H^n(F)$.  Therefore every element of $H^n(F)$ is invariant under the action of $\pi_1(B)$.
I have taken this argument from Serre's Thesis, homologie singuliere des espaces fibres,(the paper where Serre introduced spectral sequences to the world).
