Given a fibration \(F \to E \to B\), use the homotopy lifting property to define an action of \(\pi_1(E)\) on \(\pi_n(F)\), a homomorphism \(\pi_1(E) \to \operatorname{Aut}\bigl(\pi_n(F)\bigr)\), such that the composition \(\pi_1(F) \to \pi_1(E) \to \operatorname{Aut}\bigl(\pi_n(F)\bigr)\) is the usual action of \(\pi_1(F)\) on \(\pi_n(F)\). Deduce that if \(\pi_1(E) = 0\), then the action of \(\pi_1(F)\) on \(\pi_n(F)\) is trivial.
Solution
Fix a basepoint \(e_0\in F=p^{-1}(b_0)\). Let \[ [\beta]\in\pi_1(E,e_0), \qquad [\alpha]\in\pi_n(F,e_0). \] Use the relative homotopy lifting property for the cofibration \[ S^n\times\{0\}\cup\{*\}\times I \subset S^n\times I. \] On this subspace define a map to \(E\) by \(\alpha\) on \(S^n\times\{0\}\) and by \(\beta\) on \(\{*\}\times I\). Its projection extends over \(S^n\times I\) by \[ (x,t)\longmapsto p\beta(t). \] Lift this homotopy relative to the displayed subspace. At \(t=1\) the resulting map \[ \alpha_1:S^n\to E \] lands in \(F\), since \(p\beta(1)=b_0\), and is based because the basepoint track is \(\beta\). Define \[ [\beta]\cdot[\alpha]=[\alpha_1]\in\pi_n(F,e_0). \]
Homotopies of \(\alpha\) or \(\beta\) give homotopies of \(\alpha_1\), so this is well-defined. Concatenating loops concatenates the transport homotopies, hence \[ (\beta\gamma)\cdot\alpha =\beta\cdot(\gamma\cdot\alpha), \] and reversing \(\beta\) gives the inverse automorphism. Thus \[ \pi_1(E,e_0)\to\operatorname{Aut}(\pi_n(F,e_0)) \] is a homomorphism.
If \(\beta\) is a loop in \(F\), its projection is constant, and the construction is exactly the usual basepoint-change action of \(\pi_1(F)\) on \(\pi_n(F)\). Therefore the composite \[ \pi_1(F)\to\pi_1(E)\to\operatorname{Aut}(\pi_n(F)) \] is the usual action. In particular, if \(\pi_1(E)=0\), this action is trivial.