Serre Spectral Sequence and Fundamental Group Action on Homology

Solution 1:

let $p:\ E\rightarrow B$ be the fibration, and let $B$ be path-connected; then all fibers $p^{-1}(b)$ are homotopy equivalent. Furthermore, a path $f:\ [0,1]\rightarrow B$ defines a homotopy class $f_*$ of homotopy equivalence between $p^{-1}(f(0))$ and $p^{-1}(f(1))$. Even better, this only depends on the homotopy class of $f$, relative its endpoints. So specializing to the fundamental group, based at $b\in B$, we have a group homomorphism from $\pi_1(B,b)$ to $S$, where $S$ is the homotopy classes of homotopy equivalences between $p^{-1}(b)$ and $p^{-1}(b)$. If we let $F=p^{-1}(b)$, then each of these homotopy equivalences induces an automorphism of $H_n(F)$ for every $n$; that is, we get a group homomorphism from $\pi_1(B,b)$ to $Aut(H_n(F))$ for every $n$.

This is all covered well in chapters 6 and 9-10 of Kirk and Davis's "Lecture Notes in Algebraic Topology".

Solution 2:

No, it means that the homomorphism induced by following a loop is trivial. You will always get the same group back but a generator may get sent to its negative. Take a Möbius strip $\mathbb R \to M \to S^1 $ and mod out each fiber of the strip by the action of $\mathbb Z$. This is a twisted torus (might be the Klein bottle). Pick any point $p \in S^1$, and a generator $\omega$ of $H_1(\pi^{-1}(p))$. The loop around the base induces an action of $H_1(\pi^{-1}(p))$ that sends $\omega$ to $-\omega$.

Does anyone know what happens to the Serre Spectral Sequence when this is not trivial?