My goal is to give a concrete calculation with the Serre spectral sequence where the coefficient system is not simple. Here's what I have so far:
Take X = $S^1 \times S^2 / \mathbb Z_2$ where $\mathbb Z_2$ acts as the antipodal map on both factors. According to this post, the projection on the second factor induces a map $p: X \to \mathbb RP^2$ that makes $p$ into a fiber bundle with fiber $S^1$ (even a principal one).
Now I claim that the local coefficient system $H_1(\mathscr F) = \{H_1(F_b), b \in B\}$ where $F_b = p^{-1}(b)$ is the system of twisted integers, that is, the non-zero element of $\pi_1(\mathbb R P^2, *)$ acts as $-1$ on $H_1(F_*) \cong \mathbb Z$. We can deduce (see for example Elements of Homotopy Theory, p. 284) that $H_p(\mathbb RP^2; H_1(\mathscr F)) = \mathbb Z_2, 0, \mathbb Z$ for $p = 0, 1, 2$.
Accordingly, the $E^2$ term (which is also the $E^\infty$ term) of the spectral sequence looks as follows: \begin{array}{c|ccc} 1 & \mathbb Z_2 & 0 & \mathbb Z \\ 0 & \mathbb Z & \mathbb Z & 0\\\hline q/p & 0 & 1 & 2 \end{array} From this we can deduce \begin{array}{c|ccc} i & 0 & 1 & 2 & 3\\\hline H_i(X) & \mathbb Z & ? & 0 & \mathbb Z\\ \end{array} And for the missing term we can use the Hurewicz theorem with $\pi_1(X) = \mathbb Z$ (this follows from the homotopy sequence of the $S^2$-bundle $X \to S^1$ given by the projection onto the first factor).
Is my reasoning correct? Do you have any additional thoughts? Can you think of an easier/more instructive/more interesting example?
