Questions:
I think the definition of $\Phi$ relies on the assumption (b), but the left and right $\Phi$ from the commutative diagram don't have such an assumption. I wonder how is the left and right $\Phi$ defined? In particular, for the element $b\otimes i^*(c_j)\in H^*(B')\otimes H^*(F)$, I think $c_j$ here is from $H^{k_j}(E',R)$, but do we have $i^*(c_j)$, where $c_j\in H^{k_j}(E',R)$, form a basis for $H^*(F,R)$?
Why do we have $\delta c_j=0$? I think $c_j\in H^{k_j}(E,R)$ and $\delta:H^{k_j}(E')\to H^{k_j+1}(E,E')$, and I don't know how the $\delta$ act on $c_j$ and why is $p^*(b)\smile c_j\in H^*(E')$ s.t. $\delta c_j=0$.
About the claim that "The theorem for finite-dimensional $B$ will now follow by INDUCTION on $n$ and the five-lemma once we show that the left-hand $\Phi$ in the diagram is an isomorphism." I wonder for the 'induction' above means that: by $B'$ can be deformation retracted to a lower dimension CW-complex, we can use the induction so that the right $\Phi:H^*(B')\otimes H^*(F)\to H^*(E')$ is an isomorphism? And if so, then my first question arises again: i.e. for the case of $B',E'$, we don't have the assumption (b), then how do we apply the induction?
Some thoughts: I wonder if the property (b) works for the fiber bundle $F\to E'\to B'$? i.e. I wonder if there exist classes $c_j\in H^{k_j}(E'; R)$ whose restrictions $i^*(c_j)$ form a basis for $H^*(F; R)$ in each fiber $F$, where $i:F\to E$ is the inclusion. Then the map $$\Phi: H^*(B'; R)\otimes_RH^*(F;R)\to H^*(E'; R), \sum_{ij} b_i\otimes_R i^*(c_j)\mapsto \sum_{ij} p^*(b_i)\smile c_j,$$ is an isomorphism of $R$ modules?