Let $\pi\colon E \rightarrow B$ be a fiber bundle with fiber $F$. I am not assuming that $B$ is simply-connected. We then have Serre spectral sequences in both rational homology and rational cohomology:


$$E^2\_{pq} \cong H\_p(B;H\_q(F;\mathbb{Q})) \Longrightarrow H\_{p+q}(E;\mathbb{Q})$$


$$E\_2^{pq} \cong H^p(B;H^q(F;\mathbb{Q})) \Longrightarrow H^{p+q}(E;\mathbb{Q})$$


Since $B$ is not simply connected, the coefficient systems have to be regarded as local (twisted) coefficient systems on $B$. Assume that I know that one of these spectral sequences degenerates at the second page. Does it follow that the other one does as well?



Serre spectral sequence degeneration in homology vs cohomology
The answer is yes, and this follows from the comments, but it's worth spelling out the (non-obvious) details.


By Will Sawin's comment, the claim will follow if we can show that 
$$
\dim\_\mathbb{Q} H^p(B;H^q(F;\mathbb{Q})) = \dim\_\mathbb{Q} H\_p(B;H\_q(F;\mathbb{Q}))
$$
for all $p,q\ge0$. The Universal Coefficient Theorem does not hold with twisted coefficients. However, note that the vector spaces on the right are by definition the homology groups of the chain complex
$$
C\_\*(\widetilde{B})\otimes\_\pi H\_q(F;\mathbb{Q}),
$$
with differential induced by that of the chain complex $C\_\*(\widetilde{B})$ of the universal cover. Here $\pi=\pi\_1(B)$ and we are taking tensor product of $\mathbb{Z}\pi$-modules. 


Following archipelago's comment, let's see what happens if we dualize this chain complex by taking Hom into the rationals. By the [tensor-hom adjunction](https://en.wikipedia.org/wiki/Tensor-hom_adjunction) there are natural isomorphisms
$$
{\rm Hom}\_\mathbb{Z}\big(C\_\*(\widetilde{B})\otimes\_\pi H\_q(F;\mathbb{Q}),\mathbb{Q}\big)\cong {\rm Hom}\_\pi\big(C\_\*(\widetilde{B}),{\rm Hom}\_\mathbb{Z}(H\_q(F;\mathbb{Q}),\mathbb{Q})\big) \cong {\rm Hom}\_\pi\big(C\_\*(\widetilde{B}), H^q(F;\mathbb{Q})\big),$$
and so we get precisely the cochain complex whose cohomology gives the vector spaces on the left. Since everything in sight is a rational vector space, the usual algebraic Universal Coefficient Theorem implies the claim.


