# Can there be nonisomorphic functorial Serre spectral sequences?

The Serre spectral sequence is a very useful tool in algebraic topology, but as often with these beasts, the differentials can be hard to compute.

Of course, the proof that this sequence exists is usually completely constructive so in theory we should be able to give a precise account of these differentials : this is a very complicated ordeal, for various reasons.

However one often manages to compute these with various techniques : in cohomology one may use all the cohomology operations, the multiplicative structure etc.; and both in homology and cohomology one may use information we already know about the (co)homologies of the spaces but also one may use the functoriality of the spectral sequence : any map of fibrations induces a map of spectral sequences.

This often gives enough information in practice (I don't know if it's true at a research level, but at least from a student's perspective it is).

So the question is very natural : do these allow us to compute all differentials, at least in principle ? More precisely :

Suppose $$E$$ is a contravariant functor from the category of Serre fibrations to the category with objects couples where one entry is a multiplicative cohomological spectral sequence and the other entry is a filtered graded algebra together with an isomorphism witnessing that the spectral sequence converges to said algebra and with obvious arrows (maps of multiplicative spectral sequences and of graded filtered algebras that are compatible) such that for any fibration $$F\to X\to B$$, we have $$E_2 = H^*(B, H^*(F))$$ and the filtered graded algebra is some filtration of $$H^*(X)$$. Moreover assume that the multiplication on $$E_2$$ is induced by the cup product, and that the maps induced by a map of fibrations on $$E_2$$ are the obvious ones. Is it then true that $$E$$ is naturally isomorphic to the Serre spectral sequence ?

Perhaps one needs to add the cohomology operations but that's more complicated to write down. I have not given any precision on what cohomology I'm using. By default, it is singular cohomology with integer coefficients, but if the answer changes with some other coefficients (for instance $$\mathbb F_p$$ by adding that everything is compatible with the Steenrod algebra), I'm interested as well.