# Why all differentials are $0$ for Serre Spectral Sequence of trivial fibration?

Consider the fibration $F \hookrightarrow F \times B \to B$. I understand that if I take kunneth's theorem for granted that the group extensions $F_{n-i,i} \to F_{n-i+1,i-1} \to F_{n-i+1,i-1}/F_{n-i,i}$ associated to the filtration $F_{n,0} \supset ...F_{n-i,i}...$ are all trivial extensions, and that $d_r=0$ for $r>2$. How can we deduce these two facts without assuming Kunneth's theorem.

For the record, here is the reason why I get these two facts if I assume Kunneth.

The action of the fundamental group of the base on the fiber is nontrivial due to the splitting $H_*(F) \to H_*(B)$. There is always a map $E^\infty \to H_*(E)$, and by kunneth we have the surjective map $\text{a subgroup of}(E^2= H_*(B,H_*(F))) \to E^\infty \to H_*(E)=\text{(use kunneth)} H_*(B,H_*(F))$ so that the composition is the identity. Thus the first subgroup is the whole of $E^2$. Therefore it is not possible that either the series $H_n(B \times F) = \text{(by convergence of SSS)} im (H_n(B_n \times F) \mapsto H_n(B \times F))\supset im (H_n(B_{n-1} \times F) \mapsto im (H_n(B_{n-2} \times F) \mapsto H_n(B \times F)) \supset ... H_n(F)$ of group extensions be nontrivial , or $d^r$ be nonzero for $r \geq 2$.

I would like to know how to deduce two facts without using Kunneth's theorem, and hence deduce Kunneth's theorem from the SSS of this fibration.

I have a hunch that the proof that the differentials are 0 should be something like "Any map from $E^1_{p+q,q}=C_p(B) \otimes H_q(F) \to E^1_{p-2+q+1,q+1}=C_{p-2}(B) \otimes H_{q+1}(F)$ must be zero because either the domain or codomain is 0 and therefore there is no possible way that this map cannot be 0 after restricting and descending to $E^2$, and hence $d_2=0$.

To show that the series $H_n(B \times F) = \text{(by convergence of SSS)} im (H_n(B_n \times F) \mapsto H_n(B \times F))\supset im (H_n(B_{n-1} \times F) \mapsto im (H_n(B_{n-2} \times F) \mapsto H_n(B \times F)) \supset ... H_n(F)$

is trivial, I need to show that there is a splitting map $im ( H_*(X_{p}) \mapsto H_*(X)) \to im ( H_*(X_{p-1}) \mapsto H_*(X))$. I also have a hunch that this should have something to do with the topological splitting $B_n \times F \to B_n$.

Can I have some help with turning my gut feeling into a proof?

• Here is a suggestion: Build an algebraic model of what you would like your spectral sequence to look like and use the comparison theorem to show that its $E_\infty$-term coincides with that of the fibration. – iwriteonbananas Jul 21 '16 at 7:49
• Thanks for the suggestion, I am working it out now. I am using $\mathbb{Z}$ coefficients. I expected the extension problem to be trivial because if one assumes extension problem is trivial and assumes the differentials are 0, one gets that $H_n(B\times F)=\oplus_{p+q=n} E^2_{p+q,q}=\oplus_{p+q=n} H_p(B,H_q(F))$; this yields the correct answer irrespective of the coefficients used - it takes into account the $Tor$ terms. – user062295 Jul 21 '16 at 18:26
• Actually good point :) I was missing that the Tor terms are already taken into account there. – iwriteonbananas Jul 22 '16 at 7:00

The reason all the differentials are 0 is simple. I'll do it for cohomology because otherwise I need to use the coalgebra structure on homology. We know the elements of $E_r^{0,n}$ are permanent cycles for all $r,n$: there is an edge map $H^n(E)\to E_\infty^{0,n} \subset E_2^{0,n}=H_n(F)$ which has to be surjective because of the obvious splitting. Thus $E_\infty^{0,n}= E_2^{0,n}$.
The elements of $E_r^{n,0}$ are automatically permanent cycles. Thus every element $E_r^{n,m}$ that is a product of cycles is a cycle. This solves the problem for the coefficients in the rationals.
For coefficients in $\mathbb{Z}$, we have an injective map of spectral sequences $E_r \otimes \mathbb{Z}/p \to \epsilon_r$ where $\epsilon$ is the spectral sequence of the fibration with coefficients in $\mathbb{Z}/p$. Therefore the $d_r$ is 0 modulo p for all $p$ on $E^r$.
This shows that $d_r=0$ for all $r>1$.