# degeneracy of the Serre spectral sequence

The following are well-known facts on the Serre spectral sequence

For a fibration $$F \rightarrow E \rightarrow B$$ we have the Serre spectral sequence (in cohomology with a coefficients in a field $$k$$) with $$E_2$$-term

$$E_2^{p,q} = H^p(B; \mathcal{H}^q(F)) \Rightarrow H^{p+q}(E)$$

If $$B$$ is simply connected, (or $$\pi_1(B)$$ acts trivially on the cohomology of the fiber) we have that

$$E_2^{p,q} = H^p(B) \otimes H^q(F)$$

Moreover, if the spectral sequence degenerates at the $$E_2$$-term ($$d_r = 0$$ for $$r \geq 2$$),

then $$H^*(B) \otimes H^*(F) \cong H^*(E)$$.

My question is the following,

Suppose that for a fibration $$F \rightarrow E \rightarrow B$$ we know that $$H^*(E) \cong H^*(B) \otimes H^*(F)$$, does it follows that in the spectral sequence $$E_2^{p,q} = H^p(B) \otimes H^q(F)$$ ? and also that $$d_r = 0$$ for $$r \geq 2$$ ?

I assume that it is not true and a counterxample should involve a non-trivial local coefficient system, but I do not know many "computable" examples where the base space is non-simply connected.

• Your second claimed isomorpism is incorrect. If $\pi_1B$ acts trivially on $H^*F$ then the local coefficent system is trivial and you can replace local cohomology group $\mathcal{H}^*F$ with ordinary cohomology group $H^*F$, leaving $E_2^{p,q}\cong H^p(B;H^qF)$. From here you still need to apply the universal coefficent theorem. You will only get $E_2^{p,q}\cong H^pB\otimes H^qF$ if at least one of $H^qB$ or $H^q$ satisfies some flatness conditions over the ring $R$. – Tyrone Oct 30 '18 at 17:22
• @Tyrone thanks, I will edit my assumptions (maybe work with field coefficients to make things smoother) – C. Zhihao Oct 30 '18 at 17:29