Collapse of Serre spectral sequence in the presence of a cross-section

Question

I was under the impression that if a Serre fibration $f: E \rightarrow B$ has a right inverse $s: B \rightarrow E$, then the associated Serre spectral sequence would collapse on the second page. This seems to be wrong in general. Is there an additional mild condition on $f$ that would rectify this, or is this just a wrong way to think about the situation?

Answer 1

I know this is an old question, but I recently needed to know this, so I thought I would share my understanding. Ultimately, I believe that you just need field coefficients. I've sketched an argument below (where I assume there is no twisting of coefficients).

Claim. Suppose $F \overset{i}\to E \overset{p}\to B$ is fibration, with $B$ and $F$ of finite type, $\pi_1(B)$ acting trivially on $\pi_*(F)$ and $\pi_0(F)=0$. Further let $\mathbb{K}$ be a field of coefficients. If the fibration $p$ admits a section $s \colon B \to E$, then $$H^*(E;\mathbb{K}) \cong H^*(B;\mathbb{K}) \otimes_{\mathbb{K}} H^*(F;\mathbb{K}).$$

Proof. Consider the Leray-Serre cohomological spectral sequence $$E_2^{p,q} = H^p(B;H^q(F;\mathbb{K})) \implies H^*(E;\mathbb{K})$$ of the fibration $F \overset{i}\to E \overset{p}\to B$. We know that $$E_2^{p,q} = H^q(B;\mathbb{K}) \otimes_{\mathbb{K}}H^q(F;\mathbb{K})$$ since we are working with field coefficients.

In order to prove the theorem, if suffices to show that $E_2 = \dots = E_{\infty}$, i.e. that all differentials $d_2, d_3, \dots$ vanish. Indeed, since we take field coefficients, all extension problems encountered in passing from $E_\infty$ to $H^*(E;\mathbb{K})$ are trivial. In other words, we know that $$H^n(E;\mathbb{K}) \cong \bigoplus E_{\infty}^{p,q}.$$ Moreover, since the fibration $p$ admits a section, it follows from the long exact sequence in cohomology (induced by $p$) that $$ i^* \colon H^q(E;\mathbb{K}) \to H^q(F;\mathbb{K}) $$ is surjective for all $q$. But recall that $i^*$ is equal to the composition $$H^q(E;\mathbb{K}) \twoheadrightarrow E_{\infty}^{0,q} = E_{q+1}^{0,q} \subseteq E_q^{0,q} \subseteq \dots \subseteq E_2^{0,q} = H^q(F;\mathbb{K}).$$ So the surjectivity of $i^*$ forces the above inclusions to be equalities. Thus all $d_r$, when restricted to the $q$-axis, must vanish.

On the other hand, we know that $$E_2^{p,q} = E_2^{p,0} \otimes E_2^{0,q}$$ since $\mathbb{K}$ is a field and $d_2=0$ on $E_2^{p,0}$ (as a first quadrant spectral sequence). Moreover, $d_2$ is a derivation with respect to this tensor product decomposition of the $E_2$-page, thus all $d_2$ differentials vanish and $E_3 = E_2$.

We can repeat the above argument for $d_3$ differentials and, continuing in this fashion, we see that the spectral sequence collapses at $E_2$, as desired. QED