Let $p: E \to B$ be a locally trivial fibration with fiber $F$. If necessary, suppose that $B$ is simply connected. Suppose that the Serre spectral sequence leaves the term $H_p(B, H_q(F, \mathbb{Q}))$ unaltered, in the sense that all morphisms coming in an out of $E^\ell_{p,q}$ are zero for all $\ell$. Furthermore, supose that all other terms with the same total degree are zero. In this case, we get an isomorphism $$H_p(B, H_q(F, \mathbb{Q})) \cong H_{p+q}(E, \mathbb{Q}).$$ If we know explicit generators of $H_p(B, \mathbb{Q})$ and $H_q(F, \mathbb{Q})$ (in my case these are explicit submanifolds), is it possible to get the explicit generators of $H_{p+q}(E, \mathbb{Q})$?

1$\begingroup$ What does it mean to `get the explicit generators'? $\endgroup$– Nicholas KuhnNov 23, 2022 at 20:11

$\begingroup$ @NicholasKuhn I mean to get an explicit cycle whose class in homology gives a generator $\endgroup$– Eduardo de LorenzoNov 24, 2022 at 21:01
1 Answer
You have not specified your coefficients, but it sounds like you are working over the integers. In that case, if the $E^2$ term is not a free abelian group, then you only know that $H_*(E)$ has a filtration with associated graded group $H_*(B;H_*(F))$, it need not be the case that $H_*(E)\simeq H_*(B;H_*(F))$. Even if the $E^2$ page is free abelian of finite rank, it is not necessarily true that $H_*(E)\simeq H_*(B;H_*(F))$ as coalgebras, or equivalently that $H^*(E)\simeq H^*(B;H^*(F))$ as rings. A nice example involves the Milnor hypersurface $$ F = \{([u],[v])\in\mathbb{C}P^p\times\mathbb{C}P^q: \sum_{i=0}^pu_iv_i=0\} $$ (where we assume $p\leq q$). There is a fibre bundle $$ F = \mathbb{C}P^{q1} \xrightarrow{i} E \xrightarrow{f} B= \mathbb{C}P^p $$ given by $f([u],[v])=[u]$. In the associated spectral sequence we have $$ E_2^{**} = H^*(\mathbb{C}P^p)\otimes H^*(\mathbb{C}P^{q1}) = \mathbb{Z}[x,y]/(x^{p+1},y^{q}) = \mathbb{Z}\{x^iy^j:0\leq i\leq p,0\leq j<q\} $$ with $x\in E_2^{20}$ and $y\in E_2^{02}$. One can choose $y\in H^2(E)$ representing the class $y\in E_2^{02}$ but the choice is indeterminate by multiples of $x$. There is a natural choice that depends on special features of this example and does not generalise in any obvious way. With that choice, the ring structure is actually $$ H^*(E)=\mathbb{Z}[x,y]/(x^{p+1},y^qy^{q1}x+y^{q2}x^2\dotsb\pm x^q). $$ There is no possible choice of $y$ satisfying $y^q=0$.

$\begingroup$ I'm actually working with rational coefficients. I edited the question to reflect that. $\endgroup$ Nov 23, 2022 at 12:02

$\begingroup$ OP is asking about one specific degree $p+q$, which means that there is no ring structure. He also supposes that $H_{p+i}(B,H_{qi}F)=0$ for $i\neq 0$. $\endgroup$ Nov 23, 2022 at 21:07