Explicit generators from Serre spectral sequence 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 $H_{p+q}(E, \mathbb{Q})$?
 A: 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
$$ F = \mathbb{C}P^{q-1} \xrightarrow{i} E \xrightarrow{f} B= \mathbb{C}P^p $$
given by $f([u],[v])=[u]$.  In the associated spectral sequence we have
$$ E_2^{**} = H^*(\mathbb{C}P^p)\otimes H^*(\mathbb{C}P^{q-1}) = 
    \mathbb{Z}[x,y]/(x^{p+1},y^{q}) = 
    \mathbb{Z}\{x^iy^j:0\leq i\leq p,0\leq j<q\}
$$
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
$$ H^*(E)=\mathbb{Z}[x,y]/(x^{p+1},y^q-y^{q-1}x+y^{q-2}x^2-\dotsb\pm x^q). $$
There is no possible choice of $y$ satisfying $y^q=0$.
