Relative lifts produce cohomological transgressions

A degree-$m+1$ class of $B$ lifts to $H^{m+1}(B,B^m)$: the restriction to $B^m$ is zero because cellular cochains on an $m$-dimensional CW complex vanish in degree $m+1$, and the pair sequence gives the lift. Choose a relative cocycle $v$ for that lift, hence vanishing on $B^m$. Extend a cocycle representing $x$ from $F$ to an absolute cochain $u$ of $E$. The asserted relative-class equality means $du-p^{\ast }v=dt$ for a cochain $t$ vanishing on $F$. Replace $u$ by $u-t$. It still restricts to the same fiber cocycle and now satisfies $du=p^{\ast }v$ exactly.

In the decreasing skeletal filtration, $p^{\ast }v$ lies in filtration $m+1$, since it vanishes over $B^m$. Thus $u$ is an $(m+1)$-cycle representative in column zero. The filtered-complex page formula shows it survives every earlier page, and the page-$m+1$ differential is represented by its coboundary $p^{\ast }v$. On the row of fiber degree zero the projection edge identifies this class with $y$, modulo precisely the earlier incoming boundaries. This identification is the base-edge identification in the published cohomological Serre construction, not an assumption of a new operation on a page. The column-zero identification sends the restriction of $u$ to the original fiber class; simple connectivity makes its transport constant. The finite-quotient filtration comparison in the published Serre theorem identifies these representative computations with the actual Serre pages, despite the raw singular filtration being unbounded.

Finally the connector-compatibility lemma and naturality of relative squares give $\delta S{q}^{a}x=S{q}^a\delta x=S{q}^a{p}^{\ast }y=p^{\ast }S{q}^{a}y\mathrm{.}$ Apply the same representative argument in total degree $m+a+1$. This proves both the survival and the precise differential page; no rule about squares of an arbitrary spectral-sequence cycle has been assumed. Zero operations simply give zero representatives.