The first Serre differential is the cellular boundary with local coefficients

Put $n=p+q$. In the exact couple of the filtration, the first differential is $d_1=jk$: the connector $k:H_n(E_p,E_{p-1};R)⟶H_{n-1}(E_{p-1};R)$ is followed by the quotient map $j:H_{n-1}(E_{p-1};R)⟶H_{n-1}(E_{p-1},E_{p-2};R)\mathrm{.}$ By [F2] this exact-couple differential is the filtered-complex $d_1$ with no extra sign. Concretely, [F3] sends a relative cycle represented by $c$ to the class represented by $\mathrm{\partial }c$ in the preceding relative layer.

Fix oriented cells $e^p$ and $f^{p-1}$. Project the source and target of Step 1.1 to their $e$- and $f$-summands using the cell-excision maps of [F1]. Naturality of pair connectors and excision gives a commutative square whose upper map is the boundary contribution of the attaching map ${\varphi }_e{\mathrm{\mid }}_{S^{p-1}}$ to $f$, and whose vertical maps are the iterated hemisphere isomorphisms defining $\mathrm{\Psi }$. Thus the $(f,e)$ component of $\mathrm{\Psi }{d}_1{\mathrm{\Psi }}^{-1}$ is determined entirely by the oriented attaching incidence together with fiber transport along the path from $b_e$ through the attaching sphere to $b_f$.

To calculate that component, temporarily supply lifts of these two cells to a universal cover of their base component. Write $\mathrm{\partial }\tilde{e}={\sum }_f\tilde{f}{r}_{fe}$ with $r_{fe}$ the finite signed sum of deck transformations determined by the lifted attaching incidences. Finiteness follows from [F6]. For one orientation-preserving degree-one disk contribution, naturality of every connector and excision map in [F1] makes the fiber map exactly the transport along its incidence path. For an orientation-reversing contribution, reduce by naturality to a reflection of the one-dimensional disk. Its pair sequence has boundary kernel $\ker (H_q(F_{-};R)\oplus H_q(F_{+};R)\to H_q({\tilde{D}}^1;R))=\{(x,-x):x\in H_q(F_{-};R)\},$ after transport identifies the two endpoint groups. Interchanging the endpoints therefore multiplies the suspension isomorphism by $-1$. Additivity now makes a degree-$m$ incidence contribute $m$ times its transport, and summing the finitely many terms gives exactly the action of $r_{fe}$ fixed in [F4].

It follows for every supplied lift basis that $(\mathrm{\Psi }{d}_1{\mathrm{\Psi }}^{-1}{)}_f(x_e)={\sum }_e{r}_{fe}{x}_e,$ where the action notation is precisely the balanced tensor convention of [F4], hence precisely its signed transport formula. Changing lifts conjugates the incidence matrices and coefficient coordinates by the same diagonal deck transformations; changing orientations conjugates them by the same diagonal signs. Therefore [F4] descends this equality to the intrinsic cellular local complex, without any global choice of lifts. By [F5], the right side is ${\mathrm{\partial }}_{{\mathcal{H}}_q}^{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}$. This proves the asserted intertwining identity.

The next page is the homology of $(E_{\ast ,q}^1,d_1)$. Step 4.1 identifies this chain complex with $C_{\ast }^{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}(B;{\mathcal{H}}_q(p;R))$, and [F5] identifies its homology with $H_{\ast }(B;{\mathcal{H}}_q(p;R))$. Hence $E_{p,q}^2\cong H_p(B;{\mathcal{H}}_q(p;R))$.

For $p=0$ both differentials out of the left edge are zero. For $p=1$ the reflection calculation in Step 3.1 is the whole sign calculation, and for $q=0$ the same formula acts on fiber components. An empty or zero stalk, the zero ring, no incident cells, and a zero incidence coefficient contribute zero. One incidence of either sign is explicit in Step 3.1; multiple and cancelling incidences use finite additivity. Degenerate singular simplices remain in the relative chain representative of Step 1.1. Both endpoints of every incidence path occur in the typed transport $b_e\to b_f$. A cellular chain has finite support, and [F6] makes each relevant attaching support finite, so only finite supplied lift data are ever used; the intrinsic conclusion uses no AC. The statement is not an iff.