The exact couple and subquotient constructions of the filtered complex spectral sequence agree

Both initial pages identify with $H_n(F_{p}C\mathrm{/}{F}_{p-1}C)$: in the subquotient construction a cycle modulo the previous filtration is precisely a lift $c\in F_p{C}_n$ with $dc\in F_{p-1}{C}_{n-1}$, modulo $F_{p-1}{C}_n+d(F_p{C}_{n+1})$. This is the homology quotient defining the initial exact-couple page.

The connector $k$ sends this class to $[dc]\in H_{n-1}(F_{p-1}C)$ with a positive sign. Indeed the snake construction first pulls back the epic upper-row map, then factors its vertical differential through the monic lower-row map, and defines the connecting arrow by the equation $\delta \pi =qa$, where that factor $a$ satisfies inclusion composed with $a$ equal to the vertical differential. In the quotient-kernel diagram of complexes this vertical arrow is induced by $d$, so a lifted $c$ gives exactly the class of $dc$. The preconnecting and connecting definitions preserve this equation. This verifies the sign in every abelian category after epic pullback; in modules it is the stated elementwise formula.

A class represented by $c$ on the initial page lies in $N^r$ exactly when its class $[dc]$ in $H_{n-1}(F_{p-1}C)$ is induced by a cycle $z\in F_{p-r}{C}_{n-1}$. Locally this means $dc=z+db$ for $b\in F_{p-1}{C}_n$. Then $c-b\in A_{p,n}^r$ and represents the same initial-page class. Conversely $c\in A_{p,n}^r$ gives the lower-filtration cycle $dc$, so its class belongs to $N^r$. Thus $A_{p,n}^r$ maps epimorphically onto $N^r$.

The inverse image of $B^r$ under this epimorphism is $A_{p-1,n}^{r-1}+d(A_{p+r-1,n+1}^{r-1})$. To prove this, a $B^r$ class is represented by a cycle $a\in F_p{C}_n$ that becomes a boundary in $F_{p+r-1}C$: locally $a=dw$ with $w\in F_{p+r-1}{C}_{n+1}$. Equality of its initial-page class with that of $c\in A^r$ means $c-a=b+dt$ for $b\in F_{p-1}{C}_n$ and $t\in F_p{C}_{n+1}$. Hence $c=b+d(w+t)$. Now $db=dc\in F_{p-r}$, so $b\in A_{p-1,n}^{r-1}$, and $d(w+t)=c-b\in F_p$, so $w+t\in A_{p+r-1,n+1}^{r-1}$. Conversely the first denominator summand maps to zero on the initial page, while an element of the second is a cycle in $F_p$ that bounds in $F_{p+r-1}$ and therefore maps into $B^r$. These local containments descend by [F4].

The quotient comparison now identifies $N^r\mathrm{/}{B}^r$ with $A_{p,n}^r\mathrm{/}(A_{p-1,n}^{r-1}+d(A_{p+r-1,n+1}^{r-1}))$, precisely the filtered page. For $c\in A^r$, the lift of $k[c]$ through $i^{r-1}$ is the homology class of $dc$ in $F_{p-r}$. The exact-couple differential therefore sends $[c]$ to $[dc]$ on the corresponding target page, exactly the filtered differential. The target bidegree is $(p-r,q+r-1)$ on both sides.

In both constructions the next-page isomorphism is induced by including the next cycle numerator and then inverting the resulting homology isomorphism. The comparisons above come from the same chain representatives and lower-filtration corrections; hence those inclusions commute with the comparisons, and so do their inverses. Every filtered chain map preserves $A^r$, the denominator summands and the cycle/boundary comparisons, so quotient uniqueness proves naturality. At $r=1$ these are the identifications in step 1.1; zero pieces and stationary filtrations simply give zero quotients where appropriate. No global representatives, infinite sums or convergence hypotheses are used.