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Fix $n=p+q$. For r=0 the inverse image of $\ker d^0$ in $F_p{C}_n$ is $A_{p,n}^1$, since $d^0[x]=0$ exactly when $dx$ factors through $F_{p-1}{C}_{n-1}$. The incoming image is represented by $d(F_p{C}_{n+1})$. Thus [F4] and [F5] identify $H(E^0{)}_{p,q}$ with $A_{p,n}^1\mathrm{/}(F_{p-1}{C}_n+d(F_p{C}_{n+1}))=E_{p,q}^1$.

For r≥1 abbreviate $A=A_{p,n}^r$, $N=A_{p,n}^{r+1}$, $L=A_{p-1,n}^{r-1}$ and $J=d(A_{p+r,n+1}^r)$. Let K be the inverse image in A of $\ker d^r$. For a morphism x into K, [F2] says dx factors through $A_{p-r-1,n-1}^{r-1}+d(A_{p-1,n}^{r-1})$. By [F4], after an epic cover write $dx=a+db$ with a and b through those respective subobjects. Then $d(x-b)=a$ lies in $F_{p-r-1}{C}_{n-1}$ and $x-b$ lies in $F_p{C}_n$, so $x-b$ factors through N; b factors through L. Descent of subobject membership gives $K\subseteq N+L$.

Conversely, for x through N, dx lies in $F_{p-r-1}{C}_{n-1}$ and $d(dx)=0$, hence through $A_{p-r-1,n-1}^{r-1}$; for b through L, db lies in the target boundary-image summand. Therefore [F2] sends both to zero and $N+L\subseteq K$. Thus $K=N+L$. The original boundary summand $d(A_{p+r-1,n+1}^{r-1})$ lies in J, because its preimage has filtration at most p+r and differential at most p. The incoming image in [F5] is the image of J in $A\mathrm{/}{B}^r$. By [F3] it lies in the kernel. Consequently $H(E^r{)}_{p,q}\cong (N+L)\mathrm{/}(L+J)$.

Every map through J has differential zero and filtration at most p, so $J\subseteq N$. Modularity [F4] gives $N\cap (L+J)=(N\cap L)+J$. The intersection $N\cap L$ is $F_{p-1}{C}_n\cap d^{-1}(F_{p-r-1}{C}_{n-1})=A_{p-1,n}^r$. The sum/quotient formula therefore gives $(N+L)\mathrm{/}(L+J)\cong N\mathrm{/}(A_{p-1,n}^r+d(A_{p+r,n+1}^r))=E_{p,q}^{r+1}\mathrm{.}$ In particular the inclusion $N\to K$ is epic after taking homology and has exactly the displayed kernel; this verifies both surjectivity and injectivity of the induced comparison.

Define ${\alpha }_r$ to be the inverse of this inclusion-induced isomorphism $E^{r+1}\to H(E^r)$, using step 1.1 when r=0. Its local rule sends a homology class represented by x to the class of x-b constructed in step 1.2. Independence of the correction follows from the kernel calculation in step 3.1, rather than from choosing b canonically. For a filtration-preserving chain map f, the inclusions of N into K commute with f, and $f(x-b)=f(x)-f(b)$ with $d^{\mathrm{\prime }}f=fd$. The induced quotient squares commute by [F4]; their inverses do too. This proves naturality.