The row filtration spectral sequence of a first quadrant double complex

The row-$p$ quotient of total degree $p+q$ is $C_{q,p}$. The arrow $h$ stays in this row and $v$ enters the preceding row, which is zero in the quotient. Thus $E_{p,q}^0=C_{q,p}$ and $d^0=h$. This remains valid when either index is negative, as the component is then zero.

Taking homology gives $E_{p,q}^1=H_q^h(C_{\ast ,p})$. A horizontal cycle $x$ has total differential $dx=vx$, so the page-one rule gives $d^1[x]=[vx]$. This is a well-defined horizontal homology class: $h(vx)=-v(hx)=0$, and if $x$ changes by $hy$ then $vx$ changes by $vhy=-hvy$, a horizontal boundary. In an abelian category these calculations mean preservation of kernel and image subobjects; they may be checked after epic pullback and descend uniquely. No global representatives are selected.

Since $v^2=0$, the induced page-one arrows square to zero, and their homology is precisely $H_p^v(H_q^h(C))$. The next-page isomorphism therefore gives the displayed $E^2$. Both $E^0$ and all later subquotients vanish off the first quadrant. The differential bidegrees are those of the filtered construction, $(-r,r-1)$, with no extra sign in $d^1$ because $h+v$ was already the total differential.

In degree $n\ge 0$, $F_{-1}^{\mathrm{r}\mathrm{o}\mathrm{w}}{T}_n=0$ and $F_n^{\mathrm{r}\mathrm{o}\mathrm{w}}{T}_n=T_n$; negative degrees are zero. Thus the bounded-filtration theorem applies degree by degree to this spectral sequence and identifies its stationary page with the associated graded of the displayed image filtration. That filtration is zero at $p=-1$ and all of $H_n(T)$ at $p=n$ for $n\ge 0$. For $n=0$ there is only one possible quotient, and for the zero complex all pages and quotients are zero. The argument uses no infinite exactness or AC.