Row and column filtrations of a first quadrant double complex

Let $C$ be a first-quadrant homological double complex. Its totalisation $T$ exists: each nonnegative diagonal has at most $n+1$ nonzero terms and negative diagonals are zero. The finite sum/product identification is provided by Sum and product totalisations agree on finite diagonal double complexes, and The total differential squares to zero gives the chain condition.

For integers $s,n$, its column filtration and row filtration are the partial biproducts $F_s^{\mathrm{c}\mathrm{o}\mathrm{l}}{T}_n={⨁}_{p+q=n,p\le s}{C}_{p,q},F_s^{\mathrm{r}\mathrm{o}\mathrm{w}}{T}_n={⨁}_{p+q=n,q\le s}{C}_{p,q}\mathrm{.}$ Their injections into $T_n$ are split monomorphisms: projecting onto the selected summands is a left inverse. Thus these are subobjects. The selected index sets increase with $s$, giving the filtration inclusions.

Both arrows preserve each cutoff: $h$ lowers $p$ and fixes $q$, while $v$ fixes $p$ and lowers $q$. Consequently $h+v$ restricts to each partial sum, and the two families are filtered subcomplexes. They vanish for $s<0$ and equal $T_n$ for $s\ge n$ when $n\ge 0$. For $n<0$ every piece and $T_n$ are zero. In particular degree zero has just the single possible component $C_{0,0}$; both filtrations jump there at $s=0$.

The quotient $F_s\mathrm{/}{F}_{s-1}$ selects column $s$ with remaining differential $v$, or row $s$ with remaining differential $h$, respectively. In spectral coordinates $(s,t)$, total degree is $s+t$; hence these associated graded components are respectively $C_{s,t}$ and $C_{t,s}$. No sign is inserted: the original double-complex arrows already anticommute. The construction uses only the specified finite biproduct maps and no choices.