Sum and product totalisations agree on finite diagonal double complexes

Fix $n$ and let $I_n=\{(p,q):p+q=n,C_{p,q}\mathrm{\ne }0\}$. A finite biproduct of the objects indexed by $I_n$ exists. Adjoining the unique maps from and to each omitted zero object makes its coproduct and product structures satisfy the universal properties for the whole diagonal: those omitted components impose no conditions on a family of maps. This constructs $T_n$ and $P_n$, including $I_n=\mathrm{\varnothing }$, when both are zero.

Define ${\eta }_n:T_n\to P_n$ by ${\pi }_{a,b}^n{\eta }_n{\iota }_{p,q}^n=1_{C_{p,q}}$ if $(a,b)=(p,q)$ and zero otherwise. Successive coproduct and product universal properties give its existence and uniqueness. Under the identifications in the preceding step it is precisely the finite biproduct comparison, so it is invertible. If $I_n$ has one element, it is the identity on that component.

Test $d_n^P{\eta }_n$ and ${\eta }_{n-1}{d}_n^T$ by precomposing with ${\iota }_{p,q}^n$ and postcomposing with ${\pi }_{a,b}^{n-1}$. Both composites are $h_{p,q}$ for $(a,b)=(p-1,q)$, $v_{p,q}$ for $(a,b)=(p,q-1)$, and zero otherwise, by the two differential formulas. Universal-property uniqueness gives $d_n^P{\eta }_n={\eta }_{n-1}{d}_n^T$.

Multiplying this equation by the inverses gives $d_n^T{\eta }_n^{-1}={\eta }_{n-1}^{-1}{d}_n^P$. Hence the degreewise inverse is also a chain map. All components of the comparison are uniquely specified, and its inverses are unique; no simultaneous choice of lifts or representatives is involved. The conclusion holds for the zero complex and for a complex supported on one row or column as well.