Product total complex of a double complex

For a homological double complex $C$ in an abelian category, suppose each diagonal product exists. Write $P_n={\mathrm{Tot}}_n^{\mathrm{\Pi }}(C)={\prod }_{p+q=n}{C}_{p,q},{\pi }_{p,q}^n:P_n\to C_{p,q}\mathrm{.}$ Define $d_n:P_n\to P_{n-1}$ by the equations ${\pi }_{a,b}^{n-1}{d}_n=h_{a+1,b}{\pi }_{a+1,b}^n+v_{a,b+1}{\pi }_{a,b+1}^n(a+b=n-1)\mathrm{.}$ The product universal property gives a unique arrow from this family of two-term sums; no support condition is imposed on product coordinates.

Here the chain condition can be checked directly. For $a+b=n-2$, composing the coordinate formula twice gives ${\pi }_{a,b}^{n-2}{d}_{n-1}{d}_n=h_{a+1,b}{h}_{a+2,b}{\pi }_{a+2,b}^n+(h_{a+1,b}{v}_{a+1,b+1}+v_{a,b+1}{h}_{a+1,b+1}){\pi }_{a+1,b+1}^n+v_{a,b+1}{v}_{a,b+2}{\pi }_{a,b+2}^n=0.$ The three coefficients vanish respectively by $h^2=0$, anticommutation at $(a+1,b+1)$ and $v^2=0$. Product uniqueness implies $d_{n-1}{d}_n=0$.

Thus $(P,d)$ is the product total complex. An all-zero diagonal gives the zero product, and a single nonzero component gives that component. The construction and calculation apply in these cases too. Existence of the specified products is retained as a hypothesis; neither product exactness nor any choice of lifts is needed.