Bounded above flat tensor complexes preserve quasi isomorphisms

Use total differential $d(x\otimes y)=d_{P}x\otimes y+(-1{)}^{i}x\otimes d_{A}y$ for $x\in P^i$. If $P^i=0$ for $i>b$ and $A^j=0$ for $j>c$, put $p=b-i,q=c-j$. This is a first-quadrant double chain complex: both differentials lower their new index. Every total degree has a finite diagonal. Each vertical column is exact because $P^{b-p}$ is flat and $A$ is acyclic. Empty diagonals and zero terms contribute zero.

For a cycle of chain total degree $k$, choose its largest horizontal index $p$ with a nonzero component. Its component at $(p,k-p)$ is a vertical cycle, since no component at horizontal index $p+1$ contributes. Exactness of that column supplies a lift in $(p,k-p+1)$ (absorbing the invertible sign $(-1{)}^{b-p}$). Subtract its total boundary. This kills that component and can introduce only one at horizontal index $p-1$. Descending through $p,p-1,\dots ,0$ terminates; at zero the extra horizontal term is zero. The cycle is a boundary. The case $k<0$ has no terms.

This is the arrow-reversal of the finite-diagonal elimination in the published cochain column-assembly proof, with augmentation zero: reversing arrows in abelian groups interchanges kernels and cokernels, while finite products and sums agree. Interchanging the two indices gives the row version and proves the assertion for a flat left complex as well. The original assembly statements concern first-quadrant cochains; step 2.1 supplies the chain argument explicitly instead of applying those statements outside their domain.

For a quasi-isomorphism $s$, its cone is acyclic and bounded above. Tensoring with a flat complex makes this cone acyclic by step 2.1 or step 3.1. Tensor of the cone identifies with the cone of the tensored map: on a shifted second-factor summand multiply by $(-1{)}^i$ for first-factor degree $i$; a shifted first-factor summand requires no correction. Direct substitution in the differential verifies these signs. The cone criterion proves invariance. For replacements $P_N\to N$ and $P_M\to M$, both arrows $P_N\otimes P_M\to N\otimes P_M$ and $P_N\otimes P_M\to P_N\otimes M$ are quasi-isomorphisms. Their localized zigzag is the natural balancing isomorphism.