The Kunneth theorem for free complexes over a PID

Use the construction proved in The Kunneth Tor map: put $Z_p=Z_{p}C$ and $A_p=B_{p-1}C$, with zero differentials. The canonical sequence $0\to Z\otimes D\to C\otimes D\to A\otimes D\to 0$ is exact. AC is used here for freeness of the cycle and boundary submodules by Boundaries and cycles in a free complex over a PID are free and their free-module lifts which split the degree sequences, exactly as in that construction.

By The long exact sequence in homology, its relevant segment gives $0\to \mathrm{coker}{\mathrm{\partial }}_{n+1}\to H_n(C\otimes D)\to \ker {\mathrm{\partial }}_n\to 0$. The calculation in the Tor-map lemma identifies $H_n(Z\otimes D)={⨁}_{p+q=n}{Z}_{p}C\otimes H_{q}D$ and identifies ${\mathrm{\partial }}_{n+1}$ with the sum of the inclusion maps $B_{p}C\otimes H_{q}D\to Z_{p}C\otimes H_{q}D$. It also identifies $\ker {\mathrm{\partial }}_n$ and the map onto it with the displayed Tor sum and its established natural quotient.

For each $(p,q)$, right exactness in Tensoring is right exact applied to $B_{p}C\to Z_{p}C\to H_{p}C\to 0$ identifies the cokernel of that inclusion-tensor map with $H_{p}C\otimes H_{q}D$. Cokernels of direct-sum maps are computed coordinatewise, giving $\mathrm{coker}{\mathrm{\partial }}_{n+1}={⨁}_{p+q=n}{H}_{p}C\otimes H_{q}D$. The injection into the middle homology sends $[z]\otimes [y]$ to $[z\otimes y]$, since it is induced by $Z\otimes D\to C\otimes D$. Thus it is precisely the cross product in The Kunneth cross-product map is well defined and natural.

Step 2.1 and step 3.1 prove exactness for the specified maps. Both are natural by the two map lemmas, hence this is a natural short exact sequence. The tensor differential and totalization are those of The tensor product of a right and a left chain complex is totalized by direct sums with the Koszul differential; the construction works for every integer degree with the stated finite diagonals, including zero summands and empty diagonals.