PID Kunneth is a two-column collapse

The free presentation of any module by the free module on its underlying set has free kernel by F2. Under AC both free modules are projective, since one can lift the images of a basis across an epimorphism by F5. Thus every module has projective dimension at most one, and all ${\mathrm{Tor}}_p$ for $p>1$ vanish. In F1 all $d_r$ with $r\ge 2$ vanish because no two surviving columns can be joined by their first-coordinate change $-r$. The finite filtration therefore has $F_0{H}_n$ equal to the tensor sum and $H_n\mathrm{/}{F}_0{H}_n$ equal to the Tor-one sum.

The column-zero map is induced by tensors of cycles, hence is the cross product in F3. To identify the other map, use the length-one presentations $B_{i}C\to Z_{i}C\to H_{i}C$ from F2. In the construction of F1, a resolution-degree-one cycle projects to a class in $B_{i}C\otimes H_{j}D$ killed by $B_{i}C\to Z_{i}C$. Its representative is exactly the image under $d_C:C_{i+1}\to B_{i}C$ tensored with the $D$ cycle. The Koszul differential gives the positive $d_C$ term. This is precisely the cycle-boundary representative used in F4's construction of the natural Tor surjection. Consequently the spectral-sequence quotient map and the published Tor map agree on representatives, so the two exact sequences have the same arrows, not merely isomorphic endpoints.

For existence of a splitting, F2 and F5 allow choices of sections $B_{i-1}C\to C_i$ and $B_{j-1}D\to D_j$ in all degrees. They decompose each complex into the direct sum, locally finite in degree, of the two-term free presentations $B_{i}C\to Z_{i}C$ placed in degrees $i+1,i$, and similarly for $D$. Their tensor is the direct sum of tensor products of those length-one resolutions. Each such tensor contributes Tor in resolution degrees zero and one and zero in higher degrees, by step 1.1 and the balanced calculation in F1. Taking homology gives a direct-sum decomposition into the two displayed sums. Its tensor summand is the canonical cross product, and its complementary summand maps isomorphically to the quotient, so it supplies a section. This construction locates the use of AC in freeness, basis lifts and the chosen sections.

Nonnaturality already occurs over $\mathbb{Z}$. Take $C_1=\mathbb{Z}a\oplus \mathbb{Z}b$, $C_0=\mathbb{Z}c$, with $da=2c,db=0$, and $D_1=\mathbb{Z}x$, $D_0=\mathbb{Z}y$, with $dx=2y$, zero elsewhere. Degree-one cycles in the tensor are freely generated by $u=b\otimes y$ and $t=a\otimes y-c\otimes x$; the degree-two boundaries are generated by $-2u$ and $-2t$. Hence $H_1(C\otimes D)=(\mathbb{Z}\mathrm{/}2)[u]\oplus (\mathbb{Z}\mathrm{/}2)[t]$. The tensor subobject is generated by $[u]$ and the Tor quotient by the image of $[t]$. The chain automorphism $a\mapsto a+b$, $b\mapsto b$, $c\mapsto c$ acts identically on $H_{\ast }C$ and both ends of the exact sequence, but sends $[t]\mapsto [t]+[u]$. Every lift of the quotient generator is $[t]+ϵ[u]$, with $ϵ\in \mathbb{Z}\mathrm{/}2$, and this automorphism fixes neither lift. A natural section would have to fix its image, which is impossible. Empty sums and zero complexes give the zero exact sequence; the same finite-filtration proof covers the bottom degree.