The Kunneth sequence splits nonnaturally

Write the sequence of The Kunneth theorem for free complexes over a PID as $0\to U_n\stackrel{\alpha }{\to }{V}_n\stackrel{\beta }{\to }{W}_n\to 0$, where $U_n={⨁}_{p+q=n}{H}_{p}C\otimes H_{q}D$ and $V_n=H_n(C\otimes D)$. The map $\alpha$ is the cycle cross product.

For $E=C,D$, the exact sequence $0\to Z_{p}E\to E_p\to B_{p-1}E\to 0$ is supplied by The cycle-boundary short exact sequences for a free complex over a PID. Under AC its last term is free by Boundaries and cycles in a free complex over a PID are free, hence projective by Free modules are projective, with the exact choice boundary. Choose sections in every degree and both complexes, using AC for this family as well as the free-module lifts. They give retractions $r_p:E_p\to Z_{p}E$.

Let $H(E)$ denote the zero-differential complex with terms $H_{p}E$, and define ${\pi }_{E,p}(x)=[r_p(x)]$. This is a chain map: $d_{E}x$ is a boundary and a cycle, so $r_{p-1}(d_{E}x)=d_{E}x$ has zero homology class. On cycles, ${\pi }_E$ is the usual homology quotient.

Tensor the two chain maps using The tensor product of a right and a left chain complex is totalized by direct sums with the Koszul differential. The target $H(C)\otimes H(D)$ has zero differential and degree-$n$ term $U_n$, so the induced homology map is $r:V_n\to U_n$. On a cross-product generator $[z]\otimes [w]$, one has $r\alpha ([z]\otimes [w])=[z]\otimes [w]$; therefore $r\alpha =1$.

The restriction $\beta :\ker r\to W_n$ is bijective. Indeed, any lift $v$ of $w$ can be replaced by $v-\alpha r(v)\in \ker r$ without changing its image; and if $v\in \ker r\cap \ker \beta$, exactness gives $v=\alpha (u)$, whence $u=r(v)=0$. Its inverse followed by inclusion is a homomorphism $s:W_n\to V_n$ satisfying $\beta s=1$ and $s\beta =1-\alpha r$. This is the required section. It depends on the selected retractions, and no natural selection is asserted.