The PID Kunneth sequence admits a section after choices

Select the sections of [F2] for both complexes, using [F4], and denote the resulting cycle retractions by $a_p:C_p\to Z_{p}C$ and $b_q:D_q\to Z_{q}D$. Define ${\pi }_{C,p}(c)=[a_p(c)]\in H_{p}C$ and ${\pi }_{D,q}(w)=[b_q(w)]\in H_{q}D$. Since a boundary is a cycle, $a_{p-1}(d_{C}c)=d_{C}c$, whose homology class is zero. Thus ${\pi }_C{d}_C=0$; likewise ${\pi }_D{d}_D=0$. With zero differentials on $H(C)$ and $H(D)$, these are chain maps.

Define $\mathrm{\Pi }:C\otimes D\to H(C)\otimes H(D)$ by $c\otimes w\mapsto {\pi }_C(c)\otimes {\pi }_D(w)$. This descends to tensors because the formula is bilinear and balanced: replacing $rc\otimes w$ by $c\otimes rw$ gives the same tensor by linearity of ${\pi }_C,{\pi }_D$. On a homogeneous tensor, $\mathrm{\Pi }d(c\otimes w)={\pi }_C(d_{C}c)\otimes {\pi }_D(w)+(-1{)}^p{\pi }_C(c)\otimes {\pi }_D(d_{D}w)=0$. The target differential is zero, so $\mathrm{\Pi }$ is a chain map.

The target has zero differential, hence its degree-$n$ homology is exactly $K_n$, even if its modules are not free. By [F3], $\mathrm{\Pi }$ induces $r_n:H_n(C\otimes D)\to K_n$. For cycles $z,w$ the retractions fix them, so $r_n{\alpha }_n([z]\otimes [w])=[z]\otimes [w]$. Elementary tensors of homology classes generate $K_n$, proving $r_n{\alpha }_n=1_{K_n}$.

Put $L=\ker r_n$. If $x\in L$ and ${\beta }_{n}x=0$, exactness gives $x={\alpha }_{n}k$ for some $k\in K_n$. Applying $r_n$ gives $0=r_{n}x=k$, so $x=0$. Therefore ${\beta }_n{\mathrm{\mid }}_L:L\to Q_n$ is injective.

Given $q\in Q_n$, surjectivity of ${\beta }_n$ supplies $x$ with ${\beta }_{n}x=q$. Set $x^{\mathrm{\prime }}=x-{\alpha }_n{r}_{n}x$. Then $r_n{x}^{\mathrm{\prime }}=r_{n}x-r_{n}x=0$ and ${\beta }_n{x}^{\mathrm{\prime }}=q$, since ${\beta }_n{\alpha }_n=0$. Thus ${\beta }_n{\mathrm{\mid }}_L$ is surjective. Its inverse $s_n:Q_n\to L\subseteq H_n(C\otimes D)$ is linear: sums and scalar multiples of inverse images are inverse images of the corresponding sums and scalar multiples, and uniqueness identifies them. This inverse needs no further selection of representatives.

By definition ${\beta }_n{s}_n(q)=q$. For any $x\in H_n(C\otimes D)$, the element $x-{\alpha }_n{r}_{n}x$ lies in $L$ and maps to ${\beta }_{n}x$, so uniqueness gives $s_n{\beta }_{n}x=x-{\alpha }_n{r}_{n}x$. Hence both asserted composites hold. The maps $(k,q)\mapsto {\alpha }_{n}k+s_{n}q$ and $x\mapsto (r_{n}x,{\beta }_{n}x)$ are inverse: use these two identities, $r_n{\alpha }_n=1$, $r_n{s}_n=0$, and ${\beta }_n{\alpha }_n=0$.

For $n=0$, $Q_n=0$, the section is the unique zero-domain map, and ${\alpha }_n{r}_n=1$ follows from step 6.1. The same proof handles a zero complex or a zero $K_n$ or $Q_n$, including degree one. The choice of $a_p,b_q$ occurs in $r_n$ and therefore in $s_n$; no compatibility of those choices with arbitrary chain maps was imposed. This proves the stated existence without asserting naturality of the chosen section.