Over a field the cross product itself is a natural isomorphism

A field is a PID: a nonzero ideal contains $c\mathrm{\ne }0$, hence also $c^{-1}c=1$ and so is the whole field; the remaining ideal is $(0)$. The displayed complexes are nonnegative and have free terms with their stated bases, so [F1] applies. Each factor has zero differential, and the Koszul formula gives $d(a\otimes u)=d(b\otimes u)=d(a\otimes v)=d(b\otimes v)=0$. The only tensor terms are exactly those displayed in degrees $0,1,2$.

First derive the DC hypothesis of [F3] from the assumed AC. Given an entire relation $E$ on a nonempty set $S$ and $a\in S$, AC chooses for each $x\in S$ an element $f(x)$ of the nonempty successor set $E_x=\{y:xEy\}$. By [F4], iterate $f$ from $a$ to obtain $x_0=a$ and $x_{m+1}=f(x_m)$, so $x_{m}E{x}_{m+1}$ for every $m$; this is [F5]. Now for any $k$-modules $U,V$, [F2] makes both projective under AC. Supply the resolutions having $U$, respectively $V$, only in degree zero and identity augmentation. The tensor resolution then has no positive-degree terms, so its homology in every degree $i>0$ is zero. Hence [F3] gives ${\mathrm{Tor}}_1^k(U,V)=0$, including zero $U$ or $V$. In particular every summand in the Kunneth Tor term vanishes for any complexes under discussion.

These tensors are bases: for example $(\lambda b)\otimes (\mu u)\mapsto \lambda \mu$ and $\nu \mapsto \nu (b\otimes u)$ are inverse linear maps $kb\otimes ku\leftrightarrow k$, since balancing identifies $(\lambda b)\otimes (\mu u)$ with $(\lambda \mu )(b\otimes u)$. The same formulas apply to the other three one-dimensional factors. In degree one the two bidegrees form a direct sum. As all differentials vanish, every element is a cycle and the only boundary is zero; this proves the listed homology bases and vanishing elsewhere.

Exactness in [F1] now makes the cross product ${\alpha }_n$ injective and surjective for every $n$. In this instance it sends $[a]\otimes [u]$, $[b]\otimes [u]$, $[a]\otimes [v]$, $[b]\otimes [v]$ to the four homology basis elements in that order. Its inverse sends those four elements back to the four tensors of classes, extended linearly. Both composites fix a basis and therefore every element. Degree zero has an empty Tor sum; degree one has the single zero group ${\mathrm{Tor}}_1(k,k)$; in higher degrees the same vanishing argument applies, including where every term is zero.

For arbitrary maps of such complexes write $h$ for the induced map on tensor-complex homology and $k_{\ast }$ for the induced map on the tensor of homologies. Naturality in [F1] gives $h{\alpha }_n={\alpha }_n^{\mathrm{\prime }}{k}_{\ast }$. Multiplying by the inverses just proved gives ${\alpha }_n^{\mathrm{\prime }-1}h=k_{\ast }{\alpha }_n^{-1}$. Hence the inverse is natural as well. There is no selected complement in this inverse: it is the unique inverse of the canonical cross product after the Tor term vanishes.