Topological Kunneth short exact sequence for homology

Both complexes are nonnegative and free in every degree by [F3], with no finite-rank assumption. Their tensor differential is exactly that in [F1] and [F2]. Thus [F1] applies and gives $0\to K\stackrel{\alpha }{\to }V\stackrel{\beta }{\to }Q\to 0$ with the two direct sums in the statement. All tensor-degree diagonals are finite since $p,q\ge 0$.

A chain homotopy changes the image of a cycle by a boundary: if $u-v=dH+Hd$ and $dz=0$, then $uz-vz=dHz$. Hence the inverse and homotopies in [F2] give mutually inverse homology maps $s=S_{\ast }:V\to W$ and $t=T_{\ast }:W\to V$. Put $\gamma =\beta t$. Although a chain inverse was constructed, its homology map is uniquely $s^{-1}$; thus $\gamma$ is independent of any inverse choices.

Define the left arrow as $s\alpha$. It sends $[z]\otimes [w]$ to $[S(z\otimes w)]=[z\times w]$, the singular cross product of [F2]. It is injective because $s$ and $\alpha$ are injective. For $x\in W$, $\gamma x=0$ exactly when $tx\in \ker \beta =\mathrm{im}\alpha$, exactly when $x\in \mathrm{im}(s\alpha )$. Finally any $q\in Q$ equals $\beta v$ for some $v\in V$, and $\gamma (sv)=q$, proving surjectivity. This verifies the exactness of the actual displayed arrows.

For maps $f:X\to X^{\mathrm{\prime }}$ and $g:Y\to Y^{\mathrm{\prime }}$, naturality of the shuffle gives $s^{\mathrm{\prime }}(f_{\mathrm{\#}}\otimes g_{\mathrm{\#}}{)}_{\ast }=(f\times g{)}_{\ast }s$. Multiplying by the inverse maps yields $t^{\mathrm{\prime }}(f\times g{)}_{\ast }=(f_{\mathrm{\#}}\otimes g_{\mathrm{\#}}{)}_{\ast }t$. Naturality of $\alpha ,\beta$ in [F1] now gives both squares of the topological sequence. The left square also follows directly from the cycle cross-product formula. No chosen cycle projections appear in these arrows.

At $n=0$, the Tor sum is empty and the cross product is an isomorphism $H_0(X;R){\otimes }_R{H}_0(Y;R)\cong H_0(X\times Y;R)$. At $n=1$, the sole Tor index pair is $(0,0)$, as stated. Empty $X$ or $Y$ gives zero complexes and a zero sequence. On the degree-zero class of a pair of point simplices, the cross product is that point in the product, including coefficient $1$. AC is inherited from the free PID theorem [F1] for its cycle/boundary and resolution constructions; the shuffle equivalence introduces none. There is no upper-degree or dimension restriction.