Cohomological Kunneth isomorphism under finite free hypotheses

Choose the sections $s_j:B_{j-1}D\to D_j$ from [F1] and put ${\pi }_j=1-s_j{d}_j:D_j\to Z_{j}D$. Since $V_j$ is free, choose a section ${\mathrm{\ell }}_j:V_j\to Z_{j}D$ of the homology quotient $q_j:Z_{j}D\to V_j$, using [F1]. AC permits these choices in every degree. Set ${\eta }_j=q_j{\pi }_j$ and $b_j={\pi }_j-{\mathrm{\ell }}_j{\eta }_j:D_j\to B_{j}D$, where the last corestriction is valid because $q_j{b}_j=0$. We have $\eta \mathrm{\ell }=1$, $d\mathrm{\ell }=0$ and $\eta d=0$: a boundary is fixed by $\pi$ and killed by $q$. Thus $\mathrm{\ell }:V\to D$ and $\eta :D\to V$ are chain maps.

Fix $q$ and choose a finite basis $e_1,\dots ,e_m$ of $V_q$ with coordinate duals $e_i^{\ast }$. For every $R$-module $U$, the map $\mathrm{Hom}(U,R)\otimes V_q^{\ast }\to \mathrm{Hom}(U\otimes V_q,R)$ sends $\phi \otimes \lambda$ to $(u\otimes v\mapsto \phi (u)\lambda (v))$. An explicit inverse sends $w$ to ${\sum }_{i=1}^m{\phi }_i\otimes e_i^{\ast }$, where ${\phi }_i(u)=w(u\otimes e_i)$. Substitution using $v={\sum }_i{e}_i^{\ast }(v)e_i$ proves one composite is identity. For the other, write $\lambda ={\sum }_i\lambda (e_i)e_i^{\ast }$ and use tensor bilinearity. Thus this is an isomorphism for arbitrary $U$, including infinitely generated $U$. It is its formula, not the chosen basis, that specifies the map. Finite rank is used in this finite inverse sum.

Define $h_j=s_{j+1}{b}_j:D_j\to D_{j+1}$. Then $d{h}_j=b_j$, while $b_{j-1}{d}_j=d_j$ because $d_j$ lands in boundaries. Hence $h_{j-1}{d}_j=s_j{d}_j$ and $dh+hd=b+sd=1-\mathrm{\ell }\eta$. Also $\eta \mathrm{\ell }=1$ by step 1.1. This is an explicit chain deformation retraction of $D$ onto $V$, not merely an isomorphism of its homology groups. The same calculation works at $j=0$ with negative terms zero.

In degree $n$, $\mathrm{Hom}((C\otimes V{)}_n,R)$ is the finite direct sum of $\mathrm{Hom}(C_{n-q}\otimes V_q,R)$ for $0\le q\le n$. Since $d_V=0$, the differential preserves $q$ and is the positive $C$ coboundary. Step 1.2 identifies each fixed-$q$ complex with $m$ copies of the $C$ cochain complex shifted in degree by $q$. Kernels and images of maps on finitely many coordinates are taken coordinatewise, so its cohomology is $H^{n-q}(X;R)\otimes V_q^{\ast }$. Taking the finite degree diagonal gives an isomorphism from ${⨁}_{p+q=n}{H}^p(X;R)\otimes V_q^{\ast }$ to $H^n(\mathrm{Hom}(C\otimes V,R))$, represented by the evaluation functionals. The absence of a degree-$(-1)$ coordinate at $q=n$ agrees with the zero incoming coboundaries in $C^0$.

Tensor with $C$: on a homogeneous tensor define $H(c\otimes y)=(-1{)}^{\mathrm{\mid }c\mathrm{\mid }}c\otimes hy$. In $dH+Hd$, the two terms involving $dc$ cancel, while the other two give $c\otimes (dh+hd)y$. Thus $1\otimes \eta$ and $1\otimes \mathrm{\ell }$ are chain homotopy inverses between $C\otimes D$ and $C\otimes V$. Precomposition with these maps gives cochain homotopy inverses on Hom into $R$: if $u-v=dH+Hd$, the operator $K(\varphi )=\varphi H$ in degree minus one obeys $\delta K+K\delta =(u-v{)}^{\ast }$. In particular no tensor exactness or Hom exactness is needed to preserve this specified homotopy equivalence. Combining with [F3], $A=(1\otimes \eta )T:D(X\times Y)\to C\otimes V$ is a chain homotopy equivalence, so $A^{\ast }$ is a cohomology isomorphism.

On $\mathrm{Hom}(D,R)$, $K\varphi =\varphi h$ gives $\delta K+K\delta =1-(\mathrm{\ell }\eta {)}^{\ast }$ by step 2.1, while $(\eta \mathrm{\ell }{)}^{\ast }=1$. These identities identify $H^q(Y;R)$ with $V_q^{\ast }={\mathrm{Hom}}_R(V_q,R)$: a functional $\lambda$ corresponds to the cohomology class $[\lambda {\eta }_q]$, with inverse induced by $\mathrm{\ell }$. Since $V$ has zero differential, its Hom cohomology is exactly its Hom graded module.

Combine step 2.2 with $A^{\ast }$ of step 3.1 and replace $V_q^{\ast }$ by $H^q(Y;R)$ using step 3.2. A representative $\phi \otimes \lambda$ is carried to the class of the functional $J(\phi ,\lambda )(1\otimes \eta )T=J(\phi ,\lambda \eta )T$. By [F2] this is precisely $[\phi ]\times [\lambda \eta ]$. Consequently the resulting isomorphism is the cross product in the statement, independent of every auxiliary basis and section used to prove it is bijective.

For the symmetric case, carry out steps 1.1 and 2.1 on $C$ instead, obtaining a deformation retraction onto $U_p=H_p(C)$. Tensoring its homotopy with $1_D$ has $d(h\otimes 1)+(h\otimes 1)d=(1-\mathrm{\ell }\eta )\otimes 1$, with mixed terms cancelling. Thus reduce to $\mathrm{Hom}(U\otimes D,R)$. On the fixed-$p$ summand its differential is $(-1{)}^p$ times the $D$ coboundary. Multiplication by this unit changes neither kernels nor images. A finite basis of $U_p$ gives the analogue of step 1.2 with the finite free factor first, hence cohomology $U_p^{\ast }\otimes H^q(Y;R)$. Its evaluation functional is $J(\lambda \eta ,\psi )T$, exactly the same ordered cross product by [F2]. This proves the symmetric assertion without assuming a symmetry theorem for that product.

Empty $X$ or $Y$ gives zero complexes. A zero-rank $V_q$ contributes an empty basis and a zero summand in steps 1.2 and 2.2; rank one gives the evident identification with a single copy of $C^{\ast }$. At $n=0$, the only diagonal is $(0,0)$ and all constructions retain the usual product of values at vertices. Although there may be infinitely many nonzero $V_q$, each total degree uses only finitely many, so no exchange of an infinite product with tensor is asserted. AC is used for the arbitrary-rank PID cycle/boundary sections and for simultaneous homology sections and finite bases across degrees; all dual and tensor calculations are explicit after those choices.