The Kunneth Tor map

Let $Z$ and $A$ be the complexes with zero differential and $Z_p=Z_{p}C$, $A_p=B_{p-1}C$. Inclusion and the corestriction of $d_C$ give $0\to Z\to C\stackrel{\rho }{\to }A\to 0$. This is a sequence of chain complexes because $d_C$ vanishes on cycles and $\rho d_C=0$. Each degree sequence splits by [F2]. Tensoring with $D$ and taking direct-sum total complexes therefore gives a short exact sequence $0\to X=Z\otimes D\to T=C\otimes D\stackrel{r=\rho \otimes 1}{\to }Y=A\otimes D\to 0$.

Since $Z_p$ and $A_p$ are free, tensoring with either is a direct sum of copies and commutes with homology. Thus $H_{n}X={⨁}_{p+q=n}{Z}_{p}C\otimes H_{q}D$ and $H_{n}Y={⨁}_{p+q=n}{B}_{p-1}C\otimes H_{q}D$. The differential on a fixed $p$ summand is $(-1{)}^p{d}_D$, which has the same cycles and boundaries as $d_D$.

The connecting map ${\mathrm{\partial }}_n:H_{n}Y\to H_{n-1}X$ is the direct sum of the maps induced by $B_{p-1}C\hookrightarrow Z_{p-1}C$. Indeed, represent a summand by a finite sum of $b\otimes y$ with $y$ a cycle in $D$, and lift $b$ to $c\in C_p$ with $d_{C}c=b$. Then $d_T(c\otimes y)=b\otimes y$, with no second term. This is the defining connecting-map calculation, so its sign is positive.

The free presentation $0\to B_{p-1}C\to Z_{p-1}C\to H_{p-1}C\to 0$ is a length-one projective resolution. Hence [F4] identifies $\ker (B_{p-1}C\otimes H_{q}D\to Z_{p-1}C\otimes H_{q}D)$ with ${\mathrm{Tor}}_1^R(H_{p-1}C,H_{q}D)$. Therefore $\ker {\mathrm{\partial }}_n$ is exactly the displayed Tor sum after reindexing $p-1$.

By [F3], $H_n(r)$ has image $\ker {\mathrm{\partial }}_n$. Define ${\pi }_n$ as $H_n(r)$ corestricted to this kernel and followed by the identification in step 4.1. It is well defined on homology and surjective. A pair of chain maps induces maps of the sequence in step 1.1 and of the free presentations in step 4.1, so [F3] and the comparison naturality in [F4] prove naturality of ${\pi }_n$. Empty sums and zero modules cause no exception.