The cycle-boundary tensor sequence has the Kunneth kernel and cokernel

AC implies the particular DC principle needed in [F5]. For an entire relation $E$ on a nonempty set $S$, each successor set $E_x=\{y:xEy\}$ is nonempty. Choose $f(x)\in E_x$ simultaneously. Recursion from any prescribed $a\in S$ gives $x_0=a$, $x_{m+1}=f(x_m)$, hence $x_{m}E{x}_{m+1}$ for every $m$. This is [F10]. For each $H_{r}C$ and $H_{q}D$, [F1] supplies a length-one free resolution, and [F11] makes it projective. Thus [F5] applies to these actual supplied resolutions.

Choose the degreewise sections from [F1]. In bidegree $(p,q)$, tensoring $C_p\cong Z_{p}C\oplus B_{p-1}C$ with $D_q$ identifies the two maps with inclusion and projection of a direct sum. Their kernel and image therefore agree and projection is onto. Summing over the finite diagonal proves exactness of $0\to X\to T\to Y\to 0$. Inclusion and $\rho \otimes 1$ are chain maps: $d_C$ kills cycles and $\rho d_C=0$, while the terms $(-1{)}^p{d}_D$ agree on both sides.

For a free module $U={⨁}_{i\in I}R{e}_i$ placed in degree $p$, [F3] identifies $U\otimes D_q$ with ${⨁}_i{D}_q$ by $e_i\otimes y\mapsto y$ in coordinate $i$ (the maps $R\otimes D_q\to D_q$, $r\otimes y\mapsto ry$, and $y\mapsto 1\otimes y$ are inverse). Its differential is coordinatewise $(-1{)}^p{d}_D$. A finite tuple is a cycle exactly when every coordinate is a cycle. Its image consists exactly of finite tuples of boundaries: for the reverse containment choose a preimage for each of the finitely many nonzero coordinates and multiply it by $(-1{)}^p$. Quotienting therefore gives $U\otimes H_{q}D\cong H_{p+q}(U\otimes D)$ by $u\otimes [y]\mapsto [u\otimes y]$. Although a basis proves bijectivity, this formula is independent of that basis.

Apply this calculation to the free modules $U=Z_{p}C$ and $U=B_{p-1}C$ and sum in $p$. There is no differential between distinct $p$ summands, so cycles, boundaries, and homology decompose over this finite diagonal. This gives both displayed homology identifications. The summand with $p=0$ in $H_{n}Y$ is zero because $B_{-1}C=0$.

Fix $r,q\ge 0$. The resolution $P_1=B_{r}C\hookrightarrow P_0=Z_{r}C\twoheadrightarrow H_{r}C$, with $P_i=0$ for $i\ge 2$, is projective by step 1.1. Tensoring it with $H_{q}D$ computes first homology as the kernel of $j_r\otimes 1:B_{r}C\otimes H_{q}D\to Z_{r}C\otimes H_{q}D$, because the degree-two boundary is zero. By [F5] this kernel is ${\mathrm{Tor}}_1^R(H_{r}C,H_{q}D)$. Right exactness gives the cokernel as $H_{r}C\otimes H_{q}D$, using the augmentation $z\mapsto [z]$. No injectivity of $j_r\otimes 1$ is assumed.

A class in a $(p,q)$ summand of $H_{n}Y$ is a finite sum of $b\otimes [y]$ with $d_{D}y=0$. Lift its representing cycle to the corresponding sum $s_p(b)\otimes y$ in $T_n$. Its differential is $b\otimes y+(-1{)}^p{s}_p(b)\otimes d_{D}y=b\otimes y$, now in $X_{n-1}$. The lift-and-boundary construction of the connecting map in [F4] therefore gives ${\mathrm{\partial }}_n(b\otimes [y])=b\otimes [y]$ with $b$ included into $Z_{p-1}C$. The formula holds for sums, not just decomposable classes.

In step 3.1 put $r=p-1$ and discard the zero $p=0$ summand. A sum maps to zero exactly when each of its components does; its cokernel is the sum of the component cokernels, since each relation lies in its own summand. Step 2.2 thus gives the asserted kernel in degree $n-1$ and cokernel in degree $n$. At $n=0$ the kernel is zero; at $n=1$ it is precisely ${\mathrm{Tor}}_1(H_{0}C,H_{0}D)$. If $H_{q}D=0$, both tensor modules are zero. If $H_{r}C=0$, its presentation has $B_{r}C=Z_{r}C$ and inclusion the identity, so tensoring gives an isomorphism with zero kernel and cokernel.

The homology LES makes the image of $H_n(\rho \otimes 1)$ exactly $\ker {\mathrm{\partial }}_n$. Corestrict it and use the identification of step 2.2. This is the same canonical sequence, map $\rho$, positive connecting map, and free presentation used to construct the quotient in [F7], so it is that quotient, rather than merely some surjection onto an isomorphic module. On the other side, the cokernel class of $z\otimes [w]\in Z_{p}C\otimes H_{q}D$ is $[z]\otimes [w]$; its image under $H_n(X\to T)$ is $[z\otimes w]$ by step 1.3. This identifies the asserted injection formula at the level of the actual maps.