Kunneth Exactness and Splittings over Principal Ideal Domains

Fix a basis $(e_a{)}_{a\in I}$ of $F$ and well-order $I$. Put $F_{\le a}=\mathrm{span}\{e_b:b\le a\}$, $F_{<a}=\mathrm{span}\{e_b:b<a\}$, and $N_{\le a}=N\cap F_{\le a}$. The coordinate projection $p_a:F\to R$ is linear by uniqueness of basis expansions.

Its image $J_a=p_a(N_{\le a})$ is an ideal: $p_a(x)-p_a(y)=p_a(x-y)$ and $r{p}_a(x)=p_a(rx)$ for $x,y\in N_{\le a}$ and $r\in R$. Write $J_a=({\rho }_a)$. For each $a$ with $J_a\mathrm{\ne }0$, the set of pairs $(\rho ,u)$ with $J_a=(\rho )$, $u\in N_{\le a}$, and $p_a(u)=\rho$ is nonempty. AC selects such a pair $({\rho }_a,u_a)$ simultaneously for these indices. In particular ${\rho }_a\mathrm{\ne }0$. Let $S=\{a:J_a\mathrm{\ne }0\}$.

We verify the hypothesis of transfinite induction for the assertion that $N_{\le a}$ is spanned by the $u_b$ with $b\in S$, $b\le a$. Suppose the assertion holds at every $b<a$ and take $x\in N_{\le a}$. If $p_a(x)=0$, put $y=x$. Otherwise $a\in S$, and $p_a(x)=r{\rho }_a$ for some $r\in R$; put $y=x-r{u}_a$. In both situations $y\in N\cap F_{<a}$. If $y\mathrm{\ne }0$, its finite support has a greatest element $b<a$, so the assertion at $b$ expresses $y$ in the required earlier generators. If $y=0$, its expression is the empty sum. Restoring $r{u}_a$ when present proves the assertion at $a$.

For a finite relation ${\sum }_{a\in E}{r}_a{u}_a=0$ with distinct $a\in S$, suppose some coefficient is nonzero and take the greatest such index $a$. Every $u_b$ with $b<a$ has zero $a$ coordinate. Applying $p_a$ gives $r_a{\rho }_a=0$. Since $R$ is a domain and ${\rho }_a\mathrm{\ne }0$, this forces $r_a=0$, contrary to its selection. Thus all coefficients vanish.

Transfinite induction now proves the assertion at every $a$. Each nonzero $x\in N$ has a greatest support index and hence belongs to one $N_{\le a}$; therefore the $u_a$ span $N$. This also covers a limit initial segment: every one of its finite supports lies in a smaller principal initial segment, so no additional generator is needed at a limit cut.

Spanning and independence make $(u_a{)}_{a\in S}$ a basis. If $N=0$, every $J_a=0$ and this is the empty basis; if $I=\mathrm{\varnothing }$, also $F=N=0$. In rank one the same construction is simply the zero ideal or its single nonzero generator. These possibilities require no choice from an empty fiber.

Both $Z_{n}C$ and $B_{n}C$ are submodules of the free module $C_n$, the latter lying in the former since $d_n{d}_{n+1}=0$. Apply the local submodule lemma to get their freeness for every $n$. Thus the second sequence in [F1] is a length-one free presentation of $H_{n}C$.

The surjection $d_n:C_n\twoheadrightarrow B_{n-1}C$ admits a lift of the identity of its free target, hence a section $s_n$. For each $n$ the set of such sections is nonempty; AC selects one for every degree. At $n=0$, take the unique map $s_0:0\to C_0$.

Define $u_n(z,b)=z+s_n(b)$ and $v_n(c)=(c-s_n(d_{n}c),d_{n}c)$. The first component of $v_n(c)$ is a cycle because its differential is $d_{n}c-d_n{s}_n(d_{n}c)=0$. Both maps are linear. Substitution gives $u_n{v}_n(c)=c$ and $v_n{u}_n(z,b)=(z,b)$, since $d_{n}z=0$ and $d_n{s}_n(b)=b$. Thus they are inverse isomorphisms.

Compute $d_n{u}_n(z,b)=b$. Since $d_{n-1}b=0$, its image under $v_{n-1}$ is $(b,0)$. For each $p$ let $E^{(p)}$ have $B_{p}C$ in degree $p+1$, $Z_{p}C$ in degree $p$, and differential the inclusion. In degree $n$, ${⨁}_{p\ge 0}{E}^{(p)}$ is $Z_{n}C\oplus B_{n-1}C$ with precisely the differential just computed. The maps $u_n$ therefore form the claimed chain isomorphism. Only two summands occur in each degree.

Inclusion $\iota$ is a chain map because $d_n$ kills $Z_{n}C$. The map $\rho$ is a chain map to the zero-differential complex $A(C)$ because ${\rho }_{n-1}{d}_n=d_{n-1}{d}_n=0$. Its kernel and image in degree $n$ are $Z_{n}C$ and $B_{n-1}C$, respectively. This proves the canonical short exact sequence, and the selected $s_n$ prove degreewise splitting. Those sections need not be chain maps: $d_n{s}_n(b)=b$ can be nonzero. All formulas hold for the zero complex and for degree zero; replacing $C$ throughout by $D$ proves the stated second application.

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.

Write $u:H_{n}X\to H_{n}T$ and $v:H_{n}T\to H_{n}Y$ for the maps of [F1], with $T=C\otimes D$. Its LES gives $\ker u=\mathrm{im}{\mathrm{\partial }}_{n+1}$, $\ker v=\mathrm{im}u$, and $\mathrm{im}v=\ker {\mathrm{\partial }}_n$. Therefore $\bar{u}:\mathrm{coker}{\mathrm{\partial }}_{n+1}\to H_{n}T$, $[x]\mapsto u(x)$, is well defined and injective: $u(x)=0$ exactly when $x\in \mathrm{im}{\mathrm{\partial }}_{n+1}$.

Let $f:C\to C^{\mathrm{\prime }}$ and $g:D\to D^{\mathrm{\prime }}$ be chain maps between complexes satisfying the hypotheses. The relation $d_{C^{\mathrm{\prime }}}f=f{d}_C$ sends cycles to cycles and boundaries to boundaries, and gives ${\rho }^{\mathrm{\prime }}f=A(f)\rho$, where $A(f{)}_p$ is $f_{p-1}$ restricted to $B_{p-1}C$. Thus $(Z(f)\otimes g,f\otimes g,A(f)\otimes g)$ is a morphism of the canonical short exact tensor sequences. By [F4] it commutes with $u,v,\mathrm{\partial }$, hence with their induced kernel/cokernel maps.

The corestriction $\bar{v}:H_{n}T\to \ker {\mathrm{\partial }}_n$ is onto and has kernel $\mathrm{im}\bar{u}$. Thus $0\to \mathrm{coker}{\mathrm{\partial }}_{n+1}\stackrel{\bar{u}}{\to }{H}_{n}T\stackrel{\bar{v}}{\to }\ker {\mathrm{\partial }}_n\to 0$ is exact. Under the explicit identifications in [F1], $\bar{u}$ is precisely ${\alpha }_n$ of [F2], and $\bar{v}$ is precisely ${\beta }_n$ of [F3]. This proves all three exactness assertions for the maps in the statement.

The cokernel identification commutes with these maps since $z\otimes [w]$ goes to $f(z)\otimes [g(w)]$ and then to $[f(z)]\otimes [g(w)]$. On a kernel summand, the maps $B_{r}C\to B_r{C}^{\mathrm{\prime }}$ and $Z_{r}C\to Z_r{C}^{\mathrm{\prime }}$ form a map of the actual length-one resolutions lifting $H_{r}f$. Tensoring with $H_{q}g$ induces the Tor map used by the natural quotient [F3]. Hence step 1.2 gives both naturality squares for the displayed sequence. Equivalently the first square follows by evaluating [F2] on $[z]\otimes [w]$. No selected sections enter $\rho$, these resolution maps, or the two final arrows.

At $n=0$ the Tor sum is empty, so exactness makes ${\alpha }_0:H_{0}C\otimes H_{0}D\to H_{0}T$ an isomorphism. At $n=1$ the right term is ${\mathrm{Tor}}_1(H_{0}C,H_{0}D)$. If one complex is zero, $X,T,Y$ and both end terms are zero, so the same proof gives the zero exact sequence. There is no upper endpoint: for each $n$ only finitely many pairs $p+q=n$ occur, regardless of the ranks.

Select the sections of [F2] for both complexes, using [F4], and denote the resulting cycle retractions by $a_p:C_p\to Z_{p}C$ and $b_q:D_q\to Z_{q}D$. Define ${\pi }_{C,p}(c)=[a_p(c)]\in H_{p}C$ and ${\pi }_{D,q}(w)=[b_q(w)]\in H_{q}D$. Since a boundary is a cycle, $a_{p-1}(d_{C}c)=d_{C}c$, whose homology class is zero. Thus ${\pi }_C{d}_C=0$; likewise ${\pi }_D{d}_D=0$. With zero differentials on $H(C)$ and $H(D)$, these are chain maps.

Define $\mathrm{\Pi }:C\otimes D\to H(C)\otimes H(D)$ by $c\otimes w\mapsto {\pi }_C(c)\otimes {\pi }_D(w)$. This descends to tensors because the formula is bilinear and balanced: replacing $rc\otimes w$ by $c\otimes rw$ gives the same tensor by linearity of ${\pi }_C,{\pi }_D$. On a homogeneous tensor, $\mathrm{\Pi }d(c\otimes w)={\pi }_C(d_{C}c)\otimes {\pi }_D(w)+(-1{)}^p{\pi }_C(c)\otimes {\pi }_D(d_{D}w)=0$. The target differential is zero, so $\mathrm{\Pi }$ is a chain map.

The target has zero differential, hence its degree-$n$ homology is exactly $K_n$, even if its modules are not free. By [F3], $\mathrm{\Pi }$ induces $r_n:H_n(C\otimes D)\to K_n$. For cycles $z,w$ the retractions fix them, so $r_n{\alpha }_n([z]\otimes [w])=[z]\otimes [w]$. Elementary tensors of homology classes generate $K_n$, proving $r_n{\alpha }_n=1_{K_n}$.

Put $L=\ker r_n$. If $x\in L$ and ${\beta }_{n}x=0$, exactness gives $x={\alpha }_{n}k$ for some $k\in K_n$. Applying $r_n$ gives $0=r_{n}x=k$, so $x=0$. Therefore ${\beta }_n{\mathrm{\mid }}_L:L\to Q_n$ is injective.

Given $q\in Q_n$, surjectivity of ${\beta }_n$ supplies $x$ with ${\beta }_{n}x=q$. Set $x^{\mathrm{\prime }}=x-{\alpha }_n{r}_{n}x$. Then $r_n{x}^{\mathrm{\prime }}=r_{n}x-r_{n}x=0$ and ${\beta }_n{x}^{\mathrm{\prime }}=q$, since ${\beta }_n{\alpha }_n=0$. Thus ${\beta }_n{\mathrm{\mid }}_L$ is surjective. Its inverse $s_n:Q_n\to L\subseteq H_n(C\otimes D)$ is linear: sums and scalar multiples of inverse images are inverse images of the corresponding sums and scalar multiples, and uniqueness identifies them. This inverse needs no further selection of representatives.

By definition ${\beta }_n{s}_n(q)=q$. For any $x\in H_n(C\otimes D)$, the element $x-{\alpha }_n{r}_{n}x$ lies in $L$ and maps to ${\beta }_{n}x$, so uniqueness gives $s_n{\beta }_{n}x=x-{\alpha }_n{r}_{n}x$. Hence both asserted composites hold. The maps $(k,q)\mapsto {\alpha }_{n}k+s_{n}q$ and $x\mapsto (r_{n}x,{\beta }_{n}x)$ are inverse: use these two identities, $r_n{\alpha }_n=1$, $r_n{s}_n=0$, and ${\beta }_n{\alpha }_n=0$.

For $n=0$, $Q_n=0$, the section is the unique zero-domain map, and ${\alpha }_n{r}_n=1$ follows from step 6.1. The same proof handles a zero complex or a zero $K_n$ or $Q_n$, including degree one. The choice of $a_p,b_q$ occurs in $r_n$ and therefore in $s_n$; no compatibility of those choices with arbitrary chain maps was imposed. This proves the stated existence without asserting naturality of the chosen section.