Universal Coefficients and Kunneth Theorems

By definition, ${\delta }^{n+1}({\delta }^{n}f)=(f\circ d_{n+1})\circ d_{n+2}=f\circ (d_{n+1}{d}_{n+2})$.

The chain-complex identity $d_{n+1}{d}_{n+2}=0$ makes this composite zero, so the Hom groups form a cochain complex.

Corestricting $d_n:C_n\to C_{n-1}$ to its image gives a surjection $C_n\twoheadrightarrow B_{n-1}C$ whose kernel is $Z_{n}C$.

Since $d_n{d}_{n+1}=0$, $B_{n}C\subseteq Z_{n}C$; the quotient map $Z_{n}C\twoheadrightarrow Z_{n}C\mathrm{/}{B}_{n}C=H_{n}C$ has kernel $B_{n}C$, proving both sequences.

Use [L4] to well-order $W$. For $a\in W$ let $F_{\le a}$ and $F_{<a}$ be the spans of the corresponding basis initial segments. Let $I_a$ be the image of $N\cap F_{\le a}$ under the $a$-coordinate map. This image is an ideal because the coordinate map is $R$-linear.

If $I_a\mathrm{\ne }0$, [L1] supplies a nonzero generator $r_a$ and the definition of the image supplies $x_a\in N\cap F_{\le a}$ with $a$-coordinate $r_a$. The sets of such pairs $(r_a,x_a)$ are nonempty and indexed by a subset of $W$; use [L3] to choose them simultaneously. This is the second use of AC, after well-ordering the basis.

If $y\in N\cap F_{\le a}$ and $I_a=0$, then $y\in F_{<a}$. If $I_a\mathrm{\ne }0$, write its $a$-coordinate as $c{r}_a$; then $y-c{x}_a\in N\cap F_{<a}$. In either case the remainder has finite support below $a$.

Induct on $a$ using [L5] to show that $N\cap F_{\le a}$ is spanned by the chosen $x_b$ with $b\le a$. By step 3.1 it suffices to handle a finite-support remainder in $F_{<a}$. A zero remainder needs no generators; otherwise its support has a greatest element $b<a$, so the induction hypothesis at $b$ applies. This includes least and limit positions without selecting a cofinal sequence. Every nonzero element of $N$ has a greatest support element, so the chosen family spans $N$.

In a nonzero finite relation among the chosen $x_a$, take the greatest index $a$ with nonzero coefficient $c_a$. Its $a$-coordinate is $c_a{r}_a$: all smaller-index vectors have zero coordinate there. Since $R$ is a domain and $r_a\mathrm{\ne }0$, this coordinate is nonzero, a contradiction. Thus the family is independent and is a basis by step 4.1. If $W$ or $N$ is empty or zero as applicable, the same argument gives the empty basis.

Both $B_{n}C$ and $Z_{n}C$ are $R$-submodules of $C_n$.

Applying A submodule of an arbitrary-rank free module over a PID is free under the stated Choice hypothesis separately to these two inclusions proves that both modules are free.

Since $dz=0$, $d(z\otimes g)=dz\otimes g=0$, so $z\otimes g$ represents a homology class.

Replacing $z$ by $z+dw$ changes $z\otimes g$ by $d(w\otimes g)$, and $(zr)\otimes g=z\otimes (rg)$; hence the formula descends through both quotients.

AC is used in Boundaries and cycles in a free complex over a PID are free to make $B_j$ and $Z_j$ free, hence flat, even for infinitely generated $C_j$. Thus $0\to B_{n-1}\to Z_{n-1}\to H_{n-1}C\to 0$ is a supplied free resolution of length one. Tensoring it with $G$ identifies ${\mathrm{Tor}}_1^R(H_{n-1}C,G)$ with $T:=\ker (B_{n-1}{\otimes }_{R}G\to Z_{n-1}{\otimes }_{R}G)$. AC also implies the DC premise used for comparison of supplied resolutions in The balanced Tor bifunctor, so this kernel represents the stated balanced Tor group naturally.

Corestrict $d_n:C_n\twoheadrightarrow B_{n-1}$ before tensoring. Let ${\bar{d}}_n\otimes 1:C_n\otimes G\to B_{n-1}\otimes G$ denote the resulting surjection. If $c\in C_n\otimes G$ is a cycle, then the composite of ${\bar{d}}_n\otimes 1$ with $B_{n-1}\otimes G\to Z_{n-1}\otimes G\to C_{n-1}\otimes G$ is zero. The last arrow is injective: tensor $0\to Z_{n-1}\to C_{n-1}\to B_{n-2}\to 0$ and use flatness of $B_{n-2}$. Hence $({\bar{d}}_n\otimes 1)c\in T$. If $c=(d_{n+1}\otimes 1)b$, then ${\bar{d}}_n{d}_{n+1}=0$, so this element is zero. We obtain a homomorphism on homology.

A chain map $f:C\to C^{\mathrm{\prime }}$ restricts to maps of the corresponding $B$ and $Z$ modules; together with a coefficient map $G\to G^{\mathrm{\prime }}$ (over the same ring), these maps commute with the corestricted differentials and the inclusions defining $T$. Thus the cycle formula commutes with them. Naturality of the free-resolution kernel identification with Tor follows from the comparison and coherence built into The balanced Tor bifunctor. The map is surjective as well: $C_n\otimes G\twoheadrightarrow B_{n-1}\otimes G$ is surjective by right exactness, and any lift of an element of $T$ is a cycle since $Z_{n-1}\otimes G\hookrightarrow C_{n-1}\otimes G$. Surjectivity is recorded to identify the obstruction map used by the universal-coefficient exact sequence, though no splitting is claimed here.

The cycle and boundary modules are free, hence flat. Since $B_{n-1}C$ is flat, tensoring $0\to Z_{n}C\to C_n\to B_{n-1}C\to 0$ remains exact. The differential on $C_n\otimes G$ factors as the resulting surjection $C_n\otimes G\twoheadrightarrow B_{n-1}C\otimes G$ followed by the map $B_{n-1}C\otimes G\to Z_{n-1}C\otimes G\hookrightarrow C_{n-1}\otimes G$. The last inclusion stays injective because $B_{n-2}C$ is flat.

Tensoring $B_{n}C\to Z_{n}C\to H_n(C)\to 0$ gives $H_n(C)\otimes G=\mathrm{coker}(B_{n}C\otimes G\to Z_{n}C\otimes G)$. By step 1.1, $Z_{n}C\otimes G$ embeds in $C_n\otimes G$, and the image of $B_{n}C\otimes G$ there is exactly the boundary group. Thus the edge map $[z]\otimes g\mapsto [z\otimes g]$ injects $H_n(C)\otimes G$ into $H_n(C\otimes G)$.

The remaining quotient maps by $d_n\otimes 1$ onto $\ker (B_{n-1}C\otimes G\to Z_{n-1}C\otimes G)$. The free presentation $0\to B_{n-1}C\to Z_{n-1}C\to H_{n-1}C\to 0$ identifies this kernel with ${\mathrm{Tor}}_1^R(H_{n-1}C,G)$. All maps come from the natural cycle, boundary and tensor maps, so the sequence is natural in $C$ and $G$.

By Boundaries and cycles in a free complex over a PID are free under AC, $B_{n-1}$ and $B_{n-2}$ are free abelian. Choose a basis of $B_{n-1}$ and a preimage in $C_n$ of each basis element. AC supplies these choices, which extend to a homomorphism $s:B_{n-1}\to C_n$ with $d_{n}s=\mathrm{id}$. Hence $C_n=Z_n\oplus s(B_{n-1})$. The same argument gives $C_{n-1}=Z_{n-1}\oplus s^{\mathrm{\prime }}(B_{n-2})$, so the inclusion $Z_{n-1}\otimes G\to C_{n-1}\otimes G$ is injective.

The free presentation $0\to B_{n-1}\stackrel{i}{\to }{Z}_{n-1}\to H_{n-1}(C)\to 0$ identifies $T:=\ker (i\otimes 1:B_{n-1}\otimes G\to Z_{n-1}\otimes G)\cong {\mathrm{Tor}}_1^{\mathbb{Z}}(H_{n-1}(C),G)\mathrm{.}$ For $t\in T$, the element $(s\otimes 1)(t)$ is a cycle in $C_n\otimes G$: its differential is the image of $(i\otimes 1)(t)=0$ in $C_{n-1}\otimes G$. Thus $\sigma :T⟶H_n(C\otimes G),t⟼[(s\otimes 1)(t)]$ is a well-defined homomorphism. The UCT quotient map of the cited theorem first takes a cycle to its $B_{n-1}\otimes G$ component under $C_n\otimes G=(Z_n\otimes G)\oplus (B_{n-1}\otimes G)$ and then identifies that component with $T$. Consequently its composite with $\sigma$ is the identity on $T$, giving the asserted splitting. The chosen section $s$ need not be preserved by chain maps, so this construction makes no naturality claim.

AC makes $B_{n-1}C$ and $Z_{n-1}C$ free by Boundaries and cycles in a free complex over a PID are free, hence projective by Free modules are projective, with the exact choice boundary. The displayed short exact sequence is therefore a length-one projective resolution of $H_{n-1}C$. By Ext can be computed from any projective resolution of the first variable, using AC to meet its DC premise and the supplied balanced-Ext data, its degree-one cohomology identifies ${\mathrm{Ext}}_R^1(H_{n-1}C,G)$ with ${\mathrm{Hom}}_R(B_{n-1}C,G)$ modulo restrictions of maps $Z_{n-1}C\to G$.

For $\psi :B_{n-1}C\to G$, define the degree-$n$ cochain $\psi d_n:C_n\to G$. It is a cocycle because $d_n{d}_{n+1}=0$. If $\psi$ changes by $g{\mathrm{\mid }}_{B_{n-1}C}$ for $g:Z_{n-1}C\to G$, projectivity of $B_{n-2}C$ splits $C_{n-1}\twoheadrightarrow B_{n-2}C$, giving a projection ${\pi }_{n-1}:C_{n-1}\to Z_{n-1}C$. Then $(g{\mathrm{\mid }}_B)d_n=(g{\pi }_{n-1})d_n$ is a coboundary, since ${\pi }_{n-1}{d}_n=d_n$. Thus $[\psi ]\mapsto [\psi d_n]$ descends from the quotient in step 1.1 to the claimed cohomology group.

For a chain map $u:C\to D$, the pullback of $\psi d_n^D$ equals $(\psi u_{n-1}{\mathrm{\mid }}_{B_{n-1}C})d_n^C$ because $u$ commutes with differentials. This is the map induced on the two projective presentations. Postcomposition with a coefficient map $G\to G^{\mathrm{\prime }}$ also commutes exactly with the construction. Hence the resulting map is natural in the complex and in $G$; the projection used in step 2.1 establishes descent but is absent from the map itself.

Under the stated Choice hypothesis, $B_{n-1}C$ is free and projective. Choose a section $s_n:B_{n-1}C\to C_n$ of the corestricted differential and put ${\pi }_n=1-s_n{d}_n:C_n\to Z_{n}C$. For any $h:H_{n}C\to G$, the cochain $h\circ q_n\circ {\pi }_n$ restricts on $Z_{n}C$ to $h\circ q_n$, vanishes on $B_{n}C$, and hence is a cocycle. Its evaluation is $h$, proving surjectivity.

If a cocycle $f:C_n\to G$ has zero evaluation, then $f{\mathrm{\mid }}_{Z_{n}C}=0$: its restriction factors through $H_{n}C$, and that induced map is zero. Since $d_n:C_n\twoheadrightarrow B_{n-1}C$ has kernel $Z_{n}C$, there is a unique $\psi :B_{n-1}C\to G$ with $f=\psi d_n$. Conversely every such $\psi d_n$ is a cocycle because $d_n{d}_{n+1}=0$.

The cochain $\psi d_n$ is a coboundary exactly when $\psi$ extends to $Z_{n-1}C$. Indeed, if $\psi d_n=\phi d_n$ for a cochain $\phi :C_{n-1}\to G$, surjectivity of $d_n$ onto $B_{n-1}C$ gives $\psi =\phi {\mathrm{\mid }}_{B_{n-1}C}$. Conversely, if $g:Z_{n-1}C\to G$ extends $\psi$, Choice and projectivity of $B_{n-2}C$ give a projection ${\pi }_{n-1}:C_{n-1}\to Z_{n-1}C$; then $\psi d_n=(g{\pi }_{n-1})d_n$ is a coboundary. Thus $\ker {\mathrm{ev}}_n$ is ${\mathrm{Hom}}_R(B_{n-1}C,G)\mathrm{/}\mathrm{im}({\mathrm{Hom}}_R(Z_{n-1}C,G))$.

The free presentation $0\to B_{n-1}C\to Z_{n-1}C\to H_{n-1}C\to 0$ identifies this quotient with ${\mathrm{Ext}}_R^1(H_{n-1}C,G)$ by Ext via a projective resolution of the first variable. Define the extension map on a representative $\psi$ by $[\psi ]\mapsto [\psi d_n]$; step 2.1 proves it is well defined and injective, while step 1.2 identifies its image with the evaluation kernel.

Naturality requires no chosen projection: for a chain map $u:C\to D$, pullback of $\psi d_n^D$ is $(\psi u_{n-1}{\mathrm{\mid }}_{B_{n-1}C})d_n^C$, exactly the map induced on the free presentations; a map $G\to G^{\mathrm{\prime }}$ commutes by postcomposition. Evaluation also commutes with these maps. The chosen projections in steps 1.1 and 2.1 establish existence and exactness only. Thus the displayed sequence is natural.

The exact sequence $0\to Z_{n}C\to C_n\stackrel{d_n}{\to }{B}_{n-1}C\to 0$ comes from The cycle-boundary short exact sequences for a free complex over a PID. Under AC, Boundaries and cycles in a free complex over a PID are free makes $B_{n-1}C$ free and Free modules are projective, with the exact choice boundary makes it projective. Lift its identity through $d_n$ to choose a section $s$. Then $1-s{d}_n$ factors through the inclusion $Z_{n}C\to C_n$ as a projection $\pi :C_n\to Z_{n}C$ restricting to the identity on cycles.

Write $q:Z_{n}C\to H_{n}C$ for the quotient. For $f:H_{n}C\to G$, define $\sigma (f)=[fq\pi ]$. The map $fq\pi :C_n\to G$ is a cocycle: $d_{n+1}$ lands in $B_{n}C\subseteq Z_{n}C$, where $\pi$ is the identity and $q$ is zero. This assignment is $R$-linear in $f$, so it defines a homomorphism into cohomology.

For a cycle $z$, $(fq\pi )(z)=f(qz)$, so ${\mathrm{ev}}_n\sigma (f)=f$. Thus $\sigma$ is a section of the surjection in the exact UCT sequence. Its construction uses the chosen projection; it supplies existence without asserting naturality in $C$.

A divisible abelian group is injective, so ${\mathrm{Ext}}_{\mathbb{Z}}^1(H_{n-1}C,G)=0$.

The UCT evaluation map consequently has zero kernel and is already surjective, and therefore is the stated natural isomorphism.

By Every vector space has a basis, Choice supplies a basis of $V$. It identifies $V$ with a direct sum of copies of $k$, so $V$ is free and therefore projective and flat.

Under Choice a subspace has a vector-space complement. Therefore every map from a subspace into $V$ extends across an inclusion, which is the injectivity criterion.

Fix $n$. A cochain $f:C_n\to k$ is a cocycle exactly when $f(B_n)=0$, because its coboundary is $f\circ d_{n+1}$. Such an $f$ restricts to $Z_n$ and factors through $Z_n\mathrm{/}{B}_n=H_{n}C$, giving the evaluation map [L1]. A coboundary $h\circ d_n$ vanishes on $Z_n$, so evaluation is well defined on cohomology classes.

Given $\lambda \in {\mathrm{Hom}}_k(H_{n}C,k)$, compose it with $Z_n\twoheadrightarrow H_{n}C$ to get a functional on $Z_n$ that vanishes on $B_n$. Extend it to $f:C_n\to k$ by [L2]. Then $f(B_n)=0$, so $f$ is a cocycle, and its evaluation is $\lambda$. Thus evaluation is surjective.

Suppose a cocycle $f$ evaluates to zero. Then $f{\mathrm{\mid }}_{Z_n}=0$. Define $h:B_{n-1}=\mathrm{im}{d}_n\to k$ by $h(d_{n}x)=f(x)$; if $d_{n}x=d_n{x}^{\mathrm{\prime }}$, then $x-x^{\mathrm{\prime }}\in Z_n$, so this is well defined. Extend $h$ to $\tilde{h}:C_{n-1}\to k$ by [L2]. Now $f=\tilde{h}\circ d_n$, so $f$ is a coboundary and its cohomology class is zero. Thus evaluation is injective.

Steps 2.1 and 2.2 show that the evaluation map is a bijective linear map, hence the claimed isomorphism. Each basis extension involved only the two fixed finite-dimensional spaces $C_n$ and $C_{n-1}$; no choice principle or general UCT is used.

The signed tensor differential gives $d(x\otimes y)=dx\otimes y+(-1{)}^{p}x\otimes dy=0$.

Replacing $x$ or $y$ by a boundary changes $x\otimes y$ by a signed boundary; chain maps commute with the formula, proving well-definedness and naturality.

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], licensed by [A1], prove naturality of ${\pi }_n$. Empty sums and zero modules cause no exception.

Use the construction proved in The Kunneth Tor map: put $Z_p=Z_{p}C$ and $A_p=B_{p-1}C$, with zero differentials. The canonical sequence $0\to Z\otimes D\to C\otimes D\to A\otimes D\to 0$ is exact. AC is used here for freeness of the cycle and boundary submodules by Boundaries and cycles in a free complex over a PID are free and their free-module lifts which split the degree sequences, exactly as in that construction.

By The long exact sequence in homology, its relevant segment gives $0\to \mathrm{coker}{\mathrm{\partial }}_{n+1}\to H_n(C\otimes D)\to \ker {\mathrm{\partial }}_n\to 0$. The calculation in the Tor-map lemma identifies $H_n(Z\otimes D)={⨁}_{p+q=n}{Z}_{p}C\otimes H_{q}D$ and identifies ${\mathrm{\partial }}_{n+1}$ with the sum of the inclusion maps $B_{p}C\otimes H_{q}D\to Z_{p}C\otimes H_{q}D$. It also identifies $\ker {\mathrm{\partial }}_n$ and the map onto it with the displayed Tor sum and its established natural quotient.

For each $(p,q)$, right exactness in Tensoring is right exact applied to $B_{p}C\to Z_{p}C\to H_{p}C\to 0$ identifies the cokernel of that inclusion-tensor map with $H_{p}C\otimes H_{q}D$. Cokernels of direct-sum maps are computed coordinatewise, giving $\mathrm{coker}{\mathrm{\partial }}_{n+1}={⨁}_{p+q=n}{H}_{p}C\otimes H_{q}D$. The injection into the middle homology sends $[z]\otimes [y]$ to $[z\otimes y]$, since it is induced by $Z\otimes D\to C\otimes D$. Thus it is precisely the cross product in The Kunneth cross-product map is well defined and natural.

Step 2.1 and step 3.1 prove exactness for the specified maps. Both are natural by the two map lemmas, hence this is a natural short exact sequence. The tensor differential and totalization are those of The tensor product of a right and a left chain complex is totalized by direct sums with the Koszul differential; the construction works for every integer degree with the stated finite diagonals, including zero summands and empty diagonals.

Write the sequence of The Kunneth theorem for free complexes over a PID as $0\to U_n\stackrel{\alpha }{\to }{V}_n\stackrel{\beta }{\to }{W}_n\to 0$, where $U_n={⨁}_{p+q=n}{H}_{p}C\otimes H_{q}D$ and $V_n=H_n(C\otimes D)$. The map $\alpha$ is the cycle cross product.

For $E=C,D$, the exact sequence $0\to Z_{p}E\to E_p\to B_{p-1}E\to 0$ is supplied by The cycle-boundary short exact sequences for a free complex over a PID. Under AC its last term is free by Boundaries and cycles in a free complex over a PID are free, hence projective by Free modules are projective, with the exact choice boundary. Choose sections in every degree and both complexes, using AC for this family as well as the free-module lifts. They give retractions $r_p:E_p\to Z_{p}E$.

Let $H(E)$ denote the zero-differential complex with terms $H_{p}E$, and define ${\pi }_{E,p}(x)=[r_p(x)]$. This is a chain map: $d_{E}x$ is a boundary and a cycle, so $r_{p-1}(d_{E}x)=d_{E}x$ has zero homology class. On cycles, ${\pi }_E$ is the usual homology quotient.

Tensor the two chain maps using The tensor product of a right and a left chain complex is totalized by direct sums with the Koszul differential. The target $H(C)\otimes H(D)$ has zero differential and degree-$n$ term $U_n$, so the induced homology map is $r:V_n\to U_n$. On a cross-product generator $[z]\otimes [w]$, one has $r\alpha ([z]\otimes [w])=[z]\otimes [w]$; therefore $r\alpha =1$.

The restriction $\beta :\ker r\to W_n$ is bijective. Indeed, any lift $v$ of $w$ can be replaced by $v-\alpha r(v)\in \ker r$ without changing its image; and if $v\in \ker r\cap \ker \beta$, exactness gives $v=\alpha (u)$, whence $u=r(v)=0$. Its inverse followed by inclusion is a homomorphism $s:W_n\to V_n$ satisfying $\beta s=1$ and $s\beta =1-\alpha r$. This is the required section. It depends on the selected retractions, and no natural selection is asserted.

Under the stated Choice hypothesis, every $k$-module is flat by Modules over a field are projective, flat, and injective, hence ${\mathrm{Tor}}_1^k(H_{p}C,H_{q}D)=0$ for all $p,q$.

Under the same Choice hypothesis, The Kunneth theorem for free complexes over a PID supplies the natural exact sequence. Its Tor surjection has zero target by step 1.1, while cross product remains injective, giving the stated natural isomorphism.

Under Choice, the free cycle-boundary presentation $0\to B_{q}D\to Z_{q}D\to H_{q}D\to 0$ is a supplied projective resolution by Boundaries and cycles in a free complex over a PID are free. Choice implies dependent choice: for a serial relation, choose one successor for each state using Choice and iterate that successor by recursion on $\mathbb{N}$. Thus the cited flatness/Tor criterion applies to this resolution and gives ${\mathrm{Tor}}_1^R(H_{p}C,H_{q}D)=0$ for every pair $(p,q)$.