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.

Put $F_{\alpha }=⟨e_{\beta }:\beta <\alpha ⟩$ and $N_{\alpha }=N\cap F_{\alpha }$. The image of $N_{\alpha +1}$ in $F_{\alpha +1}\mathrm{/}{F}_{\alpha }\cong R$ is an ideal $I_{\alpha }$ of $R$, hence is either zero or free of rank one. Thus $0\to N_{\alpha }\to N_{\alpha +1}\to I_{\alpha }\to 0$ splits.

At each successor with $I_{\alpha }\mathrm{\ne }0$, choose a generator and a lift $x_{\alpha }\in N_{\alpha +1}$; the splitting gives $N_{\alpha +1}=N_{\alpha }\oplus R{x}_{\alpha }$. At a limit $\lambda$, every element has finite support, so $N_{\lambda }={\bigcup }_{\alpha <\lambda }{N}_{\alpha }$ and the nested union of the earlier bases is a basis. Transfinite induction through the terminal stage $\kappa$ therefore gives a basis of $N_{\kappa }=N$. For $\kappa =0$, this is the empty basis of $N=0$.

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.

Tensoring this free presentation by $G$ identifies ${\mathrm{Tor}}_1^R(H_{n-1}C,G)$ with the kernel of $B_{n-1}C\otimes G\to Z_{n-1}C\otimes G$.

A cycle in $C\otimes G$ maps under $d_n\otimes 1$ into that kernel; changing it by a boundary changes the image by zero, so this gives the asserted natural quotient map.

The cycle and boundary modules are free, hence flat, so tensoring $0\to Z_{n}C\to C_n\to B_{n-1}C\to 0$ remains exact.

The homology sequence of these degreewise exact rows has edge map $[z]\otimes g\mapsto [z\otimes g]$ and quotient map induced by $d\otimes 1$; their kernel and cokernel are respectively $H_n(C)\otimes G$ and ${\mathrm{Tor}}_1^R(H_{n-1}C,G)$.

Because $B_{n-1}C$ is free, the surjection $C_n\twoheadrightarrow B_{n-1}C$ has a chosen section, hence $C_n\cong Z_{n}C\oplus B_{n-1}C$.

After tensoring, this chosen complement identifies a complement to the edge-image in homology and supplies a section of the UCT quotient; changing the section changes that complement, so no naturality is obtained.

Applying ${\mathrm{Hom}}_R(-,G)$ to the free presentation identifies its first cohomology with ${\mathrm{Ext}}_R^1(H_{n-1}C,G)$.

Extending a map on $B_{n-1}C$ along $d_n:C_n\twoheadrightarrow B_{n-1}C$ yields a cochain; changing the lift changes it by a coboundary, which defines the claimed map.

By the boundary-and-cycle lemma under Choice, $B_{n-1}C$ is free and therefore projective. Hence $0\to Z_{n}C\to C_n\stackrel{d_n}{\to }{B}_{n-1}C\to 0$ splits. Every map $H_{n}C\to G$ pulls back to a map $Z_{n}C\to G$ vanishing on $B_{n}C$ and extends across a chosen projection $C_n\to Z_{n}C$ to a cocycle. Thus ${\mathrm{ev}}_n$ is surjective.

A cocycle lies in $\ker {\mathrm{ev}}_n$ exactly when its restriction to $Z_{n}C$ vanishes modulo $B_{n}C$. Subtracting a representative that is zero on $Z_{n}C$ shows that the kernel is ${\mathrm{Hom}}_R(B_{n-1}C,G)/\mathrm{im}({\mathrm{Hom}}_R(Z_{n-1}C,G))\mathrm{.}$ Applying ${\mathrm{Hom}}_R(-,G)$ to 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 ${\mathrm{Ext}}_R^1(H_{n-1}C,G)$. The inclusion is the extension map of the preceding lemma, and all constructions before the optional splitting are natural, proving the exact sequence.

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.

Every $k$-module is injective, so ${\mathrm{Ext}}_k^1(H_{n-1}C,k)=0$.

Thus the UCT evaluation map is an isomorphism to the algebraic dual of $H_{n}C$; finite dimensionality ensures this is the usual finite-dimensional duality convention.

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] prove naturality of ${\pi }_n$. Empty sums and zero modules cause no exception.

Every $C_p$ is free, and every boundary module $d(C_p)=B_{p-1}C$ is free by Boundaries and cycles in a free complex over a PID are free; hence all of these modules are flat.

Weibel's cited Kunneth formula for complexes applies to the right complex $C$ and left complex $D$ under exactly the flatness conditions verified in step 1.1. It gives the displayed natural short exact sequence; its left map is the cross product of The Kunneth cross-product map is well defined and natural, and its right map is the quotient of The Kunneth Tor map. The finite-diagonal hypothesis makes each displayed direct sum finite; an empty diagonal gives the zero module.

The Kunneth theorem for free abelian complexes in the cited source states that the natural short exact sequence is noncanonically split. Its proof chooses lifts in the free cycle-boundary presentations; it does not require the generally false assertion that each boundary subgroup is a direct summand of its chain group.

Choosing those lifts gives a section of the Kunneth quotient, while the source theorem makes no natural choice of them. Thus a splitting exists, but no natural splitting is claimed.

Every $k$-module is flat, hence ${\mathrm{Tor}}_1^k(H_{p}C,H_{q}D)=0$ for all $p,q$.

The Kunneth surjection therefore has zero target, while cross product remains injective, giving the stated natural isomorphism.

Flatness gives ${\mathrm{Tor}}_1^R(H_{p}C,H_{q}D)=0$ for every pair $(p,q)$.

Thus the finite direct sum of correction terms is zero, and exactness makes the cross-product injection surjective as well.

Over the fraction field, Kunneth identifies the rank of $H_n(C\otimes D)$ with ${\sum }_{p+q=n}\mathrm{rank}{H}_{p}C\mathrm{rank}{H}_{q}D$.

Taking the finite alternating sum and regrouping it gives ${\sum }_{p,q}(-1{)}^{p+q}\mathrm{rank}{H}_{p}C\mathrm{rank}{H}_{q}D=\chi (C)\chi (D)$.

The twist sends $x\otimes y$ to $(-1{)}^{pq}y\otimes x$ and intertwines the signed tensor differentials.

Passing to homology gives ${\tau }_{\ast }([x]\times [y])=(-1{)}^{pq}[y]\times [x]$, which is the asserted graded commutativity.

Since $H_1(C)=\mathbb{Z}f$, $H_0(C)=\mathbb{Z}\mathrm{/}2$, and $d\otimes G=0$, the degree-one UCT sequence is $0⟶(\mathbb{Z}\mathrm{/}2)f⟶(\mathbb{Z}\mathrm{/}2)e\oplus (\mathbb{Z}\mathrm{/}2)f\stackrel{q}{\to }\mathbb{Z}\mathrm{/}2⟶0$, with $q(ae+bf)=a$.

The chain automorphism $u_1(e)=e+f$, $u_1(f)=f$, $u_0(g)=g$ induces the identity on both integral homology groups and hence on both outer UCT terms, while its action on the middle term is $(a,b)\mapsto (a,a+b)$.

A section must send $1$ to $(1,t)$ for some $t\in \mathbb{Z}\mathrm{/}2$. Naturality with respect to $u$ would force $(1,t+1)=(1,t)$, which is impossible. Therefore a UCT splitting cannot be chosen naturally for all complexes and coefficient groups.

UCT supplies a natural short exact sequence, while a section can be constructed only after auxiliary choices. The cited counterexample uses a specific free complex, coefficient group $\mathbb{Z}\mathrm{/}2$, and a chain automorphism acting trivially on the two outer UCT terms but nontrivially on every possible section.

By A universal-coefficient splitting cannot in general be chosen naturally, naturality for that automorphism would force $(1,t+1)=(1,t)$ in $(\mathbb{Z}\mathrm{/}2{)}^2$, which is impossible. Therefore no alternative natural choice of section can exist in general, and the asserted natural direct-sum decomposition is false.

Tensoring the cycle-boundary presentation produces its first derived functor ${\mathrm{Tor}}_1$ in homology.

Applying $\mathrm{Hom}(-,G)$ produces its first right derived functor ${\mathrm{Ext}}^1$ in cohomology, so the claimed reversal is false.

Both $C$ and $D$ have $H_0\cong \mathbb{Z}\mathrm{/}2$ and $H_1=0$, while ${\mathrm{Tor}}_1^{\mathbb{Z}}(H_{0}C,H_{0}D)cong{\mathrm{Tor}}_1^{\mathbb{Z}}(\mathbb{Z}\mathrm{/}2,\mathbb{Z}\mathrm{/}2)cong\mathbb{Z}\mathrm{/}2$.

In total degree one the Kunneth correction is this nonzero Tor group, so cross product cannot always be a tensor-product isomorphism.

Their ordinary tensor complex is concentrated in degree zero, so $H_1(C{\otimes }_{\mathbb{Z}}D)=0$. On the other hand, $H_{0}C=H_{0}D=\mathbb{Z}\mathrm{/}2$ and ${\mathrm{Tor}}_1^{\mathbb{Z}}(H_{0}C,H_{0}D)=\mathbb{Z}\mathrm{/}2$.

If the classical free-complex Kunneth short exact sequence were asserted unchanged after simply deleting freeness, then in total degree one it would surject from the zero group $H_1(C\otimes D)$ onto the nonzero Tor group from step 1.1. That is impossible. Hence freeness cannot simply be dropped without replacing ordinary tensor by a derived construction or adding suitable flatness hypotheses.

We have $H_1(C)=\mathbb{Z}f$, $H_0(C)=H_0(D)=\mathbb{Z}\mathrm{/}2$, and $H_1(D)=0$. In total degree one the Kunneth sequence is therefore $0⟶\mathbb{Z}\mathrm{/}2⟶H_1(C\otimes D)⟶\mathbb{Z}\mathrm{/}2⟶0$. Writing $v=e\otimes b-g\otimes a$, a direct kernel/modulo-boundary calculation gives $H_1(C\otimes D)=(\mathbb{Z}\mathrm{/}2)v\oplus (\mathbb{Z}\mathrm{/}2)(f\otimes b)$; the left Kunneth term is generated by $f\otimes b$ and the quotient by the class of $v$.

The chain automorphism $u(e)=e+f$, $u(f)=f$, $u(g)=g$ of $C$ induces the identity on $H_{\ast }(C)$, hence on both outer Kunneth terms, but sends $v$ to $v+f\otimes b$ in the middle term.

Every section of the quotient sends its generator to $v+t(f\otimes b)$ for some $t\in \mathbb{Z}\mathrm{/}2$. Naturality under $u\otimes 1_D$ would require this lift to be fixed, but $u$ changes $t$ to $t+1$. Thus no splitting of all Kunneth sequences can be canonical or natural.