Universal Coefficients and Kunneth Theorems — Examples

Its degree-one homology is the kernel and its degree-zero homology the cokernel of multiplication by $m$.

These are respectively the $m$-torsion subgroup and quotient displayed in the statement. The calculation follows directly from the two-term tensor complex.

Multiplication by nonzero $m$ on $\mathbb{Z}$ has kernel $0$ and cokernel $\mathbb{Z}\mathrm{/}m$.

Thus the cochain calculation has the displayed cohomology; it agrees with ${\mathrm{Ext}}_{\mathbb{Z}}^1(\mathbb{Z}\mathrm{/}m,\mathbb{Z})\cong \mathbb{Z}\mathrm{/}m$.

Tensoring produces $0\to \mathbb{Z}\mathrm{/}2\stackrel{0}{\to }\mathbb{Z}\mathrm{/}2\to 0$, so $H_1(C\otimes \mathbb{Z}\mathrm{/}2)\cong \mathbb{Z}\mathrm{/}2$.

The displayed integral complex is itself a length-one free resolution of $H_{0}C=\mathbb{Z}\mathrm{/}2$. Tensoring this resolution with $\mathbb{Z}\mathrm{/}2$ makes its differential zero, so the definition of Tor (equivalently, Tor one of two cyclic abelian groups is cyclic of gcd order) gives ${\mathrm{Tor}}_1^{\mathbb{Z}}(H_{0}C,\mathbb{Z}\mathrm{/}2)=\mathbb{Z}\mathrm{/}2$. Since $H_{1}C\otimes \mathbb{Z}\mathrm{/}2=0$, the degree-one correction is exactly the nonzero group computed in step 1.1. This finite free-resolution calculation needs no choice; under AC it is also the correction term in The universal coefficient theorem for homology over a PID.

The tensor total complex has groups $\mathbb{Z}$ in degree $2$, ${\mathbb{Z}}^2$ in degree $1$, and $\mathbb{Z}$ in degree $0$. In the basis $(e_1\otimes f_0,e_0\otimes f_1)$, its differentials are $d_2(k)=(-2k,2k)$ and $d_1(a,b)=2a+2b$ by the tensor differential sign rule (The tensor product of a right and a left chain complex is totalized by direct sums with the Koszul differential). Hence $\ker d_1=\{(k,-k):k\in \mathbb{Z}\}$ and $\mathrm{im}{d}_2=\{(2k,-2k):k\in \mathbb{Z}\}$, so $H_1(C\otimes D)\cong \mathbb{Z}\mathrm{/}2$.

Tensoring the specified free resolution $Q$ with $\mathbb{Z}\mathrm{/}2$ makes its differential multiplication by $2$ on $\mathbb{Z}\mathrm{/}2$, hence zero. By the fixed-resolution definition (Tor from a projective resolution of the right module), ${\mathrm{Tor}}_1^{\mathbb{Z},Q}(\mathbb{Z}\mathrm{/}2,\mathbb{Z}\mathrm{/}2)=\ker (2:\mathbb{Z}\mathrm{/}2\to \mathbb{Z}\mathrm{/}2)\cong \mathbb{Z}\mathrm{/}2$. The map $[(k,-k)]\mapsto [k{]}_2$ identifies this group with the degree-one homology from step 1.1. This is the claimed nonzero term for these fixed complexes; neither calculation selects or compares resolutions.

The zero differential makes $H_{0}C\cong H_{1}C\cong k$ and $H_{0}D\cong k$.

The tensor complex has $k$ in total degrees zero and one with zero differential, matching the two cross-product summands; no Tor term occurs over $k$.

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

The maps $u_1(e)=e+f$, $u_1(f)=f$, and $u_0(g)=g$ define a chain automorphism inducing the identity on $H_1(C)$ and $H_0(C)$, but $u_1\otimes 1$ sends $(a,b)$ to $(a,a+b)$ on the middle UCT term.

Any section has $s(1)=(1,t)$. Naturality with respect to $u$ would require $(u_1\otimes 1)s(1)=s(1)$ because $u$ acts identically on the two outer terms, but $(1,t+1)\mathrm{\ne }(1,t)$. Hence no section for this complex and coefficient group is invariant under all chain automorphisms.

Their Euler characteristics are $3-2=1$ and $4-1=3$.

Expanding the finite total complex gives the product of the two alternating rank sums, hence $\chi (C\otimes D)=3=1\cdot 3$.