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The universal coefficient theorem for cohomology over a PID

Statement

Assume the Axiom of Choice. Let $R$ be a PID, $C$ a chain complex of free $R$ -modules, and $G$ an $R$ -module. Then naturally $0 \to {Ext}_{R}^{1} (H_{n - 1} C, G) \to H^{n} {Hom}_{R} (C, G) \overset{{ev}_{n}}{\to} {Hom}_{R} (H_{n} C, G) \to 0$ is exact.

Proof

Given: The Axiom of Choice, the free PID-complex $C$ , an $R$ -module $G$ , the evaluation map, and the cycle-boundary sequences. Choice is used to make arbitrary-rank cycle and boundary modules free and then to split the surjections with free targets (The Axiom of Choice, Boundaries and cycles in a free complex over a PID are free, Free modules are projective, with the exact choice boundary).

1.1givenchoosealgebra

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 $π_{n} = 1 - s_{n} d_{n} : C_{n} \to Z_{n} C$ . For any $h : H_{n} C \to G$ , the cochain $h \circ q_{n} \circ π_{n}$ 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.

1.2givenalgebra

If a cocycle $f : C_{n} \to G$ has zero evaluation, then $f ∣_{Z_{n} C} = 0$ : its restriction factors through $H_{n} C$ , and that induced map is zero. Since $d_{n} : C_{n} ↠ B_{n - 1} C$ has kernel $Z_{n} C$ , there is a unique $ψ : B_{n - 1} C \to G$ with $f = ψ d_{n}$ . Conversely every such $ψ d_{n}$ is a cocycle because $d_{n} d_{n + 1} = 0$ .

2.1step 1.2givenchoosealgebra

The cochain $ψ d_{n}$ is a coboundary exactly when $ψ$ extends to $Z_{n - 1} C$ . Indeed, if $ψ d_{n} = φ d_{n}$ for a cochain $φ : C_{n - 1} \to G$ , surjectivity of $d_{n}$ onto $B_{n - 1} C$ gives $ψ = φ ∣_{B_{n - 1} C}$ . Conversely, if $g : Z_{n - 1} C \to G$ extends $ψ$ , Choice and projectivity of $B_{n - 2} C$ give a projection $π_{n - 1} : C_{n - 1} \to Z_{n - 1} C$ ; then $ψ d_{n} = (g π_{n - 1}) d_{n}$ is a coboundary. Thus $\ker {ev}_{n}$ is ${Hom}_{R} (B_{n - 1} C, G) / im ({Hom}_{R} (Z_{n - 1} C, G))$ .

3.1step 1.2step 2.1algebra

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 ${Ext}_{R}^{1} (H_{n - 1} C, G)$ by Ext via a projective resolution of the first variable. Define the extension map on a representative $ψ$ by $[ψ] \mapsto [ψ d_{n}]$ ; step 2.1 proves it is well defined and injective, while step 1.2 identifies its image with the evaluation kernel.

4.1step 1.1step 1.2step 2.1step 3.1∎

Naturality requires no chosen projection: for a chain map $u : C \to D$ , pullback of $ψ d_{n}^{D}$ is $(ψ u_{n - 1} ∣_{B_{n - 1} C}) d_{n}^{C}$ , exactly the map induced on the free presentations; a map $G \to G^{'}$ 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.

Depends on

Dependency tree · 2 levels

Sources

Weibel, An Introduction to Homological Algebra