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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→Ext⁡R1(Hn−1C,G)→HnHom⁡R(C,G)→ev⁡nHom⁡R(HnC,G)→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, Bn−1C is free and projective. Choose a section sn:Bn−1C→Cn of the corestricted differential and put πn=1−sndn:Cn→ZnC. For any h:HnC→G, the cochain h∘qn∘πn restricts on ZnC to h∘qn, vanishes on BnC, and hence is a cocycle. Its evaluation is h, proving surjectivity.

1.2givenalgebra

If a cocycle f:Cn→G has zero evaluation, then f∣ZnC=0: its restriction factors through HnC, and that induced map is zero. Since dn:Cn↠Bn−1C has kernel ZnC, there is a unique ψ:Bn−1C→G with f=ψdn. Conversely every such ψdn is a cocycle because dndn+1=0.

2.1step 1.2givenchoosealgebra

The cochain ψdn is a coboundary exactly when ψ extends to Zn−1C. Indeed, if ψdn=φdn for a cochain φ:Cn−1→G, surjectivity of dn onto Bn−1C gives ψ=φ∣Bn−1C. Conversely, if g:Zn−1C→G extends ψ, Choice and projectivity of Bn−2C give a projection πn−1:Cn−1→Zn−1C; then ψdn=(gπn−1)dn is a coboundary. Thus ker⁡ev⁡n is Hom⁡R(Bn−1C,G)/im⁡(Hom⁡R(Zn−1C,G)).

3.1step 1.2step 2.1algebra

The free presentation 0→Bn−1C→Zn−1C→Hn−1C→0 identifies this quotient with Ext⁡R1(Hn−1C,G) by Ext via a projective resolution of the first variable. Define the extension map on a representative ψ by [ψ]↦[ψdn]; 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→D, pullback of ψdnD is (ψ un−1∣Bn−1C)dnC, exactly the map induced on the free presentations; a map G→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

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Dependency tree · two levels

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Sources