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Inflation-restriction exact sequence in degree one

Statement

Let N⊴G and let A be an abelian G-group. Then the sequence

0→H1(G/N,AN)→Inf⁡H1(G,A)→Res⁡H1(N,A)

is exact.

Facts & Assumptions

Given: A normal subgroup N⊴G and an abelian G-group A.

[L1]

Restriction and inflation in degree one are given by the explicit cocycle formulas of Restriction, inflation, and the quotient conjugation action on first cohomology.

[L2]

The crossed-homomorphism model agrees with the inhomogeneous degree-one model (The inhomogeneous one-cocycle model agrees with crossed homomorphisms in degree one).

Proof

technique · direct
1.1givenL1algebra

Inflation is injective. Suppose zˉ:G/N→AN inflates to a principal crossed homomorphism g↦g⋅a−a on G. For n∈N, inflation gives 0=zˉ(N)=n⋅a−a, so a∈AN. Therefore the same formula already defines a principal cocycle on G/N, and the class of zˉ is zero.

1.2L1L2choosealgebra

Conversely, let z:G→A be a crossed homomorphism whose restriction to N is trivial in H1(N,A). Then there exists a∈A such that z(n)=n⋅a−a for all n∈N. Replace z by the cohomologous cocycle z′(g)=z(g)−(g⋅a−a). Now z′∣N=0.

2.1L1step 1.1algebra

If zˉ:G/N→AN is a cocycle, then its inflation vanishes on N because nN=N. Hence Res⁡∘Inf⁡=0.

2.2step 1.2algebra

If n∈N and g∈G, then z′(ng)=z′(n)+n⋅z′(g)=n⋅z′(g), while also z′(ng)=z′(g(g−1ng))=z′(g)+g⋅z′(g−1ng)=z′(g) because g−1ng∈N and z′∣N=0. Therefore n⋅z′(g)=z′(g) for all n∈N, so z′(g)∈AN. Also z′(gn)=z′(g)+g⋅z′(n)=z′(g), so z′ is constant on cosets of N.

3.1L1step 2.2step 2.1

Define zˉ:G/N→AN by zˉ(gN)=z′(g). Step 2.2 shows this is well defined, and the crossed-homomorphism identity for z′ implies that zˉ is a cocycle on G/N. By construction, inflating zˉ gives z′, so the original class of z lies in the image of inflation. Hence ker⁡Res⁡=im⁡Inf⁡.

4.1step 1.1step 2.1step 3.1∎

Steps 1.1, 2.1, and 3.1 prove exactness of the displayed sequence.

Depends on

Used by

Dependency tree · two levels

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Sources