Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passaudited 2026-09-01
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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A six-term cohomology sequence

Example

Fix a nonzero integer m. Let A∙ be the cochain complex with A1=Z and all other terms zero, let C∙ have C0=Z and all other terms zero, and let B∙ have B0=B1=Z with dB0=×m. Then 0→A∙→B∙→C∙→0 is the short exact sequence whose component A1→B1 and component B0→C0 are the identity and whose other components are zero. Its associated long exact cohomology sequence collapses to 0→H0(A)→H0(B)→H0(C)→∂0H1(A)→H1(B)→H1(C)→0, where ∂0 is multiplication by m.

Facts & Assumptions

Given: A nonzero integer m and the componentwise maps specified in the Example.

[L1]

Short exact sequences of cochain complexes have long exact cohomology sequences (The long exact sequence in cohomology).

Verification

technique · direct
1.1L1givenalgebra

The three complexes have cohomology only in degrees 0 and 1, and the only nontrivial differential is dB0=×m. By [L1], there is a long exact cohomology sequence. The groups immediately before H0(A) and after H1(C) are zero, so this long exact sequence collapses to the six displayed terms.

2.1L1step 1.1algebra∎

Here H0(C)=Z and H1(A)=Z. The connecting map sends 1∈C0 to the class of m∈A1, so ∂0 is multiplication by m. This exhibits the boundary as a degree-raising map.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources