Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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An explicit mayer vietoris connecting form on the circle

Example

Under countable choice the circle Mayer–Vietoris connector has a nonzero representative obtained from a locally constant overlap function.

Facts & Assumptions

Given: Assume countable choice. Let U,V be complements of opposite circle points, let UV=W0W1, and let ω equal 0 on W0 and 1 on W1.

[F1]

Explicit de rham mayer vietoris connecting class: Under countable choice, for a closed k-form ω on UV, δ[ω]=[η], where ηU=d(ρVω) and ηV=d(ρUω), with products smoothly extended by zero as in the lift construction. This class is independent of partition, lift and representative.

[F2]

Zero th de rham cohomology is locally constant functions: HdR0(M) is the algebra of locally constant real functions. For nonempty connected M it is canonically R.

[F3]

Mayer vietoris sequence in de rham cohomology: Under countable choice the de Rham Mayer–Vietoris sequence is exact: Hk(M)rHk(U)Hk(V)sHk(UV)δHk+1(M), beginning with 0H0(M).

Verification

technique · direct
1.1

Take a subordinate partition ρU+ρV=1. The local forms ηU=d(ρVω) and ηV=d(ρUω) glue by the connector formula. Explicitly η=0 on W0 and η=dρU=dρV on W1, with the prescribed smooth zero extensions at the removed points. Thus [η]=δ[(0,1)].

F1given
2.1

The degree-zero groups on U,V are constants, whose difference on the two overlap components is (ba,ba). The vector (0,1) is not diagonal, since equality to (c,c) would require c=0 and c=1. Exactness of the Mayer–Vietoris sequence defining this connector therefore gives δ[(0,1)]0.

F2F3step 1.1

Source locator

Lee, Theorem 17.20, pp.449–450, proof pp.462–463; the sign here is second restriction minus first, and the connecting representative is calculated with that convention.

Depends on

Used by

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Dependency tree · two levels

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Sources