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TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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Mayer vietoris sequence in de rham cohomology

Statement

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).

Facts & Assumptions

Given: An open cover M=UV and countable choice.

[F1]

Short exact mayer vietoris sequence of de rham complexes: Under countable choice, 0Ω(M)rΩ(U)Ω(V)sΩ(UV)0 is short exact as a sequence of real cochain complexes.

[F2]

De rham cohomology: The real de Rham cohomology is HdRk(M)=Zk(M)/Bk(M), with Zk,Bk as in def-closed-and-exact-differential-forms. This is def-cohomology-object-of-a-cochain-complex in real vector spaces. Only a closed form ω represents a class [ω]. For closed forms ω,ω, equality [ω]=[ω] means precisely ωω=dη for some (k1)-form η. Addition and real scalar multiplication are induced by those of forms. All groups on the empty manifold are zero.

[F3]

The long exact sequence in cohomology: Let 0ABC0 be a short exact sequence of cochain complexes in an abelian category. Then there is a natural exact sequence Hn(A)Hn(B)Hn(C)nHn+1(A)Hn+1(B)Hn+1(C).

Proof

technique · direct
1.1

The short exact sequence of de Rham cochain complexes satisfies the hypotheses of the cohomology long exact sequence theorem in the abelian category of real vector spaces. It gives the connecting map from overlap degree k to global degree k+1, with no additional differential sign.

F1F3given
2.1

For the middle complex, d(α,β)=(dα,dβ), so its cycle space is Zk(U)Zk(V) and its boundary space is Bk(U)Bk(V). The quotient map sends [(α,β)] to ([α],[β]), bijectively: a pair maps to zero exactly when both entries have primitives. Thus its cohomology is the displayed direct sum. All negative-degree cohomology is zero by the definition of forms, giving the stated initial zero.

F2step 1.1

Source locator

Lee, Theorem 17.20, pp.449–450, and its full proof pp.462–463. This page reverses Lee’s difference convention consistently: s(α,β)=βα.

Depends on

Used by

Dependency tree · two levels

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Sources