CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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De rham cohomology of a finite disjoint union is the direct sum

Statement

For a finite disjoint union $M = ∐_{j = 1}^{m} M_{j}$ , restrictions give $H_{d R}^{k} (M) ≅ ⨁_{j = 1}^{m} H_{d R}^{k} (M_{j})$ .

Facts & Assumptions

Given: A finite family of smooth manifolds and any integer $k$ .

[F1]

De rham cohomology: The real de Rham cohomology is $H_{d R}^{k} (M) = Z^{k} (M) / B^{k} (M)$ , with $Z^{k}, B^{k}$ 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 $ω' - ω = d η$ for some $(k - 1)$ -form $η$ . Addition and real scalar multiplication are induced by those of forms. All groups on the empty manifold are zero.

Proof

technique · direct

1.1

A form on $M$ is uniquely a tuple of forms on the open components $M_{j}$ : define its value componentwise, which is smooth locally. Its derivative is componentwise too, so $Z^{k} (M) = ⨁_{j} Z^{k} (M_{j})$ . A tuple of exact forms has a tuple of primitives, obtained by finite choice, so $B^{k} (M) = ⨁_{j} B^{k} (M_{j})$ .

F1given

2.1

The resulting map on quotient classes is onto, since a finite tuple of classes has a finite tuple of closed representatives. Its kernel consists precisely of tuples with all entries exact, which step 1.1 identifies with $B^{k} (M)$ . Thus it is an isomorphism. For $m = 0$ both sides are zero, and for $m = 1$ it is the identity.

F1step 1.1∎

Source locator

Lee, Proposition 17.5, pp.442–443; the local statement is finite only, where products and sums coincide and all witness selection is finite.

Depends on

De rham cohomology

Used by

De rham cohomology of punctured euclidean space Corollary
De rham cohomology of a finite discrete manifold Example
De rham cohomology of spheres Theorem

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John M. Lee, Introduction to Smooth Manifolds, second edition (standard reference, not scraped)