Alphabeta Math
PropositionStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Simplicial homology of a disjoint union is the direct sum

Statement

If K=⨆α∈AKα is a disjoint union of simplicial complexes, then for every n, Hnsimp(K)≅⨁α∈AHnsimp(Kα).

Facts & Assumptions

Given: A disjoint union K=⨆α∈AKα.

[L1]

For each n≥0, the simplicial chain group Cn(K) is the free abelian group on the oriented nondegenerate n-simplices of K, and the boundary map is defined simplexwise (Simplicial chain groups and the boundary operator).

[L2]

Simplicial homology is the quotient of the cycle group by the boundary group: Hnsimp(K)=Zn(K)/Bn(K). (Simplicial cycles, boundaries, and homology)

Proof

technique · direct
1.1L1given

Every nonempty simplex of K lies in exactly one summand Kα, so for each n≥0 one has Cn(K)≅⨁α∈ACn(Kα), and under this identification the boundary operator acts componentwise.

2.1L2step 1.1∎

Therefore for each n≥0 the cycle groups, boundary groups, and homology groups split componentwise, giving Hnsimp(K)≅⨁α∈AHnsimp(Kα). For n<0, both sides are zero by definition. This proves the statement for every n.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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