PropositionStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Simplicial homology of a disjoint union is the direct sum

Statement

If $K = ⨆_{α \in A} K_{α}$ is a disjoint union of simplicial complexes, then for every $n$ , $H_{n}^{s i m p} (K) ≅ ⨁_{α \in A} H_{n}^{s i m p} (K_{α}) .$

Facts & Assumptions

Given: A disjoint union $K = ⨆_{α \in A} K_{α}$ .

[L1]

For each $n \geq 0$ , the simplicial chain group $C_{n} (K)$ 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: $H_{n}^{s i m p} (K) = Z_{n} (K) / B_{n} (K) .$ (Simplicial cycles, boundaries, and homology)

Proof

technique · direct

1.1

Every nonempty simplex of $K$ lies in exactly one summand $K_{α}$ , so for each $n \geq 0$ one has $C_{n} (K) ≅ ⨁_{α \in A} C_{n} (K_{α})$ , and under this identification the boundary operator acts componentwise.

L1given

2.1

Therefore for each $n \geq 0$ the cycle groups, boundary groups, and homology groups split componentwise, giving $H_{n}^{s i m p} (K) ≅ ⨁_{α \in A} H_{n}^{s i m p} (K_{α})$ . For $n < 0$ , both sides are zero by definition. This proves the statement for every $n$ .

L2step 1.1∎

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

Allen Hatcher, Algebraic Topology (standard reference, not scraped)
Vidit Nanda, Computational Algebraic Topology, Lecture 03: Homology (standard reference, not scraped)