Alphabeta Math
LemmaStatement: Literature-sourcedProof: Literature-sourcedPipeline-generatedjudge pass (gpt-6.1-sol)
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.

Long exact sequence of a triple in singular homology

Statement

For spaces B⊆A⊆X and every abelian group G there is a long exact sequence ⋯→Hn(A,B;G)→Hn(X,B;G)→Hn(X,A;G)→δHn−1(A,B;G)→⋯ , where the first two maps are induced by inclusions and δ is the connecting homomorphism of the degreewise short exact sequence 0→C∙(A,B;G)→C∙(X,B;G)→C∙(X,A;G)→0 of relative singular chain complexes. Moreover δ factors as the connecting map Hn(X,A;G)→Hn−1(A;G) of the pair (X,A) followed by the quotient map Hn−1(A;G)→Hn−1(A,B;G).

Facts & Assumptions

Given: Spaces B⊆A⊆X and an abelian group G.

[F1]

The relative singular chain group is Cn(X,A;G)=Cn(X;G)/Cn(A;G), with Cn(A;G)⊆Cn(X;G) induced by inclusion, and both A=∅ and A=X are admitted (Relative singular chain complex).

[F2]

The singular boundary descends to homomorphisms ∂ˉn:Cn(X,A;G)→Cn−1(X,A;G) with ∂ˉn−1∂ˉn=0 (Boundary on relative chains).

[F3]

A short exact sequence of chain complexes is a sequence of chain maps 0→A∙→B∙→C∙→0 that is exact in each degree (Short exact sequence of complexes), a chain complex being a graded family with dn−1dn=0 (Chain complex in an abelian category).

[F4]

A short exact sequence of complexes in an abelian category induces a long exact sequence in homology with connecting maps ∂n (The long exact sequence in homology).

[L1]

Hn(X,A;G) is by definition the homology of the complex C∙(X,A;G) (Relative singular homology).

[L2]

The connector of the pair sequence is fixed on cycles by δ[c]:=[∂c] for a relative cycle c with ∂c∈Cn−1(A;G); the quotient map is the third arrow of the chain sequence (Relative connecting homomorphism on cycles).

Proof

technique · degreewise-quotient
1.1F1givenalgebra

In each degree the sequence 0→Cn(A,B;G)→Cn(X,B;G)→Cn(X,A;G)→0 is the sequence of quotients Cn(A)/Cn(B)→Cn(X)/Cn(B)→Cn(X)/Cn(A) induced by Cn(B)⊆Cn(A)⊆Cn(X) by [F1]; it is exact because the first map is injective (the inclusion Cn(A)→Cn(X) descends injectively after dividing by the common subgroup Cn(B)), the second is surjective, and its kernel is exactly Cn(A)/Cn(B).

2.1F2F3step 1.1

By [F2] all three boundary maps descend to the quotients, so the degreewise maps of step 1.1 commute with the boundaries and form chain maps; hence they constitute a short exact sequence of complexes in the sense of [F3].

3.1F4L1step 2.1

Applying [F4] to the short exact sequence of step 2.1 gives the long exact sequence of the statement; the homology groups are Hn(X,A;G), Hn(X,B;G), Hn(A,B;G) by [L1], and the first two maps are induced by inclusions because the chain maps of step 1.1 are.

4.1L2step 3.1algebra∎

For the factorization, let c be a relative cycle for (X,A) with ∂c∈Cn−1(A;G), representing a class in Hn(X,A;G). The connecting map of the triple sequence sends its class to the class of ∂c in Hn−1(A,B;G): this is the same cycle formula as in [L2] read in the middle complex C∙(X,B;G), where the role of the subspace is played by A modulo B. The pair connector of (X,A) sends the same class to [∂c]∈Hn−1(A;G) by [L2], and the quotient map Hn−1(A;G)→Hn−1(A,B;G) is induced by the quotient chain map; composing gives the triple connector. Hence δ factors as the pair connector followed by the quotient map.

Remarks

  • The case B=∅. Then Hn(X,B;G)=Hn(X;G) and Hn(A,B;G)=Hn(A;G) are the absolute groups and the sequence is the ordinary long exact sequence of the pair; the factorization statement is vacuous, the quotient map being an isomorphism.
  • The case A=B. Then the middle complex is C∙(X,B;G) and the sequence reads ⋯→Hn(B,B;G)→Hn(X,B;G)→ ≅ Hn(X,B;G)→Hn−1(B,B;G)→⋯, consistent with the vanishing of the two outer groups.
  • This is the exact sequence used to compare successive sublevel manifolds and to compute the connecting map of a handle stage.

Depends on

Used by

Dependency tree · two levels

19 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