Alphabeta Math
PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Ordinary homology theories have mayer vietoris for cw covers

Statement

Let X=UV be a cover by CW subcomplexes and W=UV. Every ordinary homology theory has a natural exact sequence hn(W)(i,j)hn(U)hn(V)a+bhn(X)Δhn1(W), where all four maps i,j,a,b are inclusions. The same sequence holds for a CW pair (X,C) covered by (U,CU) and (V,CV), with the corresponding relative groups.

Facts & Assumptions

Given: The objects and hypotheses in the statement above.

[F1]

Ordinary unreduced theories on CW pairs and reduced ordinary theories on based CW spaces with vertex basepoints determine one another, naturally and compatibly with morphisms and coefficients. For A the correspondence gives hn(X,A)h~n(X/A); for A= it gives hn(X)h~n(X+), where X+=X{}. Under this correspondence, pair boundaries are cofiber boundaries followed by inverse suspension, and arbitrary disjoint-sum additivity corresponds to arbitrary wedge additivity. For a CW triple BAX there is a natural exact sequence hn(A,B)hn(X,B)hn(X,A)hn1(A,B), whose last map is the pair boundary followed by hn1(A)hn1(A,B). (Unreduced pair and reduced quotient axioms are equivalent on cw pairs)

Proof

1.1

Let qU:hn(U)hn(U,W) and qV:hn(X)hn(X,V) be the pair maps, and let e:hn(U,W)hn(X,V) be the excision isomorphism. F1 supplies these pair sequences and their naturality, including Ve=jU. Define Δ=Ue1qV. The three successive composites in the asserted sequence vanish by these identities and pair exactness.

F1
2.1

If Δx=0, write z=e1qVx. Then Uz=0, so z=qUu for some uhn(U). Thus qV(xau)=0, giving xau=bv for some vhn(V). This proves exactness at hn(X).

F1step 1.1
2.2

If au+bv=0, then eqUu=0, so u=iw. Now b(jw+v)=0, hence jw+v=Vt for some thn+1(X,V). Write t=ez. Then jw+v=jUz. Set w=wUz. Pair exactness gives iw=u and the displayed equation gives jw=v. This proves exactness at the direct sum.

F1step 1.1
2.3

If iw=jw=0, choose zhn+1(U,W) with Uz=w. Then Vez=jw=0, so ez=qVx for some xhn+1(X). Consequently Δx=w, proving exactness at hn(W). All the maps defining Δ are natural, including the inverse of the natural isomorphism e, so this is a natural sequence.

F1step 1.1
3.1

For a subcomplex C, work in the based quotient X+/C+ with its cover by the images of U+ and V+. These are the based quotients by (CU)+ and (CV)+; their intersection is W+/(CW)+. Use the reduced version of the same chase, or subtract the split basepoint sequence from the unreduced one. The quotient identification in F1 converts every term to the asserted relative term. Empty members, empty intersections, and C=X give the corresponding zero terms without changing the chase.

F1step 2.1step 2.2step 2.3

Depends on

Used by

Dependency tree · two levels

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Sources