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PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13
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Relative cup products are natural and connector-compatible

Statement

Relative cup products are natural for maps of excisive triples: if f:XX takes A into A and B into B, then f(uv)=fufvHp+q(X,AB;R). Use the excisive comparisons in the relative-product definition for both triples. In particular these exist when each pair of subspaces is open in its union.

Here are explicit domains for connector compatibility. Write i:AX, U=AB, and let A:Hk(A;R)Hk+1(X,A;R) be the pair connector. Let A,B:Hk(A,AB;R)Hk+1(X,U;R) be the connector obtained by extending to a cochain on X vanishing on B, taking its coboundary, and using the small-chain quotient comparison. Then A,B(xiy)=Axy(xHp(A;R), yHq(X,B;R)), and A,B(ixy)=(1)pxAy(xHp(X,B;R), yHq(A;R)). For B= these are the ordinary pair identities. All coboundaries have the positive sign convention; R is commutative unital. No AC is needed.

Facts & Assumptions

[F1]

Relative cup product for an excisive triad constructs the target through the canonical quotient C(X)/(C(A)+C(B))C(X)/C(AB), whose dual induces an isomorphism.

[F2]

Long exact sequence of a pair in singular cohomology defines A[a]=[δa~] by any extension and proves independence by changing lifts and cocycle representatives.

[F3]

Cup product Leibniz identity supplies the positive-coboundary Leibniz rule.

[F4]

Cup product is natural, unital and associative proves the termwise cochain pullback identity.

Proof

Given: The triples and coefficient ring as stated. Put N=C(A;R)+C(B;R) and Q=HomR(C(X;R)/N,R).

1.1

Pullback of a cochain vanishing on A vanishes on A, and similarly for B. It also maps Q to the primed quotient-functional complex. By [F4] it commutes with the cup formula on cochains. The quotient maps in [F1] commute with f#, since all maps are induced by the same map of chains on X. Their induced cohomology maps commute as well, and their inverses commute because they are isomorphisms. Transporting the cochain equality through these inverses proves relative naturality.

F1F4given
1.2

Restriction gives a termwise exact sequence 0QC(X,B;R)C(A,AB;R)0. To prove surjectivity, extend a cochain from simplices in A by zero off A. It already vanishes on simplices in AB, so the extension vanishes on B. Its kernel consists exactly of cochains vanishing on both A and B, namely Q. Restriction commutes with coboundary; the extension need not. For a relative cocycle z, a lift z~ has δz~Q. Changing lifts changes this by a Q-coboundary; changing z by δw and lifting w leaves the connecting class unchanged. These are exactly the lift calculations of [F2]. Composing this well-defined connecting class with [F1]'s inverse defines the displayed A,B.

F1F2given
2.1

For the first identity represent x by a cocycle α on A and y by a cocycle β on X vanishing on B. Extend α to α~ on X. The cochain α~β vanishes on B and restricts to αiβ, so it is an admissible lift for step 1.2. Its coboundary is δα~β because δβ=0. By [F2], δα~ represents Ax and vanishes on A. Hence the very same Q-cocycle computes both sides after applying [F1]'s comparison.

F1F2F3step 1.2
2.2

For the second identity represent x by a cocycle α on X vanishing on B, and y by a cocycle β on A. Extend β to β~ on X. Now αβ~ is a lift vanishing on B, and its coboundary is (1)pαδβ~, because δα=0. The last factor represents Ay and vanishes on A. Transporting this Q-cocycle gives precisely the asserted sign and factor order.

F1F2F3step 1.2
3.1

If B=, the restriction sequence in step 1.2 is the pair sequence and [F1]'s comparison is the identity. If A=, its right-hand complex is zero and both formulas have zero sides. If A=X, every A and the displayed relative target are zero. For p=0 the second sign is positive, and degree-zero cocycles/lifts require no negative cochain. Zero ring, empty space and zero inputs give zero identities. On point spaces these are the empty/full cases; degenerate simplices are included in each extension rule. All extensions can be the stated zero extensions, so no AC or family of arbitrary lift choices is used.

F1F2step 1.2step 2.1step 2.2

Depends on

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Sources