Alphabeta Math
False statementConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedprecheck passaudited 2026-09-06 rests on unproved material (inherited)
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.

Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Universal coefficients split naturally

Statement

For groups G and trivial G-modules A, the degree-two universal-coefficient short exact sequence splits naturally in G and A.

Facts & Assumptions

Given: Fix A=C2 with trivial action and V=C2×C2, written as F22. Write pG:H2(G;A)Hom(M(G),A) for the universal-coefficient map.

[L1]

The degree-two universal-coefficient sequence is natural and admits a splitting after choices (Universal coefficients in degree two).

[L2]

A cyclic group has zero Schur multiplier (Multiplier of a cyclic group).

[L3]

For an abelian group B, M(B)2B (Multiplier of an abelian group), and alternating bilinear maps factor through its exterior square (Alternating universal property).

[L4]

Classes in H2(G;A) naturally classify central extensions of G by A when the action is trivial (H^2 classifies extensions with fixed abelian kernel action). Under this classification, restriction to a subgroup pulls back the extension, and zero represents a split extension: restricting a factor set gives the pullback factor set, and a homomorphic section has zero factor set.

Refutation

technique · contradiction
1.1

Suppose there are splitting homomorphisms sG:Hom(M(G),A)H2(G;A) natural in G, with pGsG=id. For every subgroup inclusion i:LV with LC2, [L2] gives M(L)=0, so naturality forces isV(λ)=sL(λM(i))=sL(0)=0 for every λ:M(V)A.

L1L2assume-contra
2.1

The map b((x1,x2),(y1,y2))=x1y2+x2y1 is alternating and bilinear to A. It takes value 1 on the standard basis pair, so [L3] supplies a nonzero homomorphism λ:M(V)A. Put α=sV(λ). Then pV(α)=λ0.

L3step 1.1constructalgebra
3.1

Represent α by a central extension 1AEπV1. Its restriction to each of the three order-two subgroups LV is zero by step 1.1. Hence each preimage π1(L) is a split central extension, isomorphic to C2×C2, and every element of that preimage has square 1.

L4step 1.1step 2.1
4.1

Every element of E either belongs to the kernel A, or maps to a nonzero vector of V and therefore belongs to one of these three preimages. Thus every element of E has square 1. For x,yE, this gives xy=(xy)1=y1x1=yx, so E is abelian. Choose lifts u,v of the two standard basis vectors of V. Since u2=v2=1 and uv=vu, the map (a,b)uavb is a homomorphic section of π.

step 3.1choosealgebra
5.1

This section makes α=0, contradicting pV(α)=λ0. Therefore no splitting can be natural in the group variable even for the fixed coefficient group C2, and in particular none is natural in both variables. Individual sequences still split after choices by [L1].

L4L1step 2.1step 4.1discharge-contradiction

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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