Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05 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.

Extension by zero can be strictly smaller than direct image on a punctured interval

Statement refuted

For every open immersion j:UX and every sheaf of abelian groups F on U, one has j!F=jF.

Facts & Assumptions

Given: The open immersion j:U=(1,0)(0,1)X=(1,1) and the constant sheaf Z on U.

[F1]

For an open immersion, direct image is computed by intersection: (jZ)(V)=Z(VU) (Direct image along an open immersion is restriction-compatible intersection).

[F2]

Extension by zero consists of sections whose support is closed in the test open (Extension by zero for abelian sheaves on an open subspace).

Counterexample

technique · direct
1.1

By [F1], the global section set of jZ is (jZ)(X)=Z(U)Z×Z, because U has two connected components. Let s be the section that is 1 on both components.

F1givenchoose
2.1

The germ of s is nonzero at every point of U, so its support is all of U. But U=(1,0)(0,1) is not closed in X=(1,1), since its closure contains 0. Therefore [F2] shows that s(j!Z)(X).

F2step 1.1
3.1

Thus s lies in jZ but not in j!Z, so the two sheaves are not equal. This refutes the statement.

step 1.1step 2.1

Depends on

Used by

Dependency tree · two levels

5 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