Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29
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.

Exactness is detected by members

Statement

For a composable pair A→fB→gC in an abelian category, the following are equivalent:

  1. the pair is exact at B;
  2. gf=0, and for every member y:Y→B with gy≡0 there exists a member x:X→A with fx≡y.

Facts & Assumptions

Given: The composable pair A→fB→gC.

[L1]

Exactness at B is the equality [im⁡(f)]=[ker⁡(g)] (Exactness at a node).

[L2]

The subobject inequalities underlying exactness are exactly the two factorization statements for morphisms killed by g (The subobject inequalities underlying exactness).

[L3]

Members modulo equivalence correspond to subobjects (Members modulo equivalence correspond to subobjects).

Proof

technique · direct
1.1L1L2L5assume-hypchoose

Assume the pair is exact at B, and let y:Y→B satisfy gy≡0. Choose an epic u:W→Y with gyu=0. If A→eI→mB is an image factorization of f, then exactness and [L2] give a map w:W→I with mw=yu.

1.2L2L3L5assume-hyp

Assume condition 2. For any u:U→B with gu=0, we have gu≡0, so condition 2 gives a member x:X→A with fx≡u. By [L3], the members fx and u determine the same subobject of B, and since fx factors through f, that subobject lies below the image of f. Thus every morphism killed by g factors through the image of f. Together with gf=0, [L2] yields exactness at B.

2.1L4L5step 1.1constructalgebra

Pull back the epic e:A→I along w, obtaining x:P→A and an epic α:P→W by [L4]. Then fx=mex=mwα=yuα, so fx≡y.

3.1step 2.1step 1.2∎

Thus conditions 1 and 2 are equivalent.

Depends on

Used by

Dependency tree · two levels

24 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