Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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.

Degenerate exactness criteria

Statement

In an abelian category:

  1. 0KuA is exact if and only if u is monic.
  2. 0KuAvB is exact if and only if u is a kernel of v.
  3. AvBwC0 is exact if and only if w is a cokernel of v.
  4. 0AuBvC0 is exact if and only if u is monic and v is a cokernel of u, equivalently if and only if u is a kernel of v and v is epic.

Facts & Assumptions

Given: Morphisms in an abelian category as displayed in the statement.

[L1]

Exactness of the displayed sequences is the sequence notion of Exact sequence and short exact sequence in an abelian category.

[L2]

Exactness at a node means image equals kernel, equivalently cokernel equals coimage (Exactness at a node).

[L3]

Monomorphisms are exactly the zero-kernel morphisms, and epimorphisms are exactly the zero-cokernel morphisms (In an abelian category, monic means zero kernel and epic means zero cokernel).

[L4]

Equalizers are monic and coequalizers are epic (Every equalizer is a monomorphism, and every coequalizer is an epimorphism).

[L5]

Every monomorphism is the kernel of its cokernel, and dually every epimorphism is the cokernel of its kernel (Every monomorphism is the kernel of its cokernel, and dually every epimorphism is the cokernel of its kernel).

Proof

technique · direct
1.1

Exactness of 0KuA says that the kernel of u is the zero subobject, and [L3] identifies that with u being monic. This proves claim 1.

L1L2L3
1.2

For claim 3, first assume AvBwC0 is exact. Exactness at C says the cokernel of w is zero, so [L3] makes w epic. Exactness at B gives [coker(v)]=[coim(w)] by [L2]. Because w is epic, [L5] says that w itself represents coim(w). Hence w is a cokernel of v. Conversely, if w is a cokernel of v, then [L4] makes w epic, so exactness at C follows from [L3]. Also w itself represents coim(w), so exactness at B is exactly [L2]. This proves claim 3.

L2L3L4L5
2.1

For claim 2, first assume 0KuAvB is exact. By step 1.1, u is monic. Exactness at A gives [im(u)]=[ker(v)] by [L2], and for a monomorphism the image representative is u itself. Hence u represents the same subobject as a kernel of v, so u is a kernel of v. Conversely, if u is a kernel of v, then u is monic by [L4] and therefore exactness at K follows from step 1.1. Since u itself represents ker(v), exactness at A is exactly [L2]. This proves claim 2.

L2L4step 1.1
2.2

If u is monic and v is a cokernel of u, then step 1.1 gives exactness of 0AuB, and claim 3 gives exactness of AuBvC0. Hence the full sequence is short exact.

step 1.1step 1.2
3.1

If 0AuBvC0 is short exact, then claims 2 and 3 give that u is a kernel of v and v is a cokernel of u.

step 2.1step 1.2
3.2

If u is a kernel of v and v is epic, then claim 2 gives exactness of 0AuBvC, and [L3] turns the epicity of v into exactness at C. Hence the full sequence is short exact.

L3step 2.1
4.1

Steps 3.1, 2.2, and 3.2 prove claim 4.

step 3.1step 2.2step 3.2

Depends on

Used by

Dependency tree · two levels

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Sources