TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28‡ 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.

Second isomorphism theorem in an abelian category

Statement

Let $B$ and $C$ be subobjects of an object $A$ in an abelian category. Then there is a canonical isomorphism

$(B \lor C) / C ≅ B / (B \land C) .$

Facts & Assumptions

Given: Subobjects $b : B ↣ A$ and $c : C ↣ A$ .

[L1]

The join $B \lor C$ is the image of the induced map $[b, c] : B \oplus C \to A$ (The join of two subobjects in an abelian category).

[L2]

The meet $B \land C$ is represented by the pullback of $b$ and $c$ (The meet of two subobjects is their pullback).

[L3]

A morphism modulo its kernel is canonically isomorphic to its image (First isomorphism theorem in an abelian category).

Proof

technique · direct

1.1

Let $q : A \to A / C$ be the quotient map. Consider the composite $q b : B \to A / C$ . By [L2], a morphism into $B$ is killed by $q b$ exactly when its composite into $A$ factors through $C$ , which is exactly the pullback condition defining $B \land C$ . So $\ker (q b) = B \land C$ .

L2construct

2.1

By [L3], step 1.1 gives a canonical isomorphism $B / (B \land C) ≅ im (q b) .$ The map $q$ kills $C$ , so its restriction to the join $B \lor C$ factors through the quotient $(B \lor C) / C$ . Conversely, every summand used in the defining map $[b, c]$ lands in $im (q b)$ after composing with $q$ , because the $C$ -summand dies. Hence $im (q b)$ is exactly the image of $B \lor C$ in $A / C$ , namely $(B \lor C) / C$ .

L1L3step 1.1

3.1

Combining steps 1.1 and 2.1 yields the canonical isomorphism $(B \lor C) / C ≅ B / (B \land C)$ .

step 1.1step 2.1∎

Depends on

Used by

Length is additive along a subobject Theorem
Zassenhaus butterfly lemma in an abelian category Theorem

Dependency tree · two levels

13 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

Saunders Mac Lane, Categories for the Working Mathematician, Section VIII.3 (standard reference, not scraped)