LemmaStatement: AI-generatedProof: AI-generated

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.

Transporting an exact sequence through isomorphisms

PDF p. 13: Divisibility of the normalization Picard group

Statement

Suppose $H \overset{a}{\to} A \overset{b}{\to} D \to 0$ is exact, and $u : A \to A^{'}$ and $v : D \to D^{'}$ are isomorphisms of abelian groups. Then $H \overset{u a}{\to} A^{'} \overset{v b u^{- 1}}{\to} D^{'} \to 0$ is exact.

Facts & Assumptions

Given: The homomorphisms and exactness in the statement. Exactness means $b$ is surjective and $\ker b = im a$ .

Goal: Prove exactness after transport.

Proof

1.1given

The transported map $v b u^{- 1}$ is surjective because $u^{- 1}$ and $v$ are surjective and $b$ is surjective.

2.1givenstep 1.1∎

Since $v$ is injective, $a^{'} \in \ker (v b u^{- 1})$ exactly when $u^{- 1} (a^{'}) \in \ker b = im a$ . This is equivalent to $a^{'} \in im (u a)$ , as required.

Used by · 1

Picard group of an affine curve through its boundary