LemmaStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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.

Surjective module maps remain surjective after localisation

Statement

Let $f : M \to M^{''}$ be a surjective $R$ -module homomorphism. Then the induced map

S^{- 1} f : S^{- 1} M ⟶ S^{- 1} M^{''}, (m / s) ⟼ f (m) / s,

is surjective.

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , left $R$ -modules $M, M^{''}$ , and a surjective $R$ -module homomorphism $f : M \to M^{''}$ .

[L1]

A module homomorphism preserves scalar multiplication (Module homomorphism and isomorphism, kernel, image and cokernel).

[L2]

Elements of $S^{- 1} M^{''}$ are fractions $m^{''} / s$ with $m^{''} \in M^{''}$ and $s \in S$ (Localisation of a module at a multiplicative subset).

Proof

technique · direct

1.1

Let $m^{''} / s \in S^{- 1} M^{''}$ . Since $f$ is surjective, choose $m \in M$ with $f (m) = m^{''}$ .

givenL2choose

2.1

Then $(S^{- 1} f) (m / s) = f (m) / s = m^{''} / s$ , so $m^{''} / s$ lies in the image of $S^{- 1} f$ .

step 1.1L1

3.1

Since every element of $S^{- 1} M^{''}$ is hit, $S^{- 1} f$ is surjective.

step 2.1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., Theorem 12.20 (standard reference, not scraped)