PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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.

First-isomorphism factorization for Lie group homomorphisms

Statement

Assume ${A C}_{ω}$ . Every smooth Lie-group homomorphism $F : G \to H$ factors as

$G \overset{\overset{ˉ}{F}}{\to} im F \overset{j}{\to} H,$

where $\overset{ˉ}{F}$ is a surjective submersion onto the canonical immersed image and $j$ is its injective immersed-subgroup inclusion. Algebraically, the fibres are exactly the left cosets of $\ker F$ .

Facts & Assumptions

Given: ${A C}_{ω}$ and a smooth Lie-group homomorphism $F : G \to H$ .

[F1]

The kernel is a closed embedded normal Lie subgroup. The Axiom of Countable Choice ( ${A C}_{ω}$ ), Kernels are closed embedded normal Lie subgroups.

[F2]

The image has a unique immersed structure for which the corestriction is a surjective submersion. Images are immersed Lie subgroups.

Proof

technique · direct

1.1F2

Let $\overset{ˉ}{F} : G \to im F$ be the corestriction and let $j : im F ↪ H$ be inclusion. Then $F = j \circ \overset{ˉ}{F}$ set-theoretically and as homomorphisms. By [F2], $\overset{ˉ}{F}$ is a surjective submersion and $j$ is an injective immersion with the canonical immersed-subgroup structure.

1.2F1algebra

For $g, g^{'} \in G$ , $F (g) = F (g^{'}) ⟺ F (g^{- 1} g^{'}) = e_{H} ⟺ g^{- 1} g^{'} \in \ker F ⟺ g^{'} \in g \ker F .$ Thus the fibres of both $F$ and $\overset{ˉ}{F}$ are precisely the left kernel cosets; [F1] also makes right cosets equal because the kernel is normal.

2.1F1F2step 1.1step 1.2∎

Steps 1.1 and 1.2 prove the asserted differential-geometric and algebraic factorization. The zero map, trivial kernel, nonclosed image, and disconnected groups are included. No embeddedness of the image is inferred. Countable choice is inherited through [F1]–[F2].

Depends on

Used by

The kernel and image of the determinant homomorphism Example

Dependency tree · two levels

14 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

John M. Lee, Introduction to Smooth Manifolds, 2nd ed. (standard reference, not scraped)
Pavel Etingof, MIT 18.745 Lie Groups and Lie Algebras I (standard reference, not scraped)