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.
Equivariant maps descend on free proper quotients
Statement
Let and be smooth free proper -manifolds. Every smooth -equivariant map induces a unique smooth map such that
Facts & Assumptions
Given: The two free proper -manifolds, their quotient maps, and a smooth equivariant map .
Equivariance means . Equivariant maps and equivariant vector bundles.
The quotient maps are smooth surjective submersions. Free proper action quotient manifold.
A continuous map constant on quotient fibres factors uniquely and continuously through the quotient. For a quotient map , a map out of is continuous iff its composite with is; a continuous map on constant on the fibres of factors uniquely through ; and a composite of quotient maps is a quotient map.
A smooth submersion has local projection form and smooth local sections. The constant-rank theorem for manifolds.
Proof
Proof technique: quotient universality followed by local submersion sections.
If , then for some . By [F1], , so . Thus the smooth map is constant on the fibres of .
Apply [F3] to obtain a unique continuous map with .
Let . Since is a submersion by [F2], [F4] supplies a smooth local section near . On , the factorization identity gives , which is smooth. Such neighborhoods cover , so is smooth. Uniqueness as a smooth map follows from the uniqueness in [F3]. No choice principle is used.
Depends on
- Equivariant maps and equivariant vector bundles
- Free proper action quotient manifold
- For a quotient map $q : X \to Y$, a map out of $Y$ is continuous iff its composite with $q$ is; a continuous map on $X$ constant on the fibres of $q$ factors uniquely through $q$; and a composite of quotient maps is a quotient map
- The constant-rank theorem for manifolds
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
26 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)