Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21
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.

A proper Euclidean local diffeomorphism over a connected target has constant finite fibre cardinality

Statement

Under the hypotheses of A proper Euclidean local diffeomorphism has finite diffeomorphic sheets near every target point, there is a positive natural number d such that every fibre of f is equinumerous with a set of size d (Equinumerous sets, AB and AB).

Facts & Assumptions

Given: A proper regular C1 map f:UV with nonempty source and connected target V (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).

[L1]

Every yV has an open neighbourhood whose preimage is a finite disjoint union of open sets, each carried C1-diffeomorphically onto that neighbourhood by f (A proper Euclidean local diffeomorphism has finite diffeomorphic sheets near every target point).

Proof

technique · direct
1.1

Over a neighbourhood supplied by [L1], every fibre meets each sheet exactly once. Sending a fibre point to its unique sheet gives a bijection from every fibre there to the same nonempty finite sheet index set. Thus fibre cardinality is locally constant.

L1
2.1

For each positive natural d, let Vd be the set of target points with fibre cardinality d. Step 1.1 makes every Vd open, and the Vd form a disjoint cover of V. If two were nonempty, one and the union of all the others would disconnect V. Hence exactly one Vd is nonempty, and it is all of V.

step 1.1given

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

22 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