Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21
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  • 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.
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A proper Euclidean local diffeomorphism has finite diffeomorphic sheets near every target point

Statement

Let n≥1, let U,V⊆Rn be nonempty open sets, let V be connected, and let f:U→V be a proper C1 map such that Df(x) is invertible for every x∈U. Then f is surjective, every fibre is finite, and every y∈V has an open neighbourhood whose preimage is a finite disjoint union of open sets, each carried C1-diffeomorphically onto that neighbourhood by f.

Facts & Assumptions

[L1]

A continuous map f:U→V is proper when f−1[K] is compact in U for every compact subset K of V (Proper maps between Euclidean open sets).

[L2]

A C1 map with everywhere-invertible derivative maps every open subset of its domain to an open subset of Rn (A C1 map with everywhere-invertible derivative is open).

[L3]

If K is a compact subset of a metric space X and f:X→Y is continuous into a metric space Y, then f[K] is compact in Y (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset).

[L4]

A closed subset F of a compact metric space X is compact (A closed subset of a compact metric space is compact).

Proof

technique · direct
1.1L1L3L4given

The map f is closed. Indeed, let A⊆U be closed and let y lie in the closure of f[A]. Choose a compact closed target ball K about y contained in V. By [L1], f−1[K] is compact; A∩f−1[K] is compact by [L4], and its image is compact by [L3], hence closed in V. Every sufficiently small neighbourhood of y meets that image, so y∈f[A].

2.1step 1.1L1L2given

By [L2], f[U] is open, and by step 1.1 it is closed. It is nonempty, so connectedness of V gives f[U]=V. For y∈V, [L1] makes f−1(y) compact. Local injectivity makes this fibre discrete, and its cover by neighbourhoods meeting the fibre in one point has a finite subcover; hence the fibre is nonempty and finite.

3.1step 1.1step 2.1givenchoose∎

Write f−1(y)={x1,…,xs}. Choose pairwise disjoint open local-inverse neighbourhoods Oi of the xi, with open images Wi. The closed set U∖⋃iOi has closed image by step 1.1 and that image omits y. Therefore W:=(⋂iWi)∖f[U∖⋃iOi] is an open neighbourhood of y. Its preimage is the disjoint union of Oi∩f−1[W], and each restriction is a C1 diffeomorphism onto W.

Remarks

The neighbourhood property just proved is the one named by Covering maps, evenly covered neighbourhoods, fibres, sheets, and trivial coverings ↗. The simply-connected one-sheet consequence is A connected covering of a locally path-connected simply connected space is one-sheeted and trivial ↗. Neither later result is used above.

Depends on

Used by

Dependency tree · two levels

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Sources