Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedverified 2026-09-26 (gpt-6-sol)
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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.

The maps [x]↦[mx] on R/Z are m-sheeted coverings for m≥1

Example

For every integer m≥1, the map Pm:R/Z→R/Z given by Pm([x])=[mx] is a well-defined m-sheeted covering.

Facts & Assumptions

Given: The objects, hypotheses, and choice principles stated above.

[F1]

For the quotient by integer translation, q:R→R/Z is a covering map, and every deck transformation is a unique translation x↦x+n with n∈Z. (The quotient R→R/Z is a covering with integer translations as deck transformations).

[F2]

A covering map is a continuous surjection p:E→B such that every b∈B has an open neighbourhood U for which p−1(U) is a disjoint union of open sets Vj, called sheets, and each restriction p∣Vj:Vj→U is a homeomorphism (def-continuous-map-top, def-homeomorphism-and-open-maps, def-disjoint-union-topology). Such a U is evenly covered, and p−1(b) is the fibre over b. A covering is trivial when it is isomorphic over B to a product projection B×F→B with F discrete. (Covering maps, evenly covered neighbourhoods, fibres, sheets, and trivial coverings).

[F3]

For a covering p:E→B, the cardinality of p−1(b) is locally constant as a function of b∈B. If B is connected, all fibres are equinumerous. (The cardinality of a covering fibre is locally constant and is constant on a connected base).

[F4]

On the set N×N of pairs of natural numbers, define (a,b)∼(c,d)  ⟺  a+d=b+c. This is an equivalence relation (lem-int-equivalence). The integers are the quotient Z:=(N×N)/∼, and we write [(a,b)] for the equivalence class of (a,b). (The integers as equivalence classes of pairs of naturals).

[F5]

Let a,b∈Z with b>0. Then there exist integers q and r with a=qb+r and 0≤r<b, and this pair is unique (Division with remainder in Z: for a∈Z and b>0 there are unique q,r∈Z with a=qb+r and 0≤r<b).

Verification

technique · direct
1.1givenF1

If [x]=[x+n] with n∈Z, then [m(x+n)]=[mx], so Pm is well defined. Since Pm∘q=q∘(x↦mx) and q is a quotient map, Pm is continuous. It is surjective because Pm([y/m])=[y].

2.1F1F2F5step 1.1

Fix [a] and choose I=(a−δ,a+δ) with 0<δ<1/2. The quotient map q is open, since q−1(q(O))=⋃n∈Z(O+n) is open for every open O⊆R. Thus U=q(I) is open, and q∣I is a homeomorphism onto U. For 0≤r<m, put Vr=q((I+r)/m). Each Vr is open, and Pm∣Vr is a homeomorphism onto U, with inverse q(u)↦q((u+r)/m) for u∈I. If points from Vr and Vs coincide, then u−v+r−s is a multiple of m for some u,v∈I. Since ∣u−v∣<1 and r−s is an integer between −(m−1) and m−1, this forces r=s and u=v. Finally, if Pm([x])∈U, then mx=u+n for some u∈I and n∈Z; write n=km+r with 0≤r<m by [F5], giving [x]∈Vr. Hence Pm−1(U) is the disjoint union of exactly m sheets.

3.1step 2.1

At m=1 the map is the identity, so no zero-sheet or division-by-zero case is hidden.

4.1step 3.1∎

The preceding construction and implications establish the assertion.

Depends on

Used by

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Sources