Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
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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.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Covering spaces are stable under restriction, finite products, and pullback

Statement

Restrictions of coverings to open subspaces, finite products of coverings, and pullbacks of coverings are covering maps. The empty product is the identity covering of a one-point space.

Facts & Assumptions

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

[F1]

For a covering p:EB and a continuous map f:XB, define fE:={(x,e)X×E:f(x)=p(e)} with the subspace topology, and let fp:fEX be (x,e)x (def-product-topology, def-subspace-topology-top). This is the pullback covering space; its covering property is proved in prop-covering-spaces-are-stable-under-restriction-finite-products-and-pullback. (The pullback of a covering space along a continuous map).

[F2]

A covering map is a continuous surjection p:EB such that every bB has an open neighbourhood U for which p1(U) is a disjoint union of open sets Vj, called sheets, and each restriction pVj:VjU is a homeomorphism (def-continuous-map-top, def-homeomorphism-and-open-maps, def-disjoint-union-topology). Such a U is evenly covered, and p1(b) is the fibre over b. A covering is trivial when it is isomorphic over B to a product projection B×FB with F discrete. (Covering maps, evenly covered neighbourhoods, fibres, sheets, and trivial coverings).

[F3]

The product set. Let I be a set and let Xi be a set for each iI. The product is iIXi  :=  {x:x is a function with domain I and x(i)Xi for every iI}, and we write xi:=x(i), the i-th coordinate of x. Two elements of the product are equal exactly when they agree at every index, functions being equal when they have the same domain and the same values. For jI the j-th projection is πj:iIXiXj,πj(x):=xj.. The product topology TΠ on iXi is the initial topology of the projections: the topology generated by the subbasis {πi1[U]:iI, UTi}. Finite intersections of subbasic sets form a basis for it, and they are exactly the boxes iIUi with every Ui open in Xi and Ui=Xi for all but finitely many i. (The product set iIXi of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space).

[F4]

Let (X,T) be a topological space (def-topological-space) and let SX. The subspace topology (also relative topology) on S is TS:={US:UT}, the family of traces on S of the open sets of X. The pair (S,TS) is a subspace of X. A subset of S that lies in TS is said to be open in S, and relatively open where the ambient space needs emphasis. (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

Proof

technique · direct
1.1

Restrict an evenly covered neighbourhood for restriction, take products of evenly covered neighbourhoods for finite products, and identify each pullback sheet with the corresponding open subset of the new base.

givenF2F1F4F3
2.1

The preceding construction and implications establish the assertion.

step 1.1

Depends on

Used by

Nothing in the library uses this result yet.

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Sources