Alphabeta Math
Session-authored (Fable 5 assisted)
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.

8 results · all verified · 3 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 5 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Covering Spaces and Lifting — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-16Open item page →

Trivial coverings are products with a discrete fibre

Example

If X is any space and F is a nonempty discrete space, the projection X×FX is a trivial covering with fibre F. If X=, the same holds for F=; for nonempty X, an empty fibre would violate surjectivity.

Facts & Assumptions

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

[F1]

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).

[F2]

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).

[F3]

Throughout, a topology is as in def-topological-space, and finite, at most countable and uncountable are as in def-countable, so that "countable" always means "at most countable" and every finite set is countable. Let X be a set. The six families below are topologies on X; that each really satisfies (T1), (T2) and (T3) is discharged in full after the list. Among those six is the discrete topology Tdisc:=P(X), in which every subset is open and hence every subset is also closed. (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

Verification

technique · direct
1.1

For any space X and discrete set F, verify that the projection X×FX is a covering, with every open subset of the base evenly covered.

givenF1F2F3
2.1

Identify its sheets and fibre, including F=: the projection then fails surjectivity unless X=, so state the nonempty-fibre convention explicitly.

step 1.1F1F2
3.1

The preceding construction and implications establish the assertion.

step 2.1
CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-16Open item page →

A surjective local homeomorphism need not be a covering map

Statement refuted

Let E=(0,2)(1,2) and map both summands by inclusion to B=(0,2). This map is a surjective local homeomorphism but is not a covering map.

Facts & Assumptions

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

[F1]

Every covering map is a surjective local homeomorphism, and each of its fibres is discrete in the subspace topology. (Covering maps are surjective local homeomorphisms with discrete fibres).

[F2]

For a covering p:EB, the cardinality of p1(b) is locally constant as a function of bB. If B is connected, all fibres are equinumerous. (The cardinality of a covering fibre is locally constant and is constant on a connected base).

[F3]

The underlying set. Let I be a set and let Xi be a set for each iI. The disjoint union is iIXi  :=  iI(Xi×{i}), whose elements are the pairs (x,i) with iI and xXi. For jI the j-th canonical injection is κj:XjiIXi,κj(x):=(x,j). The construction is what makes the word "disjoint" honest. Each κj is injective (def-injection-surjection-bijection), since (x,j)=(x,j) forces x=x; the images κj[Xj]=Xj×{j} are pairwise disjoint, since the second coordinate determines j; and their union is the whole set. So no assumption that the Xi are disjoint as sets is needed, and none is made: the tag i separates the copies even when Xi=Xi for ii. (The disjoint union (coproduct) iXi with the final topology of the canonical injections: a set is open exactly when each of its traces is).

[F4]

Throughout, R is the complete ordered field (def-complete-ordered-field, def-ordered-field) with its order (def-real-order). A subset IR is order-convex when x,yI and xzy imply zI, and the intervals of R are the nine listed forms, among them [a,b]={x:axb} and (a,b)={x:a<x<b}. (Intervals of R: the nine order-convex forms, nondegeneracy, and length).

Counterexample

technique · direct
1.1

Let the domain be the disjoint union of (0,2) and (1,2) and map both components by inclusion onto the base (0,2).

givenF3F4
2.1

The first component makes the map surjective and each inclusion is a local homeomorphism.

step 1.1F1F3
3.1

The fibre has one point at 1 and two immediately to its right, so local constancy of sheet number rules out a covering; equivalently, the second component supplies only a one-sided partial sheet above every neighbourhood of 1.

step 2.1F1F2F3
4.1

The preceding construction and implications establish the assertion.

step 3.1
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-16Open item page →

A covering over a disconnected base can have different sheet numbers on different components

Statement refuted

There is a covering of a two-point discrete space whose fibre over one point has one element and whose fibre over the other has two elements. Thus connectedness is necessary for global constancy of sheet number.

Facts & Assumptions

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

[F1]

For a covering p:EB, the cardinality of p1(b) is locally constant as a function of bB. If B is connected, all fibres are equinumerous. (The cardinality of a covering fibre is locally constant and is constant on a connected base).

[F2]

Throughout, a topology is as in def-topological-space, and finite, at most countable and uncountable are as in def-countable, so that "countable" always means "at most countable" and every finite set is countable. Let X be a set. The six families below are topologies on X; that each really satisfies (T1), (T2) and (T3) is discharged in full after the list. Among those six is the discrete topology Tdisc:=P(X), in which every subset is open and hence every subset is also closed. (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

Counterexample

technique · direct
1.1

Map a three-point discrete space onto a two-point discrete space with one point over the first basepoint and two over the second.

givenF2
2.1

Each singleton base neighbourhood is evenly covered, while the fibre cardinal is not globally constant.

step 1.1F1F2
3.1

This isolates exactly why connectedness appears in the sheet-number theorem.

step 2.1F1
4.1

The preceding construction and implications establish the assertion.

step 3.1
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-16Open item page →

The quotient RR/Z is a covering with integer translations as deck transformations

Example

For the quotient by integer translation, q:RR/Z is a covering map, and every deck transformation is a unique translation xx+n with nZ.

Facts & Assumptions

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

[F1]

For a covering-space action of G on E, the orbit map EE/G is a covering. If E is path-connected, the deck group of this covering consists exactly of the transformations supplied by G. (The orbit map of a covering-space action is a covering, with the acting group equal to the deck group when the total space is path-connected).

[F2]

For a covering p:EB, a deck transformation is an isomorphism h:EE over B, so ph=p (def-map-and-isomorphism-of-covering-spaces). Deck transformations form the deck group Deck(p) under composition, and this group acts on E by evaluation (def-group, def-group-action). (Deck transformations and the deck-transformation group of a covering).

[F3]

The quotient topology. Let (X,T) be a topological space (def-topological-space), let Y be a set and let q:XY be a surjection (def-injection-surjection-bijection). The quotient topology on Y induced by q is the final topology of the one-element family (q) (def-initial-and-final-topology): Tq  :=  {VY:q1[V]T}. That this is a topology is discharged in def-initial-and-final-topology, where every final topology is verified to satisfy (T1), (T2) and (T3). Dually, CY is closed in Tq exactly when q1[C] is closed in X, because q1[YV]=Xq1[V]. (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection).

[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]

Identify Z with its canonical copy inside R along the embeddings NZQR; then for every real x there is exactly one integer m with mx<m+1, written x (Integer part: for every real x there is exactly one integer m with mx<m+1).

Verification

technique · direct
1.1

By [F5] the integers sit inside R as a subgroup under the ordered-field operations, so xy iff xyZ is an equivalence relation on R; [F4] supplies only the abstract construction of Z and not its copy in R, so the embedding of [F5] is what makes the relation and the translations below meaningful. Take the quotient topology of [F3] on R/Z.

givenF3F4F5
2.1

For an open interval I of length below 1 the translates I+k, kZ, are pairwise disjoint, since two points of I differ by less than 1 while distinct integer translates differ by at least 1 in the order of [F5]; the preimage q1(q[I])=kZ(I+k) is then a union of open sets, so q[I] is open and evenly covered. Openness of I is required and not cosmetic: for I=[0,12] the preimage kZ[k,k+12] is not open, so q[I] is not even a neighbourhood.

step 1.1F3F4F5
3.1

Verify directly that translations are deck transformations and that every deck transformation is the unique integer translation determined by the image of zero, without computing the fundamental group of the quotient.

step 2.1F2F3F1
4.1

The preceding construction and implications establish the assertion.

step 3.1
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16Open item page →

The projected unit interval is not nullhomotopic in R/Z

Example

The loop t[t] in R/Z is not nullhomotopic.

Facts & Assumptions

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

[F1]

For the quotient by integer translation, q:RR/Z is a covering map, and every deck transformation is a unique translation xx+n with nZ. (The quotient RR/Z is a covering with integer translations as deck transformations).

[F2]

Let p:EB be a covering, H:Y×IB a homotopy, and H~0:YE a lift of H(,0). There is a unique lift H~:Y×IE of H extending H~0. (Existence and uniqueness of homotopy lifts through a covering map).

[F3]

Endpoint-fixed homotopic paths in the base have lifts with the same endpoint whenever their lifts begin at the same point. (The endpoint of a lifted path depends only on its endpoint-fixed homotopy class).

Verification

technique · direct
1.1

The loop t[t] lifts from zero to the path tt, whose endpoint is one.

givenF3F1
2.1

If it had an endpoint-fixed nullhomotopy, homotopy lifting would keep the terminal lift in the discrete fibre while deforming the initial lift to the constant path at zero, forcing endpoints one and zero to agree.

step 1.1F3F2
3.1

This proves essentiality without asserting π1(R/Z)Z.

step 2.1F1
4.1

The preceding construction and implications establish the assertion.

step 3.1
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16Open item page →

The maps [x][mx] on R/Z are m-sheeted coverings for m1

Example

For every integer m1, the map Pm:R/ZR/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:RR/Z is a covering map, and every deck transformation is a unique translation xx+n with nZ. (The quotient RR/Z is a covering with integer translations as deck transformations).

[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]

For a covering p:EB, the cardinality of p1(b) is locally constant as a function of bB. 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,bZ with b>0. Then there exist integers q and r with a=qb+r and 0r<b, and this pair is unique (Division with remainder in Z: for aZ and b>0 there are unique q,rZ with a=qb+r and 0r<b).

Verification

technique · direct
1.1

Check well-definedness modulo integer translation.

givenF1
2.1

Around a class take an interval short enough that its m inverse branches are disjoint; these branches give the evenly covered neighbourhood of [F2]. The fibre has exactly m points because the branches are indexed by the residues of the integers modulo m, and by the division algorithm [F5] every integer has exactly one residue r with 0r<m: existence gives m distinct branch labels and uniqueness stops two labels from coinciding. [F4] constructs Z but supplies no division algorithm.

step 1.1F2F4F5
3.1

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

step 2.1F3
4.1

The preceding construction and implications establish the assertion.

step 3.1
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16Open item page →

Pulling a covering back to an evenly covered open set gives a trivial covering

Example

If U is evenly covered by p:EB, then the pullback of p along the inclusion UB is a trivial covering of U.

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]

If X is any space and F is a nonempty discrete space, the projection X×FX is a trivial covering with fibre F. If X=, the same holds for F=; for nonempty X, an empty fibre would violate surjectivity. (Trivial coverings are products with a discrete fibre).

Verification

technique · direct
1.1

For the inclusion UB of an evenly covered open set of the base, identify the pullback with the disjoint union of the sheets over U.

givenF2F1F3
2.1

Write the explicit mutually inverse maps over U and verify their continuity from the pullback subspace topology.

step 1.1F1F2
3.1

The preceding construction and implications establish the assertion.

step 2.1
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-16Open item page →

The Hawaiian earring is locally path-connected but has no universal cover

Example

The Hawaiian earring is locally path-connected but is not semilocally simply connected at its wedge point. Consequently it has no universal cover.

Facts & Assumptions

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

[F1]

The loop t[t] in R/Z is not nullhomotopic. (The projected unit interval is not nullhomotopic in R/Z).

[F2]

If a space admits a universal covering, then it is semilocally simply connected. No local path-connectedness hypothesis is required. (A space admitting a universal covering is semilocally simply connected).

[F3]

A space X is semilocally simply connected at xX when there is a neighbourhood U of x and a basepoint-preserving inclusion (U,x)(X,x) whose induced map on fundamental groups is trivial (def-neighbourhood-top, def-induced-homomorphism-on-fundamental-groups, def-based-loops-and-fundamental-group). It is semilocally simply connected when this holds at every point. The neighbourhood need not itself be simply connected. (Semilocally simply connected spaces with explicit basepoint convention).

[F4]

The quotient topology. Let (X,T) be a topological space (def-topological-space), let Y be a set and let q:XY be a surjection (def-injection-surjection-bijection). The quotient topology on Y induced by q is the final topology of the one-element family (q) (def-initial-and-final-topology): Tq  :=  {VY:q1[V]T}. That this is a topology is discharged in def-initial-and-final-topology, where every final topology is verified to satisfy (T1), (T2) and (T3). Dually, CY is closed in Tq exactly when q1[C] is closed in X, because q1[YV]=Xq1[V]. (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection).

Verification

technique · direct
1.1

Model the earring as countably many copies of R/Z with diameters tending to zero and all zero classes identified, using the standard shrinking-wedge metric.

givenF1
2.1

Away from the wedge point, sufficiently short open arcs are path-connected neighbourhoods.

step 1.1F3F2
3.1

At the wedge point, every open neighbourhood contains a smaller metric ball whose intersection with each circle is an arc through the wedge point and which contains every sufficiently small circle, so that ball is path-connected.

step 2.1F3F2F4
4.1

Retraction to one such small circle and the essential unit loop show its inclusion carries a nontrivial loop, so semilocal simple connectedness fails at the wedge point.

step 3.1F1F2F3
5.1

The necessity theorem then rules out a universal cover.

step 4.1F2
6.1

The preceding construction and implications establish the assertion.

step 5.1

Sources