Alphabeta Math
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5 results · all verified · 2 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 3 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

The Seifert–van Kampen Theorem: Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-21Open item page →

Two paths can induce distinct change-of-basepoint isomorphisms on S1S1

Example

Let W=S1S1 have wedge point w, and let zw lie on the first circle. Write a,bπ1(W,w) for the standard loops in the first and second circles. Choose a simple path ρ in the first circle from w to z, and put σ=bρ. Then ρ and σ have the same endpoints but induce distinct isomorphisms

cρ,cσ:π1(W,w)π1(W,z).

Facts & Assumptions

Given: The points, paths, and standard loop classes in the Example.

[L1]

The group π1(W,w) is the free group F(a,b) on the two standard circle loops (π1(S1S1) is the free group on two generators).

[F1]

Loop concatenation is the fundamental-group operation, and path reversal represents inversion (Loop classes form the group π1(X,x0) under concatenation).

[F2]

The reduced words on a basis and its formal inverses form the free group on that basis (Reduced words form the free group on an alphabet).

Verification

technique · direct
1.1

For a path η from w to z, define cη([α])=[ηˉαη]. Endpoint-fixed homotopies are preserved by concatenating fixed paths, while the standard cancellation homotopies for ηηˉ and ηˉη show that cη is a homomorphism with inverse cηˉ. Thus both ρ and σ define basepoint-change isomorphisms.

F1construct
2.1

Since σ=bρ, reversal of concatenation gives cσ([α])=[ρˉbˉαbρ]=cρ(b1[α]b). Hence cρˉcρ(a)=a, whereas cρˉcσ(a)=b1ab in the identification [L1].

step 1.1L1F1
3.1

The words a and b1ab are distinct reduced words by [F2]. Therefore their images under the isomorphism cρ are distinct, so cρcσ.

step 2.1F2
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

π1(R2{0})Z

Example

Point R2{0} at (1,0). Its fundamental group is infinite cyclic:

π1(R2{0},(1,0))(Z,+).

Facts & Assumptions

Given: The punctured plane P=R2{0}, its unit circle C, and the basepoint (1,0)C.

[F1]

Radial normalization is a deformation retraction of Rn{0} onto its unit sphere for every n1 (For n1, radial normalisation is a deformation retraction of Rn{0} onto Sn1).

[F2]

Induced fundamental-group maps respect identities, composition, and pointed homotopies (Induced fundamental-group maps are well defined, functorial and invariant under based homotopy).

[F3]

The map [t](cos2πt,sin2πt) is a homeomorphism from R/Z to C sending [0] to (1,0) ([t](cos2πt,sin2πt) is a homeomorphism from R/Z to the unit circle).

[F4]

The degree map is an isomorphism π1(R/Z,[0])(Z,+) (Deg:π1(R/Z,[0])(Z,+) is an isomorphism).

Verification

technique · direct
1.1

Specializing [F1] to n=2 gives a retraction r:PC and an endpoint-fixed homotopy from idP to the composite of r with the inclusion i:CP, fixing (1,0).

F1
2.1

Functoriality gives ri=id and the pointed homotopy in step 1.1 gives ir=id, so i is an isomorphism π1(C,(1,0))π1(P,(1,0)).

step 1.1F2
3.1

The pointed homeomorphism of [F3] induces an isomorphism from the quotient-circle fundamental group to π1(C,(1,0)). Composing it with [F4] and the isomorphism of step 2.1 gives π1(P,(1,0))Z.

step 2.1F2F3F4
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21Open item page →

The once-punctured two-sphere has trivial fundamental group and the twice-punctured two-sphere has fundamental group Z

Example

Let N=(0,0,1) and S=(0,0,1) in S2. Point S2{N} at any point corresponding under stereographic projection to 0R2, and point S2{N,S} at the point corresponding to (1,0). Then

π1(S2{N})=1,π1(S2{N,S})Z.

Facts & Assumptions

Given: The two punctured spaces and basepoints in the Example.

[L1]

Stereographic projection identifies a pole complement in S2 with R2 and the double pole complement with R2{0} (Antipodal complements cover Sn by simply connected sets with path-connected overlap for n2).

[F1]

Every nonempty convex subset of Euclidean space is simply connected (Every nonempty convex subset of Rn is simply connected).

[L2]

The punctured plane pointed at (1,0) has fundamental group isomorphic to Z (π1(R2{0})Z).

[F2]

A pointed homeomorphism induces a fundamental-group isomorphism (Induced fundamental-group maps are well defined, functorial and invariant under based homotopy).

Verification

technique · direct
1.1

By [L1], stereographic projection is a pointed homeomorphism S2{N}R2 for the chosen basepoints.

L1
1.2

The same stereographic projection restricts by [L1] to a pointed homeomorphism S2{N,S}R2{0}.

L1
2.1

The plane is nonempty and convex, so [F1] makes its fundamental group trivial; [F2] transports that calculation through step 1.1.

step 1.1F1F2
3.1

By [L2] the latter space has fundamental group Z, and [F2] transports this group through step 1.2.

step 1.2L2F2
False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-21Open item page →

FALSE: every fundamental group is abelian

Statement

False claim: for every pointed topological space (X,x0), the group π1(X,x0) is abelian.

Facts & Assumptions

Given: The two-circle wedge W=S1S1 with standard loop classes a and b.

[L1]

The group π1(W,w) is the free group on a and b (π1(S1S1) is the free group on two generators).

[F1]

The reduced words on a basis and its formal inverses form the free group on that basis (Reduced words form the free group on an alphabet).

Refutation

technique · direct
1.1

Under [L1], the products ab and ba are represented by the two reduced words with syllable sequences (a,b) and (b,a).

L1
2.1

These reduced words are distinct by [F1], so abba in π1(W,w).

step 1.1F1
3.1

Thus the fundamental group of the two-circle wedge is not abelian, providing a counterexample to the universal claim.

step 2.1
False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-21Open item page →

FALSE: the two-set van Kampen conclusion needs no path-connectedness hypothesis on the overlap

Statement

False claim: let X=UV with U,V open and path-connected, and let x0UV. Even when UV is not path-connected, if C is its path component containing x0, then π1(X,x0) is the pushout of

π1(U,x0)π1(C,x0)π1(V,x0).

Facts & Assumptions

Given: The quotient circle Q=R/Z, its quotient map p, the open arcs U=p((1/8,5/8)) and V=p((3/8,9/8)), and the basepoint [0].

[F1]

The quotient map is open, and every interval shorter than one maps homeomorphically to its image in Q (The quotient map is open, and every interval shorter than one embeds in R/Z).

[F2]

Every nonempty convex subset of a Euclidean space is simply connected (Every nonempty convex subset of Rn is simply connected).

[F3]

The degree map is an isomorphism π1(Q,[0])(Z,+) (Deg:π1(R/Z,[0])(Z,+) is an isomorphism).

[F4]

A pointed homeomorphism induces a fundamental-group isomorphism (Induced fundamental-group maps are well defined, functorial and invariant under based homotopy).

Refutation

technique · direct
1.1

Both defining intervals have length 3/4<1, so [F1] makes U and V open arcs. Their displayed lifts show UV=Q, while UV=p((1/8,1/8))p((3/8,5/8)), a disjoint union of two nonempty open arcs. The basepoint lies in the first component C=p((1/8,1/8)).

F1algebra
2.1

The three arcs U,V,C are homeomorphic to open intervals, which are nonempty and convex. Hence [F2] and [F4] make all three fundamental groups trivial.

step 1.1F2F4
3.1

The pushout of the two homomorphisms from the trivial group to the two trivial factor groups is itself the trivial group: for every target group there is exactly one compatible pair of homomorphisms and exactly one homomorphism from the trivial group.

step 2.1
4.1

The actual group π1(Q,[0]) is isomorphic to Z by [F3], so it is nontrivial and cannot be the pushout computed in step 3.1. Thus the false claim fails for this cover, and path-connectedness of the full overlap cannot be omitted from the two-set theorem.

step 3.1F3

Sources