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CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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Trefoil surgery with trivial first homology and nontrivial fundamental group

Statement refuted

Every connected closed oriented 3-manifold M with H1(M;Z)=0 is simply connected, meaning that π1(M) is trivial.

Facts & Assumptions

Given: ACω, the connected exterior ET of the trefoil T, the positive meridian μ=a at the first left arc and the degree-one preferred longitude λ of this same exterior, their boundary basis ([μ],[λ]), the +1 filling of T, and the two displayed supplier statements. The preferred-longitude definition supplies [λ]=0; the homology lemma derives that [μ] generates H1(ET;Z).

[A1]

ACω is the stated axiom for the chosen exterior and peripheral data. The Axiom of Countable Choice (ACω)

[F1]

First homology of knot fillings: a p/q filling has first homology Z/∣p∣Z when p≠0.

[F2]

Group presentation for positive trefoil +1 surgery: in its specified positive-braid convention, μ=a is the positive meridian at the first left arc and λ is the degree-one preferred longitude; the matching +1 filling has presentation ⟨a,b∣aba=bab,  bab−1aba−2=1⟩.

[F3]

Oriented slopes and knot Dehn filling: filling glues an oriented solid torus to the knot exterior along the specified meridian slope.

Counterexample

Assume ACω. Let T be the trefoil given by the closure of the positive two-strand braid σ13, using the positive half-twist convention for σ1. Choose its connected exterior and the same peripheral system as in the trefoil group-presentation calculation: μ=a is the positive normal-disk meridian at the first left arc, and λ is the degree-one preferred longitude of this exterior, with ([μ],[λ]) a basis of H1(∂ET;Z). Let M=ET(1/1), where +1 means filling along [μ]+[λ]. Then H1(M;Z)=0 but π1(M) is nontrivial.

1.1givenA1F1F3constructalgebra

By [F3], M is the connected closed oriented manifold obtained by filling the connected knot exterior along its boundary torus. Apply [F1] with p=q=1. Then H1(M;Z)≅Z/1Z=0.

1.2F2constructalgebra

Let A5 denote the subgroup of even permutations of {1,2,3,4,5}; every 5-cycle is even because it is a product of four transpositions. Set a=(1 2 3 4 5) and b=(1 3 5 4 2), composing right to left. Applying the factors of aba and bab from right to left gives the same image tuple (2,1,3,5,4) on inputs (1,2,3,4,5), so both words equal (1 2)(4 5). Thus the braid relator in [F2] is satisfied.

1.3F2constructalgebra

Applying the factors of bab−1aba−2 from right to left, the six successive images for inputs 1,2,3,4,5 are respectively (4,2,3,1,2,1), (5,4,5,3,4,2), (1,3,4,5,1,3), (2,1,2,4,5,4), and (3,5,1,2,3,5). Their final entries are (1,2,3,4,5), so the filling relator is the identity. The assignments therefore satisfy both relators in [F2] and induce a homomorphism π1(M)→A5.

2.1step 1.1step 1.2step 1.3algebra∎

The image of a is a nonidentity 5-cycle, so this homomorphism has nontrivial image and π1(M)≠1. Together with step 1.1, this refutes the statement.

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