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Poincare Conjecture ProofH. Vic DannonTherefore,njj,kθ θ= ∩ fθHθk = 1f BnG = ∪∪∩θ∈Θ j∈ J k=1f H θjk ,θThus,nj∈ J k= 1 θ∈Θjk ,θ= ∪∩∪ f H .θ∪ f θS θis a subbase for T , with subbase sets f θH θθ∈[0,2 π]θ∈Θ∈τ∪ .To obtain the Poincare Homeomorphism by 5.1, we need to definea functionf3 3: Σ → S which is¬ one-oneÁ one-one correspondence between a subbase set for T ,∪ , and a subbase set for T , ∪ f θH θ.H θθ∈Ωθ∈Ω5.10 Definition of f32For x ∈Σ , there are unique θ ∈ [0,2 π], and ∈Σ so thatx θθx= ( x θ, θ).We define the mapping f from3Σ onto3S by fx ( , θ) = ( fx, θ).θ θ θ18
Poincare Conjecture ProofH. Vic Dannon5.11 f is one-oneProof:LetAssume that2Since Σθare disjoint,θSince f θis a homeomorphism,Hence,x = ( x θ, θ), and y = ( y φ, φ).( f x , θ) = ( f y , φ).θ θ φ φ= φ, and f x = f y .xxθ= y .θ θ θ θθ= y.5.12f is one-one between the subbase set H in Sand the subbase set∪θ∈ΘfHθθinΣ3∪θ∈Θθ3Proof: Let H = ∪ H θ, and K = ∪ K φ.θ∈Θφ∈ΦAssume that,{( f H , θ): θ , H H} {( f K , φ): φ , K K}∈Θ ⊆ = ∈Φ ⊆ .θ θ θ φ φ φ2Since the Σθare disjoint,Θ=Φ, and fθHθ = fθKθ, for any θ ∈Θ.Since f θis a homeomorphism, for each θ ∈Θ,Hence, H= K.Hθ= K .θ19
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