Van Kampen's Theorem Proof

Algebraic Topology · Axiom Academy

LESSON Van Kampen's Theorem Proof Establishing the isomorphism via surjectivity and injectivity To prove Van Kampen's theorem, we construct a homomorphism Φ from the amalgamated product to π₁(X) and show it's an isomorphism. The inclusion maps ι₁: U₁ ↪ X and ι₂: U₂ ↪ X induce homomorphisms: By the universal property of the amalgamated product, these extend uniquely to: Φ: π₁(U₁) *_ π₁(U₁∩U₂) π₁(U₂) → π₁(X, x₀) Goal: Show Φ is bijective (surjective + injective). 2. Surjectivity: Path Decomposition Every loop in X can be broken into pieces lying in U₁ or U₂, proving Φ is surjective. Let [γ] ∈ π₁(X, x₀) be any homotopy class of loops in X. We must show [γ] is in the image of Φ. Step 1: Since X = U₁ ∪ U₂, the loop γ: [0,1] → X is covered by open sets γ⁻¹(U₁), γ⁻¹(U₂) . Step 2: By the Lebesgue number lemma, we can subdivide [0,1] = [t₀, t₁] ∪ [t₁, t₂] ∪ ... ∪ [tₙ₋₁, tₙ] where each γ([tᵢ, tᵢ₊₁]) lies entirely in U₁ or entirely in U₂. Step 3: Each segment γᵢ = γ|[tᵢ, tᵢ₊₁] is a path in either U₁ or U₂. By concatenating with paths in U₁ ∩ U₂ from γ(tᵢ) to x₀, we get loops in U₁ or U₂. Conclusion: [γ] = Φ([γ₁] * [γ₂] * ... * [γₙ]) where each [γᵢ] comes from π₁(U₁) or π₁(U₂). 3. Injectivity: Homotopy Argument To show Φ is injective, we must prove that if a word in the amalgamated product maps to the trivial loop, then it was already trivial in the amalgamated product. Suppose w = [γ₁][γ₂]...[γₙ] ∈ π₁(U₁) *_ π₁(U₁∩U₂) π₁(U₂) and Φ(w) = [e] in π₁(X).

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