Suppose that the homomorphism \(i_*\) induced by inclusion \(i: U \cap V \to X\) is trivial.
- Show that \(j_1\) and \(j_2\) induce an epimorphism
\[ h: (\pi_1(U, x_0)/N_1) * (\pi_1(V, x_0)/N_2) \to \pi_1(X, x_0), \]
where \(N_1\) is the least normal subgroup of \(\pi_1(U, x_0)\) containing image \(i_1\), and \(N_2\) is the least normal subgroup of \(\pi_1(V, x_0)\) containing image \(i_2\).
- Show that \(h\) is an isomorphism. [Hint: Use Theorem 70.1 to define a left inverse for \(h\).]
Solution
Goal: Prove that when the inclusion \(i: U \cap V \to X\) induces the trivial homomorphism on \(\pi_1\), \(\pi_1(X, x_0)\) is naturally isomorphic to the free product of the quotients \((\pi_1(U, x_0)/N_1) \ast (\pi_1(V, x_0)/N_2)\).
1. Part (a): Construction and surjectivity of the homomorphism \(h\).Proof:
1.1. Let \(i_1: U \cap V \hookrightarrow U\), \(i_2: U \cap V \hookrightarrow V\), \(j_1: U \hookrightarrow X\), \(j_2: V \hookrightarrow X\), with \(i = j_1 \circ i_1 = j_2 \circ i_2\).
1.2. For any \(\gamma \in \pi_1(U \cap V, x_0)\), \((j_1)_*((i_1)_*(\gamma)) = i_*(\gamma) = 1\) in \(\pi_1(X, x_0)\) by assumption that \(i_*\) is trivial.
1.3. Thus \(\operatorname{im}((i_1)_*) \subseteq \ker((j_1)_*)\). Since \(\ker((j_1)_*)\) is a normal subgroup of \(\pi_1(U, x_0)\), the normal closure \(N_1\) satisfies \(N_1 \subseteq \ker((j_1)_*)\).
1.4. By the universal property of quotient groups, \((j_1)_*\) factors uniquely through a homomorphism:\[\phi_1: \pi_1(U, x_0)/N_1 \to \pi_1(X, x_0), \quad \phi_1([u]) = (j_1)_*(u).\]
1.5. By the identical argument on \(V\), \((j_2)_*\) factors uniquely through:\[\phi_2: \pi_1(V, x_0)/N_2 \to \pi_1(X, x_0), \quad \phi_2([v]) = (j_2)_*(v).\]
1.6. By the universal property of the free product of groups, \(\phi_1\) and \(\phi_2\) determine a unique homomorphism:\[h: (\pi_1(U, x_0)/N_1) \ast (\pi_1(V, x_0)/N_2) \to \pi_1(X, x_0).\]
1.7. By the Seifert-van Kampen theorem, the images of \((j_1)_*\) and \((j_2)_*\) generate \(\pi_1(X, x_0)\). Since \(\operatorname{im}(h)\) contains \(\operatorname{im}(\phi_1) = \operatorname{im}((j_1)_*)\) and \(\operatorname{im}(\phi_2) = \operatorname{im}((j_2)_*)\), \(h\) is surjective (an epimorphism).
2. Part (b): \(h\) is an isomorphism (existence of an inverse via universal property).Proof:
2.1. Let \(G = (\pi_1(U, x_0)/N_1) \ast (\pi_1(V, x_0)/N_2)\). Let \(q_1: \pi_1(U, x_0) \to \pi_1(U, x_0)/N_1 \hookrightarrow G\) and \(q_2: \pi_1(V, x_0) \to \pi_1(V, x_0)/N_2 \hookrightarrow G\) be the canonical quotient-inclusion maps.
2.2. For any \(\gamma \in \pi_1(U \cap V, x_0)\):- \((i_1)_*(\gamma) \in N_1 \implies q_1((i_1)_*(\gamma)) = 1 \in G\),- \((i_2)_*(\gamma) \in N_2 \implies q_2((i_2)_*(\gamma)) = 1 \in G\).
2.3. Therefore \(q_1((i_1)_*(\gamma)) = q_2((i_2)_*(\gamma)) = 1\) for all \(\gamma \in \pi_1(U \cap V, x_0)\).
2.4. By the universal property of the Seifert-van Kampen pushout (Theorem 70.1), the homomorphisms \(q_1\) and \(q_2\) induce a unique homomorphism:\[g: \pi_1(X, x_0) \to G\]satisfying \(g \circ (j_1)_* = q_1\) and \(g \circ (j_2)_* = q_2\).
2.5. We verify that \(g \circ h = \operatorname{id}_G\):- On generators \([u] \in \pi_1(U, x_0)/N_1\): \((g \circ h)([u]) = g(\phi_1([u])) = g((j_1)_*(u)) = q_1(u) = [u]\).- On generators \([v] \in \pi_1(V, x_0)/N_2\): \((g \circ h)([v]) = g(\phi_2([v])) = g((j_2)_*(v)) = q_2(v) = [v]\).
2.6. Since \(g \circ h\) is the identity on a generating set of \(G\), \(g \circ h = \operatorname{id}_G\).
2.7. Thus \(h\) has a left inverse \(g\), so \(h\) is injective.
2.8. Being both surjective (<1>1) and injective, \(h\) is an isomorphism of groups. Q.E.D.