Show \(\pi_1(X\times Y, (x_0, y_0)) \cong \pi_1(X,x_0) \times \pi_1(Y, y_0)\).
Solution
Goal: Let \((X, x_0)\) and \((Y, y_0)\) be pointed topological spaces. Prove that the canonical map \(\Phi \colon \pi_1(X \times Y, (x_0, y_0)) \to \pi_1(X, x_0) \times \pi_1(Y, y_0)\) given by \([\gamma] \mapsto ((p_X)_*[\gamma], (p_Y)_*[\gamma])\) is an isomorphism of groups.
1. Definition of the projection maps and homomorphism \(\Phi\).
1.1. Let \(p_X \colon X \times Y \to X\) and \(p_Y \colon X \times Y \to Y\) be the standard continuous projection maps.
1.2. The induced maps on fundamental groups \((p_X)_* \colon \pi_1(X \times Y, (x_0, y_0)) \to \pi_1(X, x_0)\) and \((p_Y)_* \colon \pi_1(X \times Y, (x_0, y_0)) \to \pi_1(Y, y_0)\) are group homomorphisms by functoriality of \(\pi_1\).
1.3. Define \(\Phi \colon \pi_1(X \times Y, (x_0, y_0)) \to \pi_1(X, x_0) \times \pi_1(Y, y_0)\) by: \[\Phi([\gamma]) = ((p_X)_*[\gamma], (p_Y)_*[\gamma]) = ([p_X \circ \gamma], [p_Y \circ \gamma]).\]
1.4. Since both components are homomorphisms, \(\Phi\) is a group homomorphism.
1.5. The induced maps \((p_X)_*\) and \((p_Y)_*\) are homomorphisms by functoriality of \(\pi_1\), and the product of two homomorphisms is a homomorphism, so \(\Phi\) is a group homomorphism.
2. Prove that \(\Phi\) is surjective.
2.1. Let \(([\alpha], [\beta]) \in \pi_1(X, x_0) \times \pi_1(Y, y_0)\), where \(\alpha \colon [0, 1] \to X\) is a loop based at \(x_0\) and \(\beta \colon [0, 1] \to Y\) is a loop based at \(y_0\).
2.2. Define \(\gamma \colon [0, 1] \to X \times Y\) by \(\gamma(t) = (\alpha(t), \beta(t))\).
2.3. By the universal property of the product topology, \(\gamma\) is continuous since its component paths \(p_X \circ \gamma = \alpha\) and \(p_Y \circ \gamma = \beta\) are continuous.
2.4. Furthermore, \(\gamma(0) = (\alpha(0), \beta(0)) = (x_0, y_0)\) and \(\gamma(1) = (\alpha(1), \beta(1)) = (x_0, y_0)\), so \(\gamma\) is a loop in \(X \times Y\) based at \((x_0, y_0)\).
2.5. Then \(\Phi([\gamma]) = ([p_X \circ \gamma], [p_Y \circ \gamma]) = ([\alpha], [\beta])\).
2.6. The loop \(\gamma(t) = (\alpha(t), \beta(t))\) is continuous by the universal property of the product topology, is based at \((x_0, y_0)\) because \(\alpha\) and \(\beta\) are based at \(x_0\) and \(y_0\), and satisfies \(p_X \circ \gamma = \alpha\) and \(p_Y \circ \gamma = \beta\); hence \(\Phi([\gamma]) = ([\alpha], [\beta])\).
3. Prove that \(\Phi\) is injective.
3.1. Suppose \([\gamma] \in \ker(\Phi)\), so \(\Phi([\gamma]) = ([c_{x_0}], [c_{y_0}])\), where \(c_{x_0}, c_{y_0}\) are constant loops.
3.2. This means there exist path homotopies:
\(F \colon [0, 1] \times [0, 1] \to X\) between \(p_X \circ \gamma\) and \(c_{x_0}\) relative to \(\{0, 1\}\),
\(G \colon [0, 1] \times [0, 1] \to Y\) between \(p_Y \circ \gamma\) and \(c_{y_0}\) relative to \(\{0, 1\}\). <2>3. Define \(H \colon [0, 1] \times [0, 1] \to X \times Y\) by \(H(t, s) = (F(t, s), G(t, s))\). <2>4. \(H\) is continuous because its component functions \(F\) and \(G\) are continuous. <2>5. Check boundary conditions:
\(H(t, 0) = (F(t, 0), G(t, 0)) = (p_X(\gamma(t)), p_Y(\gamma(t))) = \gamma(t)\),
\(H(t, 1) = (F(t, 1), G(t, 1)) = (x_0, y_0) = c_{(x_0, y_0)}(t)\),
For all \(s \in [0, 1]\), \(H(0, s) = (F(0, s), G(0, s)) = (x_0, y_0)\) and \(H(1, s) = (F(1, s), G(1, s)) = (x_0, y_0)\). <2>6. Thus \(H\) is a path homotopy in \(X \times Y\) between \(\gamma\) and the constant loop \(c_{(x_0, y_0)}\), so \([\gamma] = 1 \in \pi_1(X \times Y, (x_0, y_0))\).
3.7. The homotopy \(H(t, s) = (F(t, s), G(t, s))\) is continuous because \(F\) and \(G\) are, and its boundary conditions in 5 show it is a path homotopy from \(\gamma\) to the constant loop; hence \(\ker(\Phi) = \{1\}\), so \(\Phi\) is injective.
4. Q.E.D.
4.1. \(\Phi\) is a bijective group homomorphism, hence an isomorphism: \(\pi_1(X \times Y, (x_0, y_0)) \cong \pi_1(X, x_0) \times \pi_1(Y, y_0)\).