Show directly that if \(X\) is a topological group with identity element \(x_0\), then any two maps \(f, g: (Z, z_0) \to (X, x_0)\) which are homotopic are homotopic through basepoint-preserving maps.
Solution
1. Track of the basepoint under the unbased homotopy:
1.1. Let \(H: Z \times I \to X\) be an unbased homotopy between \(f\) and \(g\), so \(H(z, 0) = f(z)\) and \(H(z, 1) = g(z)\) for all \(z \in Z\).
1.2. Define the path \(\gamma: I \to X\) traced by the basepoint \(z_0\):\[ \gamma(t) = H(z_0, t). \]Because \(f, g: (Z, z_0) \to (X, x_0)\) are pointed maps, the endpoints of \(\gamma\) satisfy:\[ \gamma(0) = H(z_0, 0) = f(z_0) = x_0, \qquad \gamma(1) = H(z_0, 1) = g(z_0) = x_0. \]Thus \(\gamma\) is a loop in \(X\) based at the identity element \(x_0\).
2. Construction of the basepoint-preserving homotopy:
2.1. Since \(X\) is a topological group, multiplication \(m: X \times X \to X\) and inversion \(i: X \to X\) are continuous.Define \(F: Z \times I \to X\) by:\[ F(z, t) = H(z, t) \cdot (\gamma(t))^{-1}. \]As a composition of continuous functions, \(F\) is continuous.
2.2. Check the initial and final time slices:\[ F(z, 0) = H(z, 0) \cdot (\gamma(0))^{-1} = f(z) \cdot x_0^{-1} = f(z), \]\[ F(z, 1) = H(z, 1) \cdot (\gamma(1))^{-1} = g(z) \cdot x_0^{-1} = g(z). \]
2.3. Check basepoint preservation for all \(t \in [0, 1]\):\[ F(z_0, t) = H(z_0, t) \cdot (\gamma(t))^{-1} = \gamma(t) \cdot (\gamma(t))^{-1} = x_0. \]Thus \(F(z_0, t) = x_0\) for every \(t \in I\).
3. Conclusion:\(F: (Z \times I, \{z_0\} \times I) \to (X, x_0)\) is a pointed homotopy between \(f\) and \(g\). Q.E.D.