Show that if \(h, h': X \to Y\) are homotopic and \(k, k': Y \to Z\) are homotopic, then \(k \circ h\) and \(k' \circ h'\) are homotopic.
Solution
1. Setting and given homotopies:Let \(F: X \times I \to Y\) be a continuous homotopy between \(h\) and \(h'\), so that:\[F(x, 0) = h(x) \quad \text{and} \quad F(x, 1) = h'(x) \quad \text{for all } x \in X.\]Let \(G: Y \times I \to Z\) be a continuous homotopy between \(k\) and \(k'\), so that:\[G(y, 0) = k(y) \quad \text{and} \quad G(y, 1) = k'(y) \quad \text{for all } y \in Y.\]
2. Construction of the composite homotopy:Define the map \(H: X \times I \to Z\) by:\[H(x, t) = G(F(x, t), t).\]
3. Verification of continuity and boundary values:Proof:
3.1. Continuity:Consider the map \(\Phi: X \times I \to Y \times I\) defined by \(\Phi(x, t) = (F(x, t), t)\).Since \(F\) and the coordinate projection \((x, t) \mapsto t\) are continuous, \(\Phi\) is continuous.Because \(H = G \circ \Phi\) is the composition of continuous maps, \(H\) is continuous.
3.2. Evaluation at \(t = 0\):\[H(x, 0) = G(F(x, 0), 0) = G(h(x), 0) = k(h(x)) = (k \circ h)(x).\]
3.3. Evaluation at \(t = 1\):\[H(x, 1) = G(F(x, 1), 1) = G(h'(x), 1) = k'(h'(x)) = (k' \circ h')(x).\]
4. Q.E.D.By step 3, \(H\) is a continuous homotopy from \(k \circ h\) to \(k' \circ h'\), so \(k \circ h \simeq k' \circ h'\).