Given a map \(f: A \to B\) and a homotopy equivalence \(g: C \to A\), show that the fibrations \(E_f \to B\) and \(E_{fg} \to B\) are fiber homotopy equivalent.
Solution
Let \(g:C\to A\) be a homotopy equivalence with homotopy inverse \(h:A\to C\), and choose homotopies \[ gh\simeq1_A, \qquad hg\simeq1_C. \] There is an evident fiber-preserving map \[ \Phi:E_{fg}\to E_f, \qquad (c,\gamma)\longmapsto(g(c),\gamma). \]
For the reverse direction, use a chosen homotopy \(K:gh\simeq1_A\). For \((a,\gamma)\in E_f\), the path \[ t\longmapsto f(K(a,t)) \] runs from \(f(gh(a))\) to \(f(a)\). Prepending it to \(\gamma\) defines a fiber-preserving map \[ \Psi:E_f\to E_{fg}, \qquad \Psi(a,\gamma)=\bigl(h(a),fK_a*\gamma\bigr). \] The homotopies \(gh\simeq1_A\) and \(hg\simeq1_C\), together with contraction of the resulting backtracking paths, give fiber-preserving homotopies \[ \Phi\Psi\simeq1_{E_f}, \qquad \Psi\Phi\simeq1_{E_{fg}}. \] Thus \[ \boxed{E_f\to B\simeq_f E_{fg}\to B.} \]