If the fibration \(p: E \to B\) is a homotopy equivalence, show that \(p\) is a fiber homotopy equivalence of \(E\) with the trivial fibration \(\mathbb{1}: B \to B\).
Solution
The fibration \(p:E\to B\) and the identity fibration \[ 1_B:B\to B \] represent the same element of \(\mathcal M(B)\): the map \[ p:E\to B \] is itself a homotopy equivalence and satisfies \[ p=1_B\circ p. \] By the bijection of Exercise 14, equal classes in \(\mathcal M(B)\) correspond to the same class in \(\mathcal F(B)\). Thus \[ \boxed{p:E\to B\text{ is fiber homotopy equivalent to }1_B:B\to B.} \]