For a space \(B\), let \(\mathcal{M}(B)\) denote the set of equivalence classes of maps \(f: A \to B\) where \(f_1: A_1 \to B\) is equivalent to \(f_2: A_2 \to B\) if there exists a homotopy equivalence \(g: A_1 \to A_2\) such that \(f_1 \simeq f_2 g\). Show the natural map \(\mathcal{F}(B) \to \mathcal{M}(B)\) is a bijection.
Solution
Send a fibration \(p:E\to B\) to its class as a map in \(\mathcal M(B)\). A fiber homotopy equivalence is in particular a homotopy equivalence of total spaces commuting with the projections up to homotopy, so this gives a well-defined map \[ \mathcal F(B)\to\mathcal M(B). \]
It is surjective. Given any map \(f:A\to B\), its mapping-path replacement \[ p_f:E_f\to B \] is a fibration, and the inclusion \[ A\to E_f, \qquad a\mapsto(a,\text{constant path at }f(a)) \] is a homotopy equivalence with \(f\simeq p_fi\). Hence \(f\) and \(p_f\) define the same element of \(\mathcal M(B)\).
For injectivity, suppose fibrations \(p_1:E_1\to B\) and \(p_2:E_2\to B\) represent the same element of \(\mathcal M(B)\). Then there is a homotopy equivalence \[ g:E_1\to E_2 \] with \[ p_1\simeq p_2g. \] Exercises 12 and 13 imply \[ E_{p_1}\simeq_f E_{p_2g}\simeq_f E_{p_2}. \] Each fibration \(p_i\) is fiber homotopy equivalent to its mapping-path replacement \(E_{p_i}\to B\). Hence \(p_1\) and \(p_2\) are fiber homotopy equivalent. Therefore \[ \boxed{\mathcal F(B)\xrightarrow{\cong}\mathcal M(B).} \]