For a space \(B\), let \(\mathcal{F}(B)\) be the set of fiber homotopy equivalence classes of fibrations \(E \to B\). Show that a map \(f: B_1 \to B_2\) induces \(f^*: \mathcal{F}(B_2) \to \mathcal{F}(B_1)\) depending only on the homotopy class of \(f\), with \(f^*\) a bijection if \(f\) is a homotopy equivalence.
Solution
For a fibration \(p:E\to B_2\) and a map \(f:B_1\to B_2\), let \[ f^*E=\{(b,e):f(b)=p(e)\}\to B_1 \] be the pullback fibration. Pullback carries fiber-preserving maps and fiber homotopies to fiber-preserving maps and fiber homotopies, so it defines \[ f^*:\mathcal F(B_2)\to\mathcal F(B_1). \]
If \(f_0\simeq f_1\) through \(H:B_1\times I\to B_2\), pull back \(E\) along \(H\). The resulting fibration over \(B_1\times I\) restricts at the two ends to \(f_0^*E\) and \(f_1^*E\). Fiber transport along the interval, obtained from the homotopy lifting property, gives mutually inverse fiber homotopy equivalences between these endpoint restrictions. Hence \(f^*\) depends only on the homotopy class of \(f\).
Functoriality of pullback gives \[ (fg)^*=g^*f^*. \] If \(f:B_1\to B_2\) is a homotopy equivalence with homotopy inverse \(g\), then \[ gf\simeq1_{B_1}, \qquad fg\simeq1_{B_2}, \] so homotopy invariance gives \[ f^*g^*=1, \qquad g^*f^*=1. \] Thus \[ \boxed{f^*:\mathcal F(B_2)\xrightarrow{\cong}\mathcal F(B_1)} \] is a bijection.