Show that a map \(p: E \to B\) is a fibration if and only if the map \(\pi: E^I \to E_p\), \(\pi(\gamma) = (\gamma(0), p\gamma)\) has a section, that is, a map \(s: E_p \to E^I\) such that \(\pi s = \mathbb{1}\).
Solution
Write \[ E_p=\{(e,\alpha)\in E\times B^I:p(e)=\alpha(0)\}, \qquad \pi:E^I\to E_p, \quad \pi(\gamma)=(\gamma(0),p\gamma). \]
Suppose first that \(\pi\) has a section \(s\). Given a lifting problem \[ f:X\to E, \qquad H:X\times I\to B, \qquad H(x,0)=pf(x), \] let \(H_x(t)=H(x,t)\). Then \[ x\longmapsto(f(x),H_x) \] is a map \(X\to E_p\), and \[ \widetilde H(x,t)=s(f(x),H_x)(t) \] satisfies \[ \widetilde H(x,0)=f(x), \qquad p\widetilde H(x,t)=H(x,t). \] Thus \(p\) has the homotopy lifting property and is a fibration.
Conversely, suppose \(p\) is a fibration. Over \(E_p\), consider the map \[ q:E_p\to E, \qquad q(e,\alpha)=e, \] and the homotopy \[ K:E_p\times I\to B, \qquad K((e,\alpha),t)=\alpha(t). \] At time \(0\), \[ K(-,0)=pq. \] By the homotopy lifting property there is \[ \widetilde K:E_p\times I\to E \] with \(\widetilde K(-,0)=q\) and \(p\widetilde K=K\). Define \[ s(e,\alpha)(t)=\widetilde K((e,\alpha),t). \] Then \[ s(e,\alpha)(0)=e, \qquad p\,s(e,\alpha)=\alpha, \] so \(\pi s=\operatorname{id}_{E_p}\). Hence \[ \boxed{p\text{ is a fibration iff }\pi\text{ has a section}.} \]