Proof.  [#247]

Let \(p : E \to B\) be exponentiable, that is \(- \times p : \mathbb {C} / B \to \mathbb {C} / B\) (given by pullbacks in \(\mathbb {C}\)) has a right adjoint \((-)^p\).

Define \(\Pi _p : \mathbb {C} / E \to \mathbb {C} / B : g \mapsto (p \circ g)^p\), then by the given adjunction it suffices to show that maps \(p^\ast f \to g\) in \(\mathbb {C} / E\) are in natural bijection with maps \(f \times p \to p \circ g\) in \(\mathbb {C} / B\), which is easily checked by unwinding the data of such maps.

Conversely let \(p^\ast \) have a right adjoint \(\Pi _p\). Note that \(- \times p\) factors as \(p^\ast \) followed by \(p \circ -\) and moreover \((p \circ -) \dashv p^\ast \). Then composing adjunctions gives \(\Pi _p \circ p^\ast \) as a right adjoint to \(- \times p\).