Proof.
According to the discussion in Algebraic Stacks, Section 94.6 we see that f is representable if and only if for every y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y}) lying over U the 2-fibre product (\mathit{Sch}/U)_{fppf} \times _{y, \mathcal{Y}} \mathcal{X} is representable, i.e., of the form (\mathit{Sch}/V_ y)_{fppf} for some scheme V_ y over U. Objects in this 2-fibre products are triples (h : V \to U, x, \alpha : f(x) \to h^*y) where \alpha lies over \text{id}_ V. Dropping the h from the notation we see that this is equivalent to the data of an object x of \mathcal{X} and a morphism f(x) \to y. Hence the 2-fibre product is representable by V_ y and f(x_ y) \to y where x_ y is an object of \mathcal{X} over V_ y if and only if the functor in (2) is representable by x_ y with universal object a map f(x_ y) \to y.
\square
Comments (0)