The Stacks project

Proof. Note that the quadruple $(f : T \to S, d, \mathcal{L}, \psi )$ defines an element of $F_ d(T)$. Let $U_ d \subset \underline{\text{Proj}}_ S(\mathcal{A})$ be the locus where the sheaf $\mathcal{O}_{\underline{\text{Proj}}_ S(\mathcal{A})}(d)$ is invertible and generated by the image of $\psi _{univ} : \pi ^*\mathcal{A}_ d \to \mathcal{O}_{\underline{\text{Proj}}_ S(\mathcal{A})}(d)$. Recall that $U_ d$ represents the functor $F_ d$, see the proof of Lemma 27.16.5. Hence the result will follow if we can show the quadruple $(U_ d \to S, d, \mathcal{O}_{U_ d}(d), \psi _{univ}|_{\mathcal{A}^{(d)}})$ is the universal family, i.e., the representing object in $F_ d(U_ d)$. We may do this after restricting to an affine open of $S$ because (a) the formation of the functors $F_ d$ commutes with base change (see Lemma 27.16.1), and (b) the pair $(\bigoplus _{n \in \mathbf{Z}} \mathcal{O}_{\underline{\text{Proj}}_ S(\mathcal{A})}(n), \psi _{univ})$ is constructed by glueing over affine opens in $S$ (see Lemma 27.15.5). Hence we may assume that $S$ is affine. In this case the functor of quadruples $F_ d$ and the functor of triples $F_ d$ agree (see proof of Lemma 27.16.2) and moreover Lemma 27.12.2 shows that $(d, \mathcal{O}_{U_ d}(d), \psi ^ d)$ is the universal triple over $U_ d$. Going backwards through the identifications in the proof of Lemma 27.16.2 shows that $(U_ d \to S, d, \mathcal{O}_{U_ d}(d), \psi _{univ}|_{\mathcal{A}^{(d)}})$ is the universal quadruple as desired. $\square$