Lemma 78.20.2. Assume $B \to S$ and $(U, R, s, t, c)$ as in Definition 78.20.1 (1). There are canonical $1$-morphisms $\pi : \mathcal{S}_ U \to [U/R]$, and $[U/R] \to \mathcal{S}_ B$ of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$. The composition $\mathcal{S}_ U \to \mathcal{S}_ B$ is the $1$-morphism associated to the structure morphism $U \to B$.

**Proof.**
During this proof let us denote $[U/_{\! p}R]$ the category fibred in groupoids associated to the presheaf in groupoids (78.20.0.1). By construction of the stackification there is a $1$-morphism $[U/_{\! p}R] \to [U/R]$. The $1$-morphism $\mathcal{S}_ U \to [U/R]$ is simply the composition $\mathcal{S}_ U \to [U/_{\! p}R] \to [U/R]$, where the first arrow associates to the scheme $S'/S$ and morphism $x : S' \to U$ over $S$ the object $x \in U(S')$ of the fibre category of $[U/_{\! p}R]$ over $S'$.

To construct the $1$-morphism $[U/R] \to \mathcal{S}_ B$ it is enough to construct the $1$-morphism $[U/_{\! p}R] \to \mathcal{S}_ B$, see Stacks, Lemma 8.9.2. On objects over $S'/S$ we just use the map

coming from the structure morphism $U \to B$. And clearly, if $a \in R(S')$ is an “arrow” with source $s(a) \in U(S')$ and target $t(a) \in U(S')$, then since $s$ and $t$ are morphisms *over* $B$ these both map to the same element $\overline{a}$ of $B(S')$. Hence we can map an arrow $a \in R(S')$ to the identity morphism of $\overline{a}$. (This is good because the fibre category $(\mathcal{S}_ B)_{S'}$ only contains identities.) We omit the verification that this rule is compatible with pullback on these split fibred categories, and hence defines a $1$-morphism $[U/_{\! p}R] \to \mathcal{S}_ B$ as desired.

We omit the verification of the last statement. $\square$

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