Lemma 91.15.2. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of algebraic stacks over $S$. The following are equivalent:

1. for $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ the functor $f : \mathcal{X}_ U \to \mathcal{Y}_ U$ is faithful,

2. the functor $f$ is faithful, and

3. $f$ is representable by algebraic spaces.

Proof. Parts (1) and (2) are equivalent by general properties of $1$-morphisms of categories fibred in groupoids, see Categories, Lemma 4.34.8. We see that (3) implies (2) by Lemma 91.9.2. Finally, assume (2). Let $U$ be a scheme. Let $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y}_ U)$. We have to prove that

$\mathcal{W} = (\mathit{Sch}/U)_{fppf} \times _{y, \mathcal{Y}} \mathcal{X}$

is representable by an algebraic space over $U$. Since $(\mathit{Sch}/U)_{fppf}$ is an algebraic stack we see from Lemma 91.14.3 that $\mathcal{W}$ is an algebraic stack. On the other hand the explicit description of objects of $\mathcal{W}$ as triples $(V, x, \alpha : y(V) \to f(x))$ and the fact that $f$ is faithful, shows that the fibre categories of $\mathcal{W}$ are setoids. Hence Proposition 91.13.3 guarantees that $\mathcal{W}$ is representable by an algebraic space. $\square$

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