Removing the hypothesis that $j$ is a monomorphism was observed in an email from Matthew Emerton dates June 15, 2016

Lemma 91.15.5. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $j : \mathcal X \to \mathcal Y$ be a $1$-morphism of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Assume $j$ is representable by algebraic spaces. Then, if $\mathcal{Y}$ is a stack in groupoids (resp. an algebraic stack), so is $\mathcal{X}$.

Proof. The statement on algebraic stacks will follow from the statement on stacks in groupoids by Lemma 91.15.4. If $j$ is representable by algebraic spaces, then $j$ is faithful on fibre categories and for each $U$ and each $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y}_ U)$ the presheaf

$(h : V \to U) \longmapsto \{ (x, \phi ) \mid x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ V), \phi : h^*y \to f(x)\} /\cong$

is an algebraic space over $U$. See Lemma 91.9.2. In particular this presheaf is a sheaf and the conclusion follows from Stacks, Lemma 8.6.11. $\square$

Comment #4911 by Robot0079 on

If I didn't make any mistakes, the following general argument illustrates why this is true. We will show that two definitions of being representable by stack coincide.

Note that effective descent can be easily proved by the notion of sieve and properties of 2-pullback. So what we really need to show is the following fact.

Let $j: X \to Y$ be a morphism such that the pullback by any scheme is stack. If diagonal $\Delta_Y: Y \to Y\times Y$ is representable by sheaf, then $\Delta_X$ is also representable by sheaf.

Fix $T \to X\times X$, then $u: Isom_X \to Isom_Y$ is pull back of $\Delta_{X/Y}$. Given any scheme U, the fact that $X_U$ is a stack tells us u is representable by sheaf after fibered producting with U over Y. However, the target of this morphism is sheaf, hence $Isom_X \times_Y U$ is also a sheaf. In other words, $Isom_X \to Y$ is representable by sheaf. Note that this morphism factor through sheaf $Isom_Y$, we conclude that $Isom_X$ is sheaf.

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