The Stacks project

Lemma 98.8.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent:

  1. $f$ is a monomorphism,

  2. $f$ is fully faithful,

  3. the diagonal $\Delta _ f : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ is an equivalence, and

  4. there exists an algebraic space $W$ and a surjective, flat morphism $W \to \mathcal{Y}$ which is locally of finite presentation such that $V = \mathcal{X} \times _\mathcal {Y} W$ is an algebraic space, and the morphism $V \to W$ is a monomorphism of algebraic spaces.

Proof. The equivalence of (1) and (4) follows from the general discussion in Section 98.3 and in particular Lemmas 98.3.1 and 98.3.3.

The equivalence of (2) and (3) is Categories, Lemma 4.35.9.

Assume the equivalent conditions (2) and (3). Then $f$ is representable by algebraic spaces according to Algebraic Stacks, Lemma 92.15.2. Moreover, the $2$-Yoneda lemma combined with the fully faithfulness implies that for every scheme $T$ the functor

\[ \mathop{Mor}\nolimits (T, \mathcal{X}) \longrightarrow \mathop{Mor}\nolimits (T, \mathcal{Y}) \]

is fully faithful. Hence given a morphism $y : T \to \mathcal{Y}$ there exists up to unique $2$-isomorphism at most one morphism $x : T \to \mathcal{X}$ such that $y \cong f \circ x$. In particular, given a morphism of schemes $h : T' \to T$ there exists at most one lift $\tilde h : T' \to T \times _\mathcal {Y} \mathcal{X}$ of $h$. Thus $T \times _\mathcal {Y} \mathcal{X} \to T$ is a monomorphism of algebraic spaces, which proves that (1) holds.

Finally, assume that (1) holds. Then for any scheme $T$ and morphism $y : T \to \mathcal{Y}$ the fibre product $T \times _\mathcal {Y} \mathcal{X}$ is an algebraic space, and $T \times _\mathcal {Y} \mathcal{X} \to T$ is a monomorphism. Hence there exists up to unique isomorphism exactly one pair $(x, \alpha )$ where $x : T \to \mathcal{X}$ is a morphism and $\alpha : f \circ x \to y$ is a $2$-morphism. Applying the $2$-Yoneda lemma this says exactly that $f$ is fully faithful, i.e., that (2) holds. $\square$


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