**Proof.**
The equivalence of (1) and (4) follows from the general discussion in Section 100.3 and in particular Lemmas 100.3.1 and 100.3.3.

The equivalence of (2) and (3) is Categories, Lemma 4.35.10.

Assume the equivalent conditions (2) and (3). Then $f$ is representable by algebraic spaces according to Algebraic Stacks, Lemma 94.15.2. Moreover, the $2$-Yoneda lemma combined with the fully faithfulness implies that for every scheme $T$ the functor

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

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.
$\square$

## Comments (0)