The Stacks project

Proof. This follows from the general discussion in Section 100.3. $\square$

Proof. This follows from the general discussion in Section 100.3 and Morphisms of Spaces, Lemma 67.10.4. $\square$

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

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. $\square$

Proof. Let $f : \mathcal{X} \to \mathcal{Y}$ be a monomorphism of algebraic stacks. Suppose that $x_ i : \mathop{\mathrm{Spec}}(K_ i) \to \mathcal{X}$ be morphisms such that $f \circ x_1$ and $f \circ x_2$ define the same element of $|\mathcal{Y}|$. Applying the definition we find a common extension $\Omega$ with corresponding morphisms $c_ i : \mathop{\mathrm{Spec}}(\Omega ) \to \mathop{\mathrm{Spec}}(K_ i)$ and a $2$-isomorphism $\beta : f \circ x_1 \circ c_1 \to f \circ x_1 \circ c_2$. As $f$ is fully faithful, see Lemma 100.8.4, we can lift $\beta$ to an isomorphism $\alpha : x_1 \circ c_1 \to x_1 \circ c_2$. Hence $x_1$ and $x_2$ define the same point of $|\mathcal{X}|$ as desired. $\square$

Proof. We have $\mathcal{X} = \mathcal{X} \times _{\mathcal{X}'} \mathcal{X}$ by Lemma 100.8.4. Thus the result by applying Categories, Lemma 4.31.13. $\square$