The Stacks project

Proof. Proof of (1), (2), and (3). Choose an algebraic space $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Then $G = U \times _\mathcal {X} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is an algebraic space over $U$ (Lemma 101.5.1). In fact, $G$ is a group algebraic space over $U$ by the group law on relative inertia constructed in Remark 101.5.2. Moreover, $G \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is surjective and smooth as a base change of $U \to \mathcal{X}$. Finally, the base change of $e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ by $G \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is the identity $U \to G$ of $G/U$. Thus the equivalence of (a) and (c) follows from Groupoids in Spaces, Lemma 78.6.1. Since $\Delta _{f, 2}$ is the diagonal of $\Delta _ f$ we have (b) $\Leftrightarrow$ (c) by definition.

Proof of (4) and (5). Recall that (4)(b) means $\Delta _ f$ is unramified and (5)(b) means that $\Delta _ f$ is locally quasi-finite. Choose a scheme $Z$ and a morphism $a : Z \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$. Then $a = (x_1, x_2, \alpha )$ where $x_ i : Z \to \mathcal{X}$ and $\alpha : f \circ x_1 \to f \circ x_2$ is a $2$-morphism. Recall that

are cartesian squares. By Lemma 101.5.4 the algebraic space $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha (x_1, x_2)$ is a pseudo torsor for $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2)$ over $Z$. Thus the equivalences in (4) and (5) follow from Groupoids in Spaces, Lemma 78.9.5. $\square$

Proof. We first prove the equivalence of (1) and (2). Namely, $f$ is representable by algebraic spaces if and only if $f$ is faithful, see Algebraic Stacks, Lemma 94.15.2. On the other hand, $f$ is faithful if and only if for every object $x$ of $\mathcal{X}$ over a scheme $T$ the functor $f$ induces an injection $\mathit{Isom}_\mathcal {X}(x, x) \to \mathit{Isom}_\mathcal {Y}(f(x), f(x))$, which happens if and only if the kernel $K$ is trivial, which happens if and only if $e : T \to K$ is an isomorphism for every $x : T \to \mathcal{X}$. Since $K = T \times _{x, \mathcal{X}} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ as discussed above, this proves the equivalence of (1) and (2). To prove the equivalence of (2) and (3), by the discussion above, it suffices to note that a group stack is trivial if and only if its identity section is an isomorphism. Finally, the equivalence of (3) and (4) follows from the definitions: in the proof of Lemma 101.5.1 we have seen that the kernel in (4) corresponds to the fibre product $T \times _{x, \mathcal{X}} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ over $T$. $\square$

Proof. Recall from Properties of Stacks, Lemma 100.8.4 that a morphism of algebraic stacks is a monomorphism if and only if its diagonal is an isomorphism of stacks. Thus Lemma 101.6.2 can be rephrased as saying that a morphism is representable by algebraic spaces if the diagonal is a monomorphism. In particular, it shows that condition (3) of Lemma 101.3.4 is actually an if and only if, i.e., a morphism of algebraic stacks is representable by algebraic spaces if and only if its diagonal is a monomorphism. $\square$

Proof. Follows from Lemmas 101.3.5, 101.3.6, and 101.3.7 applied to $\Delta _{f, 1}$. $\square$

Proof. Proof of (1). Assume that $\Delta _{f, 2}$ and $\Delta _{f, 1}$ are universally closed. Then $\Delta _{f, 1}$ is separated and universally closed by Lemma 101.6.4. By Morphisms of Spaces, Lemma 67.9.7 and Algebraic Stacks, Lemma 94.10.9 we see that $\Delta _{f, 1}$ is quasi-compact. Hence it is quasi-compact, separated, universally closed and locally of finite type (by Lemma 101.3.3) so proper. This proves “$\Leftarrow$” of (1). The proof of the implication in the other direction is omitted.

Proof of (2). This follows immediately from Lemma 101.6.4.

Proof of (3). This follows from the fact that $\Delta _{f, 2}$ is always locally quasi-finite by Lemma 101.3.4 applied to $\Delta _ f = \Delta _{f, 1}$.

Proof of (4). This follows from the fact that $\Delta _{f, 2}$ is always unramified as Lemma 101.3.4 applied to $\Delta _ f = \Delta _{f, 1}$ shows that $\Delta _{f, 2}$ is locally of finite type and a monomorphism. See More on Morphisms of Spaces, Lemma 76.14.8. $\square$

Proof. Proof of (1). Observe that the diagram

is cartesian. Hence this follows from the fact that $f$ is separated (resp. quasi-separated, resp. quasi-DM, resp. DM) if and only if the diagonal is proper (resp. quasi-compact and quasi-separated, resp. locally quasi-finite, resp. unramified).

Proof of (2). This is true because

hence the morphism in (2) is a base change of the morphism in (1). $\square$