## 99.8 Monomorphisms of algebraic stacks

We define a monomorphism of algebraic stacks in the following way. We will see in Lemma 99.8.4 that this is compatible with the corresponding $2$-category theoretic notion.

Definition 99.8.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. We say $f$ is a monomorphism if it is representable by algebraic spaces and a monomorphism in the sense of Section 99.3.

First some basic lemmas.

Lemma 99.8.2. Let $\mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $\mathcal{Z} \to \mathcal{Y}$ be a monomorphism. Then $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X}$ is a monomorphism.

Proof. This follows from the general discussion in Section 99.3. $\square$

Lemma 99.8.3. Compositions of monomorphisms of algebraic stacks are monomorphisms.

Proof. This follows from the general discussion in Section 99.3 and Morphisms of Spaces, Lemma 66.10.4. $\square$

Lemma 99.8.4. 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 99.3 and in particular Lemmas 99.3.1 and 99.3.3.

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 93.15.2. 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. $\square$

Lemma 99.8.5. A monomorphism of algebraic stacks induces an injective map of sets of points.

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 99.8.4, 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. $\square$

Lemma 99.8.6. Let $\mathcal{X} \to \mathcal{X}' \to \mathcal{Y}$ be morphisms of algebraic stacks. If $\mathcal{X} \to \mathcal{X}'$ is a monomorphism then the canonical diagram

$\xymatrix{ \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \times _\mathcal {Y} \mathcal{X} \ar[d] \\ \mathcal{X}' \ar[r] & \mathcal{X}' \times _\mathcal {Y} \mathcal{X}' }$

is a fibre product square.

Proof. We have $\mathcal{X} = \mathcal{X} \times _{\mathcal{X}'} \mathcal{X}$ by Lemma 99.8.4. Thus the result by applying Categories, Lemma 4.31.13. $\square$

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