Definition 66.10.1. Let $S$ be a scheme. A morphism of algebraic spaces over $S$ is called a *monomorphism* if it is an injective map of sheaves, i.e., a monomorphism in the category of sheaves on $(\mathit{Sch}/S)_{fppf}$.

## 66.10 Monomorphisms

A representable morphism $X \to Y$ of algebraic spaces is a monomorphism according to Section 66.3 if for every scheme $Z$ and morphism $Z \to Y$ the morphism $Z \times _ Y X \to Z$ is representable by a monomorphism of schemes. This means exactly that $Z \times _ Y X \to Z$ is an injective map of sheaves on $(\mathit{Sch}/S)_{fppf}$. Since this is supposed to hold for all $Z$ and all maps $Z \to Y$ this is in turn equivalent to the map $X \to Y$ being an injective map of sheaves on $(\mathit{Sch}/S)_{fppf}$. Thus we may define a monomorphism of a (possibly nonrepresentable^{1}) morphism of algebraic spaces as follows.

The following lemma shows that this also means that it is a monomorphism in the category of algebraic spaces over $S$.

Lemma 66.10.2. Let $S$ be a scheme. Let $j : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:

$j$ is a monomorphism (as in Definition 66.10.1),

$j$ is a monomorphism in the category of algebraic spaces over $S$, and

the diagonal morphism $\Delta _{X/Y} : X \to X \times _ Y X$ is an isomorphism.

**Proof.**
Note that $X \times _ Y X$ is both the fibre product in the category of sheaves on $(\mathit{Sch}/S)_{fppf}$ and the fibre product in the category of algebraic spaces over $S$, see Spaces, Lemma 64.7.3. The equivalence of (1) and (3) is a general characterization of injective maps of sheaves on any site. The equivalence of (2) and (3) is a characterization of monomorphisms in any category with fibre products.
$\square$

Lemma 66.10.3. A monomorphism of algebraic spaces is separated.

**Proof.**
This is true because an isomorphism is a closed immersion, and Lemma 66.10.2 above.
$\square$

Lemma 66.10.4. A composition of monomorphisms is a monomorphism.

**Proof.**
True because a composition of injective sheaf maps is injective.
$\square$

Lemma 66.10.5. The base change of a monomorphism is a monomorphism.

**Proof.**
This is a general fact about fibre products in a category of sheaves.
$\square$

Lemma 66.10.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent

$f$ is a monomorphism,

for every scheme $Z$ and morphism $Z \to Y$ the base change $Z \times _ Y X \to Z$ of $f$ is a monomorphism,

for every affine scheme $Z$ and every morphism $Z \to Y$ the base change $Z \times _ Y X \to Z$ of $f$ is a monomorphism,

there exists a scheme $V$ and a surjective étale morphism $V \to Y$ such that the base change $V \times _ Y X \to V$ is a monomorphism, and

there exists a Zariski covering $Y = \bigcup Y_ i$ such that each of the morphisms $f^{-1}(Y_ i) \to Y_ i$ is a monomorphism.

**Proof.**
We will use without further mention that a base change of a monomorphism is a monomorphism, see Lemma 66.10.5. In particular it is clear that (1) $\Rightarrow $ (2) $\Rightarrow $ (3) $\Rightarrow $ (4) (by taking $V$ to be a disjoint union of affine schemes étale over $Y$, see Properties of Spaces, Lemma 65.6.1). Let $V$ be a scheme, and let $V \to Y$ be a surjective étale morphism. If $V \times _ Y X \to V$ is a monomorphism, then it follows that $X \to Y$ is a monomorphism. Namely, given any cartesian diagram of sheaves

if $c$ is a surjection of sheaves, and $a$ is injective, then also $d$ is injective. Thus (4) implies (1). Proof of the equivalence of (5) and (1) is omitted. $\square$

Lemma 66.10.7. An immersion of algebraic spaces is a monomorphism. In particular, any immersion is separated.

**Proof.**
Let $f : X \to Y$ be an immersion of algebraic spaces. For any morphism $Z \to Y$ with $Z$ representable the base change $Z \times _ Y X \to Z$ is an immersion of schemes, hence a monomorphism, see Schemes, Lemma 26.23.8. Hence $f$ is representable, and a monomorphism.
$\square$

We will improve on the following lemma in Decent Spaces, Lemma 67.19.1.

Lemma 66.10.8. Let $S$ be a scheme. Let $k$ be a field and let $Z \to \mathop{\mathrm{Spec}}(k)$ be a monomorphism of algebraic spaces over $S$. Then either $Z = \emptyset $ or $Z = \mathop{\mathrm{Spec}}(k)$.

**Proof.**
By Lemmas 66.10.3 and 66.4.9 we see that $Z$ is a separated algebraic space. Hence there exists an open dense subspace $Z' \subset Z$ which is a scheme, see Properties of Spaces, Proposition 65.13.3. By Schemes, Lemma 26.23.11 we see that either $Z' = \emptyset $ or $Z' \cong \mathop{\mathrm{Spec}}(k)$. In the first case we conclude that $Z = \emptyset $ and in the second case we conclude that $Z' = Z = \mathop{\mathrm{Spec}}(k)$ as $Z \to \mathop{\mathrm{Spec}}(k)$ is a monomorphism which is an isomorphism over $Z'$.
$\square$

Lemma 66.10.9. Let $S$ be a scheme. If $X \to Y$ is a monomorphism of algebraic spaces over $S$, then $|X| \to |Y|$ is injective.

**Proof.**
Immediate from the definitions.
$\square$

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