## 67.10 Monomorphisms

A representable morphism $X \to Y$ of algebraic spaces is a monomorphism according to Section 67.3 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 nonrepresentable1) morphism of algebraic spaces as follows.

Definition 67.10.1. 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 $(\mathit{Sch}/S)_{fppf}$.

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

Lemma 67.10.2. Let $S$ be a scheme. Let $j : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:

1. $j$ is a monomorphism (as in Definition 67.10.1),

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

3. 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 65.7.3. 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. $\square$

Proof. This is true because an isomorphism is a closed immersion, and Lemma 67.10.2 above. $\square$

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

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

Lemma 67.10.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent

1. $f$ is a monomorphism,

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

3. 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,

4. 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

5. 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 67.10.5. 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 66.6.1). 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

$\vcenter { \xymatrix{ \mathcal{F} \ar[r]_ a \ar[d]_ b & \mathcal{G} \ar[d]^ c \\ \mathcal{H} \ar[r]^ d & \mathcal{I} } } \quad \quad \mathcal{F} = \mathcal{H} \times _\mathcal {I} \mathcal{G}$

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 67.10.7. 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 68.19.1.

Lemma 67.10.8. 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 67.10.3 and 67.4.9 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 66.13.3. 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 67.10.9. 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. $\square$

 We do not know whether any monomorphism of algebraic spaces is representable. For a discussion see More on Morphisms of Spaces, Section 76.4.

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