Lemma 66.19.1. Let $S$ be a scheme. Let $f : X \to Y$ be a representable morphism of algebraic spaces over $S$. Then $f$ is universally injective (in the sense of Section 66.3) if and only if for all fields $K$ the map $X(K) \to Y(K)$ is injective.

## 66.19 Universally injective morphisms

We have already defined in Section 66.3 what it means for a representable morphism of algebraic spaces to be universally injective. For a field $K$ over $S$ (recall this means that we are given a structure morphism $\mathop{\mathrm{Spec}}(K) \to S$) and an algebraic space $X$ over $S$ we write $X(K) = \mathop{\mathrm{Mor}}\nolimits _ S(\mathop{\mathrm{Spec}}(K), X)$. We first translate the condition for representable morphisms into a condition on the functor of points.

**Proof.**
We are going to use Morphisms, Lemma 29.10.2 without further mention. Suppose that $f$ is universally injective. Then for any field $K$ and any morphism $\mathop{\mathrm{Spec}}(K) \to Y$ the morphism of schemes $\mathop{\mathrm{Spec}}(K) \times _ Y X \to \mathop{\mathrm{Spec}}(K)$ is universally injective. Hence there exists at most one section of the morphism $\mathop{\mathrm{Spec}}(K) \times _ Y X \to \mathop{\mathrm{Spec}}(K)$. Hence the map $X(K) \to Y(K)$ is injective. Conversely, suppose that for every field $K$ the map $X(K) \to Y(K)$ is injective. Let $T \to Y$ be a morphism from a scheme into $Y$, and consider the base change $f_ T : T \times _ Y X \to T$. For any field $K$ we have

by definition of the fibre product, and hence the injectivity of $X(K) \to Y(K)$ guarantees the injectivity of $(T \times _ Y X)(K) \to T(K)$ which means that $f_ T$ is universally injective as desired. $\square$

Next, we translate the property that the transformation between field valued points is injective into something more geometric.

Lemma 66.19.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:

the map $X(K) \to Y(K)$ is injective for every field $K$ over $S$

for every morphism $Y' \to Y$ of algebraic spaces over $S$ the induced map $|Y' \times _ Y X| \to |Y'|$ is injective, and

the diagonal morphism $X \to X \times _ Y X$ is surjective.

**Proof.**
Assume (1). Let $g : Y' \to Y$ be a morphism of algebraic spaces, and denote $f' : Y' \times _ Y X \to Y'$ the base change of $f$. Let $K_ i$, $i = 1, 2$ be fields and let $\varphi _ i : \mathop{\mathrm{Spec}}(K_ i) \to Y' \times _ Y X$ be morphisms such that $f' \circ \varphi _1$ and $f' \circ \varphi _2$ define the same element of $|Y'|$. By definition this means there exists a field $\Omega $ and embeddings $\alpha _ i : K_ i \subset \Omega $ such that the two morphisms $f' \circ \varphi _ i \circ \alpha _ i : \mathop{\mathrm{Spec}}(\Omega ) \to Y'$ are equal. Here is the corresponding commutative diagram

In particular the compositions $g \circ f' \circ \varphi _ i \circ \alpha _ i$ are equal. By assumption (1) this implies that the morphism $g' \circ \varphi _ i \circ \alpha _ i$ are equal, where $g' : Y' \times _ Y X \to X$ is the projection. By the universal property of the fibre product we conclude that the morphisms $\varphi _ i \circ \alpha _ i : \mathop{\mathrm{Spec}}(\Omega ) \to Y' \times _ Y X$ are equal. In other words $\varphi _1$ and $\varphi _2$ define the same point of $Y' \times _ Y X$. We conclude that (2) holds.

Assume (2). Let $K$ be a field over $S$, and let $a, b : \mathop{\mathrm{Spec}}(K) \to X$ be two morphisms such that $f \circ a = f \circ b$. Denote $c : \mathop{\mathrm{Spec}}(K) \to Y$ the common value. By assumption $|\mathop{\mathrm{Spec}}(K) \times _{c, Y} X| \to |\mathop{\mathrm{Spec}}(K)|$ is injective. This means there exists a field $\Omega $ and embeddings $\alpha _ i : K \to \Omega $ such that

is commutative. Composing with the projection to $\mathop{\mathrm{Spec}}(K)$ we see that $\alpha _1 = \alpha _2$. Denote the common value $\alpha $. Then we see that $\{ \alpha : \mathop{\mathrm{Spec}}(\Omega ) \to \mathop{\mathrm{Spec}}(K)\} $ is a fpqc covering of $\mathop{\mathrm{Spec}}(K)$ such that the two morphisms $a, b$ become equal on the members of the covering. By Properties of Spaces, Proposition 65.17.1 we conclude that $a = b$. We conclude that (1) holds.

Assume (3). Let $x, x' \in |X|$ be a pair of points such that $f(x) = f(x')$ in $|Y|$. By Properties of Spaces, Lemma 65.4.3 we see there exists a $x'' \in |X \times _ Y X|$ whose projections are $x$ and $x'$. By assumption and Properties of Spaces, Lemma 65.4.4 there exists a $x''' \in |X|$ with $\Delta _{X/Y}(x''') = x''$. Thus $x = x'$. In other words $f$ is injective. Since condition (3) is stable under base change we see that $f$ satisfies (2).

Assume (2). Then in particular $|X \times _ Y X| \to |X|$ is injective which implies immediately that $|\Delta _{X/Y}| : |X| \to |X \times _ Y X|$ is surjective, which implies that $\Delta _{X/Y}$ is surjective by Properties of Spaces, Lemma 65.4.4. $\square$

By the two lemmas above the following definition does not conflict with the already defined notion of a universally injective representable morphism of algebraic spaces.

Definition 66.19.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. We say $f$ is *universally injective* if for every morphism $Y' \to Y$ the induced map $|Y' \times _ Y X| \to |Y'|$ is injective.

To be sure this means that any or all of the equivalent conditions of Lemma 66.19.2 hold.

Remark 66.19.4. A universally injective morphism of schemes is separated, see Morphisms, Lemma 29.10.3. This is not the case for morphisms of algebraic spaces. Namely, the algebraic space $X = \mathbf{A}^1_ k/\{ x \sim -x \mid x \not= 0\} $ constructed in Spaces, Example 64.14.1 comes equipped with a morphism $X \to \mathbf{A}^1_ k$ which maps the point with coordinate $x$ to the point with coordinate $x^2$. This is an isomorphism away from $0$, and there is a unique point of $X$ lying above $0$. As $X$ isn't separated this is a universally injective morphism of algebraic spaces which is not separated.

Lemma 66.19.5. The base change of a universally injective morphism is universally injective.

**Proof.**
Omitted. Hint: This is formal.
$\square$

Lemma 66.19.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:

$f$ is universally injective,

for every scheme $Z$ and any morphism $Z \to Y$ the morphism $Z \times _ Y X \to Z$ is universally injective,

for every affine scheme $Z$ and any morphism $Z \to Y$ the morphism $Z \times _ Y X \to Z$ is universally injective,

there exists a scheme $Z$ and a surjective morphism $Z \to Y$ such that $Z \times _ Y X \to Z$ is universally injective, and

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

**Proof.**
We will use that being universally injective is preserved under base change (Lemma 66.19.5) without further mention in this proof. It is clear that (1) $\Rightarrow $ (2) $\Rightarrow $ (3) $\Rightarrow $ (4).

Assume $g : Z \to Y$ as in (4). Let $y : \mathop{\mathrm{Spec}}(K) \to Y$ be a morphism from the spectrum of a field into $Y$. By assumption we can find an extension field $\alpha : K \subset K'$ and a morphism $z : \mathop{\mathrm{Spec}}(K') \to Z$ such that $y \circ \alpha = g \circ z$ (with obvious abuse of notation). By assumption the morphism $Z \times _ Y X \to Z$ is universally injective, hence there is at most one lift of $g \circ z : \mathop{\mathrm{Spec}}(K') \to Y$ to a morphism into $X$. Since $\{ \alpha : \mathop{\mathrm{Spec}}(K') \to \mathop{\mathrm{Spec}}(K)\} $ is a fpqc covering this implies there is at most one lift of $y : \mathop{\mathrm{Spec}}(K) \to Y$ to a morphism into $X$, see Properties of Spaces, Proposition 65.17.1. Thus we see that (1) holds.

We omit the verification that (5) is equivalent to (1). $\square$

Lemma 66.19.7. A composition of universally injective morphisms is universally injective.

**Proof.**
Omitted.
$\square$

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