## 101.14 Universally injective morphisms

Let $f$ be a morphism of algebraic stacks which is representable by algebraic spaces. In Properties of Stacks, Section 100.3 we have defined what it means for $f$ to be universally injective. Here is another characterization.

Lemma 101.14.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent

1. $f$ is universally injective (as in Properties of Stacks, Section 100.3), and

2. for every morphism of algebraic stacks $\mathcal{Z} \to \mathcal{Y}$ the map $|\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \to |\mathcal{Z}|$ is injective.

Proof. Assume (1), and let $\mathcal{Z} \to \mathcal{Y}$ be as in (2). Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Z}$. By assumption the morphism $V \times _\mathcal {Y} \mathcal{X} \to V$ of algebraic spaces is universally injective, in particular the map $|V \times _\mathcal {Y} \mathcal{X}| \to |V|$ is injective. By Properties of Stacks, Section 100.4 in the commutative diagram

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

the horizontal arrows are open and surjective, and moreover

$|V \times _\mathcal {Y} \mathcal{X}| \longrightarrow |V| \times _{|\mathcal{Z}|} |\mathcal{Z} \times _\mathcal {Y} \mathcal{X}|$

is surjective. Hence as the left vertical arrow is injective it follows that the right vertical arrow is injective. This proves (2). The implication (2) $\Rightarrow$ (1) follows from the definitions. $\square$

Thus we may use the following natural definition.

Definition 101.14.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. We say $f$ is universally injective if for every morphism of algebraic stacks $\mathcal{Z} \to \mathcal{Y}$ the map

$|\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \to |\mathcal{Z}|$

is injective.

Lemma 101.14.3. The base change of a universally injective morphism of algebraic stacks by any morphism of algebraic stacks is universally injective.

Proof. This is immediate from the definition. $\square$

Lemma 101.14.4. The composition of a pair of universally injective morphisms of algebraic stacks is universally injective.

Proof. Omitted. $\square$

Lemma 101.14.5. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent

1. $f$ is universally injective,

2. $\Delta : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ is surjective, and

3. for an algebraically closed field, for $x_1, x_2 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$, and for a $2$-arrow $\beta : f \circ x_1 \to f \circ x_2$ there is a $2$-arrow $\alpha : x_1 \to x_2$ with $\beta = \text{id}_ f \star \alpha$.

Proof. (1) $\Rightarrow$ (2). If $f$ is universally injective, then the first projection $|\mathcal{X} \times _\mathcal {Y} \mathcal{X}| \to |\mathcal{X}|$ is injective, which implies that $|\Delta |$ is surjective.

(2) $\Rightarrow$ (1). Assume $\Delta$ is surjective. Then any base change of $\Delta$ is surjective (see Properties of Stacks, Section 100.5). Since the diagonal of a base change of $f$ is a base change of $\Delta$, we see that it suffices to show that $|\mathcal{X}| \to |\mathcal{Y}|$ is injective. If not, then by Properties of Stacks, Lemma 100.4.3 we find that the first projection $|\mathcal{X} \times _\mathcal {Y} \mathcal{X}| \to |\mathcal{X}|$ is not injective. Of course this means that $|\Delta |$ is not surjective.

(3) $\Rightarrow$ (2). Let $t \in |\mathcal{X} \times _\mathcal {Y} \mathcal{X}|$. Then we can represent $t$ by a morphism $t : \mathop{\mathrm{Spec}}(k) \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ with $k$ an algebraically closed field. By our construction of $2$-fibre products we can represent $t$ by $(x_1, x_2, \beta )$ where $x_1, x_2 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ and $\beta : f \circ x_1 \to f \circ x_2$ is a $2$-morphism. Then (3) implies that there is a $2$-morphism $\alpha : x_1 \to x_2$ mapping to $\beta$. This exactly means that $\Delta (x_1) = (x_1, x_1, \text{id})$ is isomorphic to $t$. Hence (2) holds.

(2) $\Rightarrow$ (3). Let $x_1, x_2 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be morphisms with $k$ an algebraically closed field. Let $\beta : f \circ x_1 \to f \circ x_2$ be a $2$-morphism. As in the previous paragraph, we obtain a morphism $t = (x_1, x_2, \beta ) : \mathop{\mathrm{Spec}}(k) \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$. By Lemma 101.3.3

$T = \mathcal{X} \times _{\Delta , \mathcal{X} \times _\mathcal {Y} \mathcal{X}, t} \mathop{\mathrm{Spec}}(k)$

is an algebraic space locally of finite type over $\mathop{\mathrm{Spec}}(k)$. Condition (2) implies that $T$ is nonempty. Then since $k$ is algebraically closed, there is a $k$-point in $T$. Unwinding the definitions this means there is a morphism $\alpha : x_1 \to x_2$ in $\mathop{\mathrm{Mor}}\nolimits (\mathop{\mathrm{Spec}}(k), \mathcal{X})$ such that $\beta = \text{id}_ f \star \alpha$. $\square$

Lemma 101.14.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a universally injective morphism of algebraic stacks. Let $y : \mathop{\mathrm{Spec}}(k) \to \mathcal{Y}$ be a morphism where $k$ is an algebraically closed field. If $y$ is in the image of $|\mathcal{X}| \to |\mathcal{Y}|$, then there is a morphism $x : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ with $y = f \circ x$.

Proof. We first remark this lemma is not a triviality, because the assumption that $y$ is in the image of $|f|$ means only that we can lift $y$ to a morphism into $\mathcal{X}$ after possibly replacing $k$ by an extension field. To prove the lemma we may base change $f$ by $y$, hence we may assume we have a nonempty algebraic stack $\mathcal{X}$ and a universally injective morphism $\mathcal{X} \to \mathop{\mathrm{Spec}}(k)$ and we want to find a $k$-valued point of $\mathcal{X}$. We may replace $\mathcal{X}$ by its reduction. We may choose a field $k'$ and a surjective, flat, locally finite type morphism $\mathop{\mathrm{Spec}}(k') \to \mathcal{X}$, see Properties of Stacks, Lemma 100.11.2. Since $\mathcal{X} \to \mathop{\mathrm{Spec}}(k)$ is universally injective, we find that

$\mathop{\mathrm{Spec}}(k') \times _\mathcal {X} \mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k' \otimes _ k k')$

is surjective as the base change of the surjective morphism $\Delta : \mathcal{X} \to \mathcal{X} \times _{\mathop{\mathrm{Spec}}(k)} \mathcal{X}$ (Lemma 101.14.5). Since $k$ is algebraically closed $k' \otimes _ k k'$ is a domain (Algebra, Lemma 10.49.4). Let $\xi \in \mathop{\mathrm{Spec}}(k') \times _\mathcal {X} \mathop{\mathrm{Spec}}(k')$ be a point mapping to the generic point of $\mathop{\mathrm{Spec}}(k' \otimes _ k k')$. Let $U$ be the reduced induced closed subscheme structure on the connected component of $\mathop{\mathrm{Spec}}(k') \times _\mathcal {X} \mathop{\mathrm{Spec}}(k')$ containing $\xi$. Then the two projections $U \to \mathop{\mathrm{Spec}}(k')$ are locally of finite type, as this was true for the projections $\mathop{\mathrm{Spec}}(k') \times _\mathcal {X} \mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k')$ as base changes of the morphism $\mathop{\mathrm{Spec}}(k') \to \mathcal{X}$. Applying Varieties, Proposition 33.31.1 we find that the integral closures of the two images of $k'$ in $\Gamma (U, \mathcal{O}_ U)$ are equal. Looking in $\kappa (\xi )$ means that any element of the form $\lambda \otimes 1$ is algebraically dependend on the subfield

$1 \otimes k' \subset (\text{fraction field of }k' \otimes _ k k') \subset \kappa (\xi ).$

Since $k$ is algebraically closed, this is only possible if $k' = k$ and the proof is complete. $\square$

Lemma 101.14.7. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent:

1. $f$ is universally injective,

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

Proof. The implication (1) $\Rightarrow$ (2) is immediate. Assume (2) holds. We will show that $\Delta _ f : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ is surjective, which implies (1) by Lemma 101.14.5. Consider an affine scheme $V$ and a smooth morphism $V \to \mathcal{Y}$. Since $g : V \times _\mathcal {Y} \mathcal{X} \to V$ is universally injective by (2), we see that $\Delta _ g$ is surjective. However, $\Delta _ g$ is the base change of $\Delta _ f$ by the smooth morphism $V \to \mathcal{Y}$. Since the collection of these morphisms $V \to \mathcal{Y}$ are jointly surjective, we conclude $\Delta _ f$ is surjective. $\square$

Lemma 101.14.8. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $W \to \mathcal{Y}$ be surjective, flat, and locally of finite presentation where $W$ is an algebraic space. If the base change $W \times _\mathcal {Y} \mathcal{X} \to W$ is universally injective, then $f$ is universally injective.

Proof. Observe that the diagonal $\Delta _ g$ of the morphism $g : W \times _\mathcal {Y} \mathcal{X} \to W$ is the base change of $\Delta _ f$ by $W \to \mathcal{Y}$. Hence if $\Delta _ g$ is surjective, then so is $\Delta _ f$ by Properties of Stacks, Lemma 100.3.3. Thus the lemma follows from the characterization (2) in Lemma 101.14.5. $\square$

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