The Stacks project

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

the horizontal arrows are open and surjective, and moreover

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. $\square$

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

Proof. Omitted. $\square$

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

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 $\beta = \text{id}_ f \star \alpha$. $\square$

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

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 dependent on the subfield

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

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. $\square$

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. $\square$