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The Stacks project

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


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