Lemma 100.14.5. 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 99.5). 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 99.4.3 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 100.3.3

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

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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