Monomorphisms of stacks are injective on points.

Lemma 99.8.5. A monomorphism of algebraic stacks induces an injective map of sets of points.

Proof. Let $f : \mathcal{X} \to \mathcal{Y}$ be a monomorphism of algebraic stacks. Suppose that $x_ i : \mathop{\mathrm{Spec}}(K_ i) \to \mathcal{X}$ be morphisms such that $f \circ x_1$ and $f \circ x_2$ define the same element of $|\mathcal{Y}|$. Applying the definition we find a common extension $\Omega$ with corresponding morphisms $c_ i : \mathop{\mathrm{Spec}}(\Omega ) \to \mathop{\mathrm{Spec}}(K_ i)$ and a $2$-isomorphism $\beta : f \circ x_1 \circ c_1 \to f \circ x_1 \circ c_2$. As $f$ is fully faithful, see Lemma 99.8.4, we can lift $\beta$ to an isomorphism $\alpha : x_1 \circ c_1 \to x_1 \circ c_2$. Hence $x_1$ and $x_2$ define the same point of $|\mathcal{X}|$ as desired. $\square$

Comment #823 by on

Suggested slogan: A monomorphism of algebraic stacks induces an injective map of sets of points.

Comment #6501 by Mitchell Faulk on

It seems to me that we would like, using that $f$ is fully faithful, to lift $\beta$ to an isomorphism $\alpha : x_1 \circ c_1 \to x_2 \circ c_2$, not $\alpha : f \circ x_1 \circ c_1 \to f \circ x_1 \circ c_2$.

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