Lemma 76.3.2. A radicial morphism of algebraic spaces is universally injective.
Proof. Let $S$ be a scheme. Let $f : X \to Y$ be a radicial morphism of algebraic spaces over $S$. It is clear from the definition that given a morphism $\mathop{\mathrm{Spec}}(K) \to Y$ there is at most one lift of this morphism to a morphism into $X$. Hence we conclude that $f$ is universally injective by Morphisms of Spaces, Lemma 67.19.2. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)