Lemma 66.19.1. Let $S$ be a scheme. Let $f : X \to Y$ be a representable morphism of algebraic spaces over $S$. Then $f$ is universally injective (in the sense of Section 66.3) if and only if for all fields $K$ the map $X(K) \to Y(K)$ is injective.

**Proof.**
We are going to use Morphisms, Lemma 29.10.2 without further mention. Suppose that $f$ is universally injective. Then for any field $K$ and any morphism $\mathop{\mathrm{Spec}}(K) \to Y$ the morphism of schemes $\mathop{\mathrm{Spec}}(K) \times _ Y X \to \mathop{\mathrm{Spec}}(K)$ is universally injective. Hence there exists at most one section of the morphism $\mathop{\mathrm{Spec}}(K) \times _ Y X \to \mathop{\mathrm{Spec}}(K)$. Hence the map $X(K) \to Y(K)$ is injective. Conversely, suppose that for every field $K$ the map $X(K) \to Y(K)$ is injective. Let $T \to Y$ be a morphism from a scheme into $Y$, and consider the base change $f_ T : T \times _ Y X \to T$. For any field $K$ we have

by definition of the fibre product, and hence the injectivity of $X(K) \to Y(K)$ guarantees the injectivity of $(T \times _ Y X)(K) \to T(K)$ which means that $f_ T$ is universally injective as desired. $\square$

## 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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)