The Stacks project

Remark 66.19.4. A universally injective morphism of schemes is separated, see Morphisms, Lemma 29.10.3. This is not the case for morphisms of algebraic spaces. Namely, the algebraic space $X = \mathbf{A}^1_ k/\{ x \sim -x \mid x \not= 0\} $ constructed in Spaces, Example 64.14.1 comes equipped with a morphism $X \to \mathbf{A}^1_ k$ which maps the point with coordinate $x$ to the point with coordinate $x^2$. This is an isomorphism away from $0$, and there is a unique point of $X$ lying above $0$. As $X$ isn't separated this is a universally injective morphism of algebraic spaces which is not separated.


Comments (1)

Comment #210 by Rex on

Typo: "equipped with a morphsm"


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