Processing math: 100%

The Stacks project

Lemma 35.13.4. Let \{ f_ i : X_ i \to X\} _{i \in I} be a family of morphisms of schemes. If for every morphism Y \to X with Y affine the family of base changes g_ i : Y_ i \to Y forms an effective epimorphism, then the family of f_ i forms a universal effective epimorphism in the category of schemes.

Proof. Let Y \to X be a morphism of schemes. We have to show that the base changes g_ i : Y_ i \to Y form an effective epimorphism. To do this, assume given a scheme T and morphisms h_ i : Y_ i \to T with h_ i \circ \text{pr}_1 = h_ j \circ \text{pr}_2 on Y_ i \times _ Y Y_ j. Choose an affine open covering Y = \bigcup V_\alpha . Set V_{\alpha , i} equal to the inverse image of V_\alpha in Y_ i. Then we see that V_{\alpha , i} \to V_\alpha is the base change of f_ i by V_\alpha \to X. Thus by assumption the family of restrictions h_ i|_{V_{\alpha , i}} come from a morphism of schemes h_\alpha : V_\alpha \to T. We leave it to the reader to show that these agree on overlaps and define the desired morphism Y \to T. See discussion in Schemes, Section 26.14. \square


Comments (2)

Comment #6531 by Jonas Ehrhard on

This might be nitpicking, but for consistency reasons I think it should be "universal effective epimorphism" instead of "universally".

There are also:

  • 2 comment(s) on Section 35.13: Fpqc coverings are universal effective epimorphisms

Your email address will not be published.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.