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. $\square$
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Your email address will not be published.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 (2)
Comment #6531 by Jonas Ehrhard on
Comment #6583 by Johan on
There are also: