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. $\square$

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".

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