Lemma 35.13.5. Let $\{ f_ i : X_ i \to X\} _{i \in I}$ be a family of morphisms of affine schemes. Assume the equivalent assumption of Lemma 35.13.2 hold and that moreover for any morphism of affines $Y \to X$ the map

$\coprod X_ i \times _ X Y \longrightarrow Y$

is a submersive map of topological spaces (Topology, Definition 5.6.3). Then our family of morphisms is a universal effective epimorphism in the category of schemes.

Proof. By Lemma 35.13.4 it suffices to base change our family of morphisms by $Y \to X$ with $Y$ affine. Set $Y_ i = X_ i \times _ X Y$. Let $T$ be a scheme and let $h_ i : Y_ i \to Y$ be a family of morphisms such that $h_ i \circ \text{pr}_1 = h_ j \circ \text{pr}_2$ on $Y_ i \times _ Y Y_ j$. Note that $Y$ as a set is the coequalizer of the two maps from $\coprod Y_ i \times _ Y Y_ j$ to $\coprod Y_ i$. Namely, surjectivity by the affine case of Lemma 35.13.3 and injectivity by Lemma 35.13.1. Hence there is a set map of underlying sets $h : Y \to T$ compatible with the maps $h_ i$. By the second condition of the lemma we see that $h$ is continuous! Thus if $y \in Y$ and $U \subset T$ is an affine open neighbourhood of $h(y)$, then we can find an affine open $V \subset Y$ such that $h(V) \subset U$. Setting $V_ i = Y_ i \times _ Y V = X_ i \times _ X V$ we can use the result proved in Lemma 35.13.2 to see that $h|_ V : V \to U \subset T$ comes from a unique morphism of affine schemes $h_ V : V \to U$ agreeing with $h_ i|_{V_ i}$ as morphisms of schemes for all $i$. Glueing these $h_ V$ (see Schemes, Section 26.14) gives a morphism $Y \to T$ as desired. $\square$

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