Proposition 69.15.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume that $X$ is affine and $f$ is surjective and universally closed^{1}. Then $Y$ is affine.

**Proof.**
We may and do view $f : X \to Y$ as a morphism of algebraic spaces over $\mathop{\mathrm{Spec}}(\mathbf{Z})$ (see Spaces, Definition 64.16.2). By Morphisms of Spaces, Lemma 66.9.8 we see that $Y$ is a separated algebraic space. Then by Morphisms of Spaces, Lemma 66.20.11 we find that $f$ is affine. Whereupon by Morphisms of Spaces, Lemma 66.45.7 we see that $f$ is integral.

By the preceding paragraph, we may assume $f : X \to Y$ is surjective and integral, $X$ is affine, and $Y$ is separated. Since $f$ is surjective and $X$ is quasi-compact we also deduce that $Y$ is quasi-compact.

Consider the sheaf $\mathcal{A} = f_*\mathcal{O}_ X$. This is a quasi-coherent sheaf of $\mathcal{O}_ Y$-algebras, see Morphisms of Spaces, Lemma 66.11.2. By Lemma 69.9.1 we can write $\mathcal{A} = \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i$ as a filtered colimit of finite type $\mathcal{O}_ Y$-modules. Let $\mathcal{A}_ i \subset \mathcal{A}$ be the $\mathcal{O}_ Y$-subalgebra generated by $\mathcal{F}_ i$. Since the map of algebras $\mathcal{O}_ Y \to \mathcal{A}$ is integral, we see that each $\mathcal{A}_ i$ is a finite quasi-coherent $\mathcal{O}_ Y$-algebra. Hence

is a finite morphism of algebraic spaces. Here $\underline{\mathop{\mathrm{Spec}}}$ is the construction of Morphisms of Spaces, Lemma 66.20.7. It is clear that $X = \mathop{\mathrm{lim}}\nolimits _ i X_ i$. Hence by Lemma 69.5.10 we see that for $i$ sufficiently large the scheme $X_ i$ is affine. Moreover, since $X \to Y$ factors through each $X_ i$ we see that $X_ i \to Y$ is surjective. Hence we conclude that $Y$ is affine by Lemma 69.15.1. $\square$

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## Comments (2)

Comment #1921 by Matthieu Romagny on

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