Lemma 69.15.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume that $f$ is surjective and finite, and assume that $X$ is affine. Then $Y$ is affine.

**Proof.**
We may and do view $f : X \to Y$ as a morphism of algebraic space over $\mathop{\mathrm{Spec}}(\mathbf{Z})$ (see Spaces, Definition 64.16.2). Note that a finite morphism is affine and universally closed, see Morphisms of Spaces, Lemma 66.45.7. By Morphisms of Spaces, Lemma 66.9.8 we see that $Y$ is a separated algebraic space. As $f$ is surjective and $X$ is quasi-compact we see that $Y$ is quasi-compact.

By Lemma 69.11.3 we can write $X = \mathop{\mathrm{lim}}\nolimits X_ a$ with each $X_ a \to Y$ finite and of finite presentation. By Lemma 69.5.10 we see that $X_ a$ is affine for $a$ large enough. Hence we may and do assume that $f : X \to Y$ is finite, surjective, and of finite presentation.

By Proposition 69.8.1 we may write $Y = \mathop{\mathrm{lim}}\nolimits Y_ i$ as a directed limit of algebraic spaces of finite presentation over $\mathbf{Z}$. By Lemma 69.7.1 we can find $0 \in I$ and a morphism $X_0 \to Y_0$ of finite presentation such that $X_ i = X_0 \times _{Y_0} Y_ i$ for $i \geq 0$ and such that $X = \mathop{\mathrm{lim}}\nolimits _ i X_ i$. By Lemma 69.6.7 we see that $X_ i \to Y_ i$ is finite for $i$ large enough. By Lemma 69.6.4 we see that $X_ i \to Y_ i$ is surjective for $i$ large enough. By Lemma 69.5.10 we see that $X_ i$ is affine for $i$ large enough. Hence for $i$ large enough we can apply Cohomology of Spaces, Lemma 68.17.3 to conclude that $Y_ i$ is affine. This implies that $Y$ is affine and we conclude. $\square$

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## Comments (1)

Comment #8260 by Matthieu Romagny on

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