Lemma 85.15.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of affine formal algebraic spaces which is representable by algebraic spaces. Then $f$ is representable (by schemes) and affine.

**Proof.**
We will show that $f$ is affine; it will then follow that $f$ is representable and affine by Morphisms of Spaces, Lemma 65.20.3. Write $Y = \mathop{\mathrm{colim}}\nolimits Y_\mu $ and $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as in Definition 85.5.1. Let $T \to Y$ be a morphism where $T$ is a scheme over $S$. We have to show that $X \times _ Y T \to T$ is affine, see Bootstrap, Definition 78.4.1. To do this we may assume that $T$ is affine and we have to prove that $X \times _ Y T$ is affine. In this case $T \to Y$ factors through $Y_\mu \to Y$ for some $\mu $, see Lemma 85.5.4. Since $f$ is quasi-compact we see that $X \times _ Y T$ is quasi-compact (Lemma 85.13.3). Hence $X \times _ Y T \to X$ factors through $X_\lambda $ for some $\lambda $. Similarly $X_\lambda \to Y$ factors through $Y_\mu $ after increasing $\mu $. Then $X \times _ Y T = X_\lambda \times _{Y_\mu } T$. We conclude as fibre products of affine schemes are affine.
$\square$

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