Lemma 69.5.12. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $X = \mathop{\mathrm{lim}}\nolimits X_ i$ be a directed limit of algebraic spaces over $B$ with affine transition morphisms. Let $Y \to X$ be a morphism of algebraic spaces over $B$.

1. If $Y \to X$ is a closed immersion, $X_ i$ quasi-compact, and $Y \to B$ locally of finite type, then $Y \to X_ i$ is a closed immersion for $i$ large enough.

2. If $Y \to X$ is an immersion, $X_ i$ quasi-separated, $Y \to B$ locally of finite type, and $Y$ quasi-compact, then $Y \to X_ i$ is an immersion for $i$ large enough.

3. If $Y \to X$ is an isomorphism, $X_ i$ quasi-compact, $X_ i \to B$ locally of finite type, the transition morphisms $X_{i'} \to X_ i$ are closed immersions, and $Y \to B$ is locally of finite presentation, then $Y \to X_ i$ is an isomorphism for $i$ large enough.

4. If $Y \to X$ is a monomorphism, $X_ i$ quasi-separated, $Y \to B$ locally of finite type, and $Y$ quasi-compact, then $Y \to X_ i$ is a monomorphism for $i$ large enough.

Proof. Proof of (1). Choose $0 \in I$. As $X_0$ is quasi-compact, we can choose an affine scheme $W$ and an étale morphism $W \to B$ such that the image of $|X_0| \to |B|$ is contained in $|W| \to |B|$. Choose an affine scheme $U_0$ and an étale morphism $U_0 \to X_0 \times _ B W$ such that $U_0 \to X_0$ is surjective. (This is possible by our choice of $W$ and the fact that $X_0$ is quasi-compact; details omitted.) Let $V \to Y$, resp. $U \to X$, resp. $U_ i \to X_ i$ be the base change of $U_0 \to X_0$ (for $i \geq 0$). It suffices to prove that $V \to U_ i$ is a closed immersion for $i$ sufficiently large. Thus we reduce to proving the result for $V \to U = \mathop{\mathrm{lim}}\nolimits U_ i$ over $W$. This follows from the case of schemes, which is Limits, Lemma 32.4.16.

Proof of (2). Choose $0 \in I$. Choose a quasi-compact open subspace $X'_0 \subset X_0$ such that $Y \to X_0$ factors through $X'_0$. After replacing $X_ i$ by the inverse image of $X'_0$ for $i \geq 0$ we may assume all $X_ i'$ are quasi-compact and quasi-separated. Let $U \subset X$ be a quasi-compact open such that $Y \to X$ factors through a closed immersion $Y \to U$ ($U$ exists as $Y$ is quasi-compact). By Lemma 69.5.7 we may assume that $U = \mathop{\mathrm{lim}}\nolimits U_ i$ with $U_ i \subset X_ i$ quasi-compact open. By part (1) we see that $Y \to U_ i$ is a closed immersion for some $i$. Thus (2) holds.

Proof of (3). Choose $0 \in I$. Choose an affine scheme $U_0$ and a surjective étale morphism $U_0 \to X_0$. Set $U_ i = X_ i \times _{X_0} U_0$, $U = X \times _{X_0} U_0 = Y \times _{X_0} U_0$. Then $U = \mathop{\mathrm{lim}}\nolimits U_ i$ is a limit of affine schemes, the transition maps of the system are closed immersions, and $U \to U_0$ is of finite presentation (because $U \to B$ is locally of finite presentation and $U_0 \to B$ is locally of finite type and Morphisms of Spaces, Lemma 66.28.9). Thus we've reduced to the following algebra fact: If $A = \mathop{\mathrm{lim}}\nolimits A_ i$ is a directed colimit of $R$-algebras with surjective transition maps and $A$ of finite presentation over $A_0$, then $A = A_ i$ for some $i$. Namely, write $A = A_0/(f_1, \ldots , f_ n)$. Pick $i$ such that $f_1, \ldots , f_ n$ map to zero under the surjective map $A_0 \to A_ i$.

Proof of (4). Set $Z_ i = Y \times _{X_ i} Y$. As the transition morphisms $X_{i'} \to X_ i$ are affine hence separated, the transition morphisms $Z_{i'} \to Z_ i$ are closed immersions, see Morphisms of Spaces, Lemma 66.4.5. We have $\mathop{\mathrm{lim}}\nolimits Z_ i = Y \times _ X Y = Y$ as $Y \to X$ is a monomorphism. Choose $0 \in I$. Since $Y \to X_0$ is locally of finite type (Morphisms of Spaces, Lemma 66.23.6) the morphism $Y \to Z_0$ is locally of finite presentation (Morphisms of Spaces, Lemma 66.28.10). The morphisms $Z_ i \to Z_0$ are locally of finite type (they are closed immersions). Finally, $Z_ i = Y \times _{X_ i} Y$ is quasi-compact as $X_ i$ is quasi-separated and $Y$ is quasi-compact. Thus part (3) applies to $Y = \mathop{\mathrm{lim}}\nolimits _{i \geq 0} Z_ i$ over $Z_0$ and we conclude $Y = Z_ i$ for some $i$. This proves (4) and the lemma. $\square$

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