The Stacks project

Proof. To prove part (a) we have to check conditions (1), (2), (3) of Categories, Definition 4.20.1. Condition (1) holds exactly because we assumed that $X$ is compactifyable. Let $j_ i : X \to \overline{X}_ i$, $i = 1, 2$ be two compactifications. Then we can consider the scheme theoretic image $\overline{X}$ of $(j_1, j_2) : X \to \overline{X}_1 \times _ S \overline{X}_2$. This determines a third compactification $j : X \to \overline{X}$ which dominates both $j_ i$:

Thus (2) holds. Let $f_1, f_2 : \overline{X}_1 \to \overline{X}_2$ be two morphisms between compactifications $j_ i : X \to \overline{X}_ i$, $i = 1, 2$. Let $\overline{X} \subset \overline{X}_1$ be the equalizer of $f_1$ and $f_2$. As $\overline{X}_2 \to S$ is separated, we see that $\overline{X}$ is a closed subscheme of $\overline{X}_1$ and hence proper over $S$. Moreover, we obtain an open immersion $X \to \overline{X}$ because $f_1|_ X = f_2|_ X = \text{id}_ X$. The morphism $(X \to \overline{X}) \to (j_1 : X \to \overline{X}_1)$ given by the closed immersion $\overline{X} \to \overline{X}_1$ equalizes $f_1$ and $f_2$ which proves condition (3).

Proof of (b). Let $j : X \to \overline{X}$ be a compactification. If $\overline{X}'$ denotes the scheme theoretic closure of $X$ in $\overline{X}$, then $X$ is dense and scheme theoretically dense in $\overline{X}'$ by Morphisms, Lemma 29.7.7. This proves the first condition of Categories, Definition 4.17.3. Since we have already shown the category of compactifications of $X$ is cofiltered, the second condition of Categories, Definition 4.17.3 follows from the first (we omit the solution to this categorical exercise).

Proof of (c). After replacing $\overline{X}'$ with the scheme theoretic closure of $j'(X)$ (which doesn't change the underlying topological space) this follows from Morphisms, Lemma 29.6.8. $\square$

Proof. Let $j : X \to \overline{X}$ be a compactification of $X$ over $S$. Then we let $\overline{X}'$ be the scheme theoretic image of $(j, f) : X \to \overline{X} \times _ S Y$. The morphism $\overline{X}' \to Y$ is proper because $\overline{X} \times _ S Y \to Y$ is proper as a base change of $\overline{X} \to S$. On the other hand, since $Y$ is separated over $S$, the morphism $(1, f) : X \to X \times _ S Y$ is a closed immersion (Schemes, Lemma 26.21.10) and hence $X \to \overline{X}'$ is an open immersion by Morphisms, Lemma 29.6.8 applied to the “partial section” $s = (j, f)$ to the projection $\overline{X} \times _ S Y \to \overline{X}$. $\square$

Proof. Axiom RMS1 is trivial to verify. Let us check RMS2 holds. Suppose given a diagram

with $u : X' \to X$ an isomorphism. Then we let $Y' = Y \times _ X X'$ with the projection map $v : Y' \to Y$ (an isomorphism). We also set $\overline{Y}' = \overline{Y} \times _{\overline{X}} \overline{X}'$ with the projection map $\overline{v} : \overline{Y}' \to \overline{Y}$ It is clear that $Y' \to \overline{Y}'$ is an open immersion. The diagram

shows that axiom RMS2 holds.

Let us check RMS3 holds. Suppose given a pair of morphisms $(f, \overline{f}), (g, \overline{g}) : (X, \overline{X}) \to (Y, \overline{Y})$ of compactifications and a morphism $(v, \overline{v}) : (Y, \overline{Y}) \to (Y', \overline{Y}')$ such that $v$ is an isomorphism and such that $(v, \overline{v}) \circ (f, \overline{f}) = (v, \overline{v}) \circ (g, \overline{g})$. Then $f = g$. Hence if we let $\overline{X}' \subset \overline{X}$ be the equalizer of $\overline{f}$ and $\overline{g}$, then $(u, \overline{u}) : (X, \overline{X}') \to (X, \overline{X})$ will be a morphism of the category of compactifications such that $(f, \overline{f}) \circ (u, \overline{u}) = (g, \overline{g}) \circ (u, \overline{u})$ as desired. $\square$

Proof. Denote $\mathcal{C}$ the category of compactifications and denote $Q : \mathcal{C} \to \mathcal{C}'$ the localization functor of Categories, Lemma 4.27.16. Denote $\mathcal{D}$ the category of compactifyable schemes over $S$. It is clear from the lemma just cited and our choice of multiplicative system that we obtain a functor $\mathcal{C}' \to \mathcal{D}$. This functor is clearly essentially surjective. If $f : X \to Y$ is a morphism of compactifyable schemes, then we choose an open immersion $Y \to \overline{Y}$ into a scheme proper over $S$, and then we choose an embedding $X \to \overline{X}$ into a scheme $\overline{X}$ proper over $\overline{Y}$ (possible by Lemma 38.32.2 applied to $X \to \overline{Y}$). This gives a morphism $(X, \overline{X}) \to (Y, \overline{Y})$ of compactifications which produces our given morphism $X \to Y$. Finally, suppose given a pair of morphisms in the localized category with the same source and target: say

and

which produce the same morphism $X \to Y$ over $S$, in other words $f \circ u^{-1} = g \circ v^{-1}$. By Categories, Lemma 4.27.13 we may assume that $(X', \overline{X}') = (X'', \overline{X}'')$ and $(u, \overline{u}) = (v, \overline{v})$. In this case we can consider the equalizer $\overline{X}''' \subset \overline{X}'$ of $\overline{f}$ and $\overline{g}$. The morphism $(w, \overline{w}) : (X', \overline{X}''') \to (X', \overline{X}')$ is in the multiplicative subset and we see that $a = b$ in the localized category by precomposing with $(w, \overline{w})$. $\square$