## 86.10 Algebraization

In this section we prove a generalization of the result on dilatations from the paper of Artin [ArtinII]. We first reformulate the algebra results proved above into the language of formal algebraic spaces.

Let $S$ be a scheme. Let $V$ be a locally Noetherian formal algebraic space over $S$. We denote $\mathcal{C}_ V$ the category of formal algebraic spaces $W$ over $V$ such that the structure morphism $W \to V$ is rig-étale.

Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $T \subset |X|$ be a closed subset. Recall that $X_{/T}$ denotes the formal completion of $X$ along $T$, see Formal Spaces, Section 85.9. More generally, for any algebraic space $Y$ over $X$ we denote $Y_{/T}$ the completion of $Y$ along the inverse image of $T$ in $|Y|$, so that $Y_{/T}$ is a formal algebraic space over $X_{/T}$.

Lemma 86.10.1. Let $S$ be a scheme. Let $X$ be a locally Noetherian algebraic space over $S$. Let $T \subset |X|$ be a closed subset. If $Y \to X$ is morphism of algebraic spaces which is locally of finite type and étale over $X \setminus T$, then $Y_{/T} \to X_{/T}$ is rig-étale, i.e., $Y_{/T}$ is an object of $\mathcal{C}_{X_{/T}}$ defined above.

Proof. Choose a surjective étale morphism $U \to X$ with $U = \coprod U_ i$ a disjoint union of affine schemes, see Properties of Spaces, Lemma 64.6.1. For each $i$ choose a surjective étale morphism $V_ i \to Y \times _ X U_ i$ where $V_ i = \coprod V_{ij}$ is a disjoint union of affines. Write $U_ i = \mathop{\mathrm{Spec}}(A_ i)$ and $V_{ij} = \mathop{\mathrm{Spec}}(B_{ij})$. Let $I_ i \subset A_ i$ be an ideal cutting out the inverse image of $T$ in $U_ i$. Then we may apply Lemma 86.4.2 to see that the map of $I_ i$-adic completions $A_ i^\wedge \to B_{ij}^\wedge$ has the property $P$ of Lemma 86.5.1. Since $\{ \text{Spf}(A_ i^\wedge ) \to X_{/T}\}$ and $\{ \text{Spf}(B_{ij}) \to Y_{/T}\}$ are coverings as in Formal Spaces, Definition 85.7.1 we see that $Y_{/T} \to X_{/T}$ is rig-étale by definition. $\square$

Lemma 86.10.2. Let $X$ be a Noetherian affine scheme. Let $T \subset X$ be a closed subset. Let $U$ be an affine scheme and let $U \to X$ a finite type morphism étale over $X \setminus T$. Let $V$ be a Noetherian affine scheme over $X$. For any morphism $c' : V_{/T} \to U_{/T}$ over $X_{/T}$ there exists an étale morphism $b : V' \to V$ of affine schemes which induces an isomorphism $b_{/T} : V'_{/T} \to V_{/T}$ and a morphism $a : V' \to U$ such that $c' = a_{/T} \circ b_{/T}^{-1}$.

Proof. This is a reformulation of Lemma 86.4.4. $\square$

Lemma 86.10.3. Let $X$ be a Noetherian affine scheme. Let $T \subset X$ be a closed subset. Let $W \to X_{/T}$ be a rig-étale morphism of formal algebraic spaces with $W$ an affine formal algebraic space. Then there exists an affine scheme $U$, a finite type morphism $U \to X$ étale over $X \setminus T$ such that $W \cong U_{/T}$. Moreover, if $W \to X_{/T}$ is étale, then $U \to X$ is étale.

Proof. The existence of $U$ is a restatement of Lemma 86.7.4. The final statement follows from More on Morphisms, Lemma 37.12.3. $\square$

Let $S$ be a scheme. Let $X$ be a locally Noetherian algebraic space over $S$ and let $T \subset |X|$ be a closed subset. Let us denote $\mathcal{C}_{X, T}$ the category of algebraic spaces $Y$ over $X$ such that the structure morphism $f : Y \to X$ is locally of finite type and an isomorphism over the complement of $T$. Formal completion defines a functor

86.10.3.1
$$\label{restricted-equation-completion-functor} F_{X, T} : \mathcal{C}_{X, T} \longrightarrow \mathcal{C}_{X_{/T}},\quad (f : Y \to X) \longmapsto (f_{/T} : Y_{/T} \to X_{/T})$$

see Lemma 86.10.1.

Lemma 86.10.4. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Z \to Y$ be morphisms of algebraic spaces. Let $T \subset |X|$ be closed. Assume that

1. $X$ is locally Noetherian,

2. $g$ is a monomorphism and locally of finite type,

3. $f|_{X \setminus T} : X \setminus T \to Y$ factors through $g$, and

4. $f_{/T} : X_{/T} \to Y$ factors through $g$,

then $f$ factors through $g$.

Proof. Consider the fibre product $E = X \times _ Y Z \to X$. By assumption the open immersion $X \setminus T \to X$ factors through $E$ and any morphism $\varphi : X' \to X$ with $|\varphi |(|X'|) \subset T$ factors through $E$ as well, see Formal Spaces, Section 85.9. By More on Morphisms of Spaces, Lemma 74.20.3 this implies that $E \to X$ is étale at every point of $E$ mapping to a point of $T$. Hence $E \to X$ is an étale monomorphism, hence an open immersion (Morphisms of Spaces, Lemma 65.51.2). Then it follows that $E = X$ since our assumptions imply that $|X| = |E|$. $\square$

Lemma 86.10.5. Let $S$ be a scheme. Let $X$, $Y$ be locally Noetherian algebraic spaces over $S$. Let $T \subset |X|$ and $T' \subset |Y|$ be closed subsets. Let $a, b : X \to Y$ be morphisms of algebraic spaces over $S$ such that $a|_{X \setminus T} = b|_{X \setminus T}$, such that $|a|(T) \subset T'$ and $|b|(T) \subset T'$, and such that $a_{/T} = b_{/T}$ as morphisms $X_{/T} \to Y_{/T'}$. Then $a = b$.

Proof. Let $E$ be the equalizer of $a$ and $b$. Then $E$ is an algebraic space and $E \to X$ is locally of finite type and a monomorphism, see Morphisms of Spaces, Lemma 65.4.1. Our assumptions imply we can apply Lemma 86.10.4 to the two morphisms $f = \text{id} : X \to X$ and $g : E \to X$ and the closed subset $T$ of $|X|$. $\square$

Lemma 86.10.6. Let $S$ be a scheme. Let $X$ be a locally Noetherian algebraic space over $S$. Let $T \subset |X|$ be a closed subset. Let $s, t : R \to U$ be two morphisms of algebraic spaces over $X$. Assume

1. $R$, $U$ are locally of finite type over $X$,

2. the base change of $s$ and $t$ to $X \setminus T$ is an étale equivalence relation, and

3. the formal completion $(t_{/T}, s_{/T}) : R_{/T} \to U_{/T} \times _{X_{/T}} U_{/T}$ is an equivalence relation too.

Then $(t, s) : R \to U \times _ X U$ is an étale equivalence relation.

Proof. The morphisms $s, t : R \to U$ are étale over $X \setminus T$ by assumption. Since the formal completions of the maps $s, t : R \to U$ are étale, we see that $s$ and $t$ are étale for example by More on Morphisms, Lemma 37.12.3. Applying Lemma 86.10.4 to the morphisms $\text{id} : R \times _{U \times _ X U} R \to R \times _{U \times _ X U} R$ and $\Delta : R \to R \times _{U \times _ X U} R$ we conclude that $(t, s)$ is a monomorphism. Applying it again to $(t \circ \text{pr}_0, s \circ \text{pr}_1) : R \times _{s, U, t} R \to U \times _ X U$ and $(t, s) : R \to U \times _ X U$ we find that “transitivity” holds. We omit the proof of the other two axioms of an equivalence relation. $\square$

Remark 86.10.7. Let $S$, $X$, and $T \subset |X|$ be as in (86.10.3.1). Let $U \to X$ be an algebraic space over $X$ such that $U \to X$ is locally of finite type and étale outside of $T$. We will construct a factorization

$U \longrightarrow Y \longrightarrow X$

with $Y$ in $\mathcal{C}_{X, T}$ such that $U_{/T} \to Y_{/T}$ is an isomorphism. We may assume the image of $U \to X$ contains $X \setminus T$, otherwise we replace $U$ by $U \amalg (X \setminus T)$. For an algebraic space $Z$ over $X$, let us denote $Z^\circ$ the open subspace which is the inverse image of $X \setminus T$. Let

$R = U \amalg _{U^\circ } (U \times _ X U)^\circ$

be the pushout of $U^\circ \to U$ and the diagonal morphism $U^\circ \to U^\circ \times _ X U^\circ = (U \times _ X U)^\circ$. Since $U^\circ \to X$ is étale, the diagonal is an open immersion and we see that $R$ is an algebraic space (this follows for example from Spaces, Lemma 63.8.5). The two projections $(U \times _ X U)^\circ \to U$ extend to $R$ and we obtain two étale morphisms $s, t : R \to U$. Checking on each piece separatedly we find that $R$ is an étale equivalence relation on $U$. Set $Y = U/R$ which is an algebraic space by Bootstrap, Theorem 78.10.1. Since $U^\circ \to X \setminus T$ is a surjective étale morphism and since $R^\circ = U^\circ \times _{X \setminus T} U^\circ$ we see that $Y^\circ \to X \setminus T$ is an isomorphism. In other words, $Y \to X$ is an object of $\mathcal{C}_{X, T}$. On the other hand, the morphism $U \to Y$ induces an isomorphism $U_{/T} \to Y_{/T}$. Namely, the formal completion of $R$ along the inverse image of $T$ is equal to the formal completion of $U$ along the inverse image of $T$ by our choice of $R$. By our construction of the formal completion in Formal Spaces, Section 85.9 we conclude that $U_{/T} = Y_{/T}$.

Lemma 86.10.8. Let $S$ be a scheme. Let $X$ be a Noetherian affine algebraic space over $S$. Let $T \subset |X|$ be a closed subset. Then the functor $F_{X, T}$ is an equivalence.

Before we prove this lemma let us discuss an example. Suppose that $S = \mathop{\mathrm{Spec}}(k)$, $X = \mathbf{A}^1_ k$, and $T = \{ 0\}$. Then $X_{/T} = \text{Spf}(k[[x]])$. Let $W = \text{Spf}(k[[x]] \times k[[x]])$. Then the corresponding $Y$ is the affine line with zero doubled (Schemes, Example 26.14.3). Moreover, this is the output of the construction in Remark 86.10.7 starting with $U = X \amalg X$.

Proof. For any scheme or algebraic space $Z$ over $X$, let us denote $Z_0 \subset Z$ the inverse image of $T$ with the induced reduced closed subscheme or subspace structure. Note that $Z_0 = (Z_{/T})_{red}$ is the reduction of the formal completion.

The functor $F_{X, T}$ is faithful by Lemma 86.10.5.

Let $Y, Y'$ be objects of $\mathcal{C}_{X, T}$ and let $a' : Y_{/T} \to Y'_{/T}$ be a morphism in $\mathcal{C}_{X_{/T}}$. To prove $F_{X, T}$ is fully faithful, we will construct a morphism $a : Y \to Y'$ in $\mathcal{C}_{X, T}$ such that $a' = a_{/T}$.

Let $U$ be an affine scheme and let $U \to Y$ be an étale morphism. Because $U$ is affine, $U_0$ is affine and the image of $U_0 \to Y_0 \to Y'_0$ is a quasi-compact subspace of $|Y'_0|$. Thus we can choose an affine scheme $V$ and an étale morphism $V \to Y'$ such that the image of $|V_0| \to |Y'_0|$ contains this quasi-compact subset. Consider the formal algebraic space

$W = U_{/T} \times _{Y'_{/T}} V_{/T}$

By our choice of $V$ the above, the map $W \to U_{/T}$ is surjective. Thus there exists an affine formal algebraic space $W'$ and an étale morphism $W' \to W$ such that $W' \to W \to U_{/T}$ is surjective. Then $W' \to U_{/T}$ is étale. By Lemma 86.10.3 $W' = U'_{/T}$ for $U' \to U$ étale and $U'$ affine. Write $V = \mathop{\mathrm{Spec}}(C)$. By Lemma 86.10.2 there exists an étale morphism $U'' \to U'$ of affines which is an isomorphism on completions and a morphism $U'' \to V$ whose completion is the composition $U''_{/T} \to U'_{/T} \to W \to V_{/T}$. Thus we get

$Y \longleftarrow U'' \longrightarrow Y'$

over $X$ agreeing with the given map on formal completions such that the image of $U''_0 \to Y_0$ is the same as the image of $U_0 \to Y_0$.

Taking a disjoint union of $U''$ as constructed in the previous paragraph, we find a scheme $U$, an étale morphism $U \to Y$, and a morphism $b : U \to Y'$ over $X$, such that the diagram

$\xymatrix{ U_{/T} \ar[d] \ar[rd]^{b_{/T}} \\ Y_{/T} \ar[r]^{a'} & Y'_{/T} }$

is commutative and such that $U_0 \to Y_0$ is surjective. Taking a disjoint union with the open $X \setminus T$ (which is also open in $Y$ and $Y'$), we find that we may even assume that $U \to Y$ is a surjective étale morphism. Let $R = U \times _ Y U$. Then the two compositions $R \to U \to Y'$ agree both over $X \setminus T$ and after formal completion along $T$, whence are equal by Lemma 86.10.5. This means exactly that $b$ factors as $U \to Y \to Y'$ to give us our desired morphism $a : Y \to Y'$.

Essential surjectivity. Let $W$ be an object of $\mathcal{C}_{X_{/T}}$. We prove $W$ is in the essential image in a number of steps.

Step 1: $W$ is an affine formal algebraic space. Then we can find $U \to X$ of finite type and étale over $X \setminus T$ such that $U_{/T}$ is isomorphic to $W$, see Lemma 86.10.3. Thus we see that $W$ is in the essential image by the construction in Remark 86.10.7.

Step 2: $W$ is separated. Choose $\{ W_ i \to W\}$ as in Formal Spaces, Definition 85.7.1. By Step 1 the formal algebraic spaces $W_ i$ and $W_ i \times _ W W_ j$ are in the essential image. Say $W_ i = (Y_ i)_{/T}$ and $W_ i \times _ W W_ j = (Y_{ij})_{/T}$. By fully faithfulness we obtain morphisms $t_{ij} : Y_{ij} \to Y_ i$ and $s_{ij} : Y_{ij} \to Y_ j$ matching the projections $W_ i \times _ W W_ j \to W_ i$ and $W_ i \times _ W W_ j \to W_ j$. Set $R = \coprod Y_{ij}$ and $U = \coprod Y_ i$ and denote $s = \coprod s_{ij} : R \to U$ and $t = \coprod t_{ij} : R \to U$. Applying Lemma 86.10.6 we find that $(t, s) : R \to U \times _ X U$ is an étale equivalence relation. Thus we can take the quotient $Y = U/R$ and it is an algebraic space, see Bootstrap, Theorem 78.10.1. Since completion commutes with fibre products and taking quotient sheaves, we find that $Y_{/T} \cong W$ in $\mathcal{C}_{X_{/T}}$.

Step 3: $W$ is general. Choose $\{ W_ i \to W\}$ as in Formal Spaces, Definition 85.7.1. The formal algebraic spaces $W_ i$ and $W_ i \times _ W W_ j$ are separated. Hence by Step 2 the formal algebraic spaces $W_ i$ and $W_ i \times _ W W_ j$ are in the essential image. Then we argue exactly as in the previous paragraph to see that $W$ is in the essential image as well. This concludes the proof. $\square$

Theorem 86.10.9. Let $S$ be a scheme. Let $X$ be a locally Noetherian algebraic space over $S$. Let $T \subset |X|$ be a closed subset. The functor $F_{X, T}$ (86.10.3.1)

$\left\{ \begin{matrix} \text{algebraic spaces }Y\text{ locally of finite} \\ \text{type over }X\text{ such that }Y \to X \\ \text{is an isomorphism over }X \setminus T \end{matrix} \right\} \longrightarrow \left\{ \begin{matrix} \text{formal algebraic spaces }W\text{ endowed} \\ \text{with a rig-étale morphism }W \to X_{/T} \end{matrix} \right\}$

given by formal completion is an equivalence.

Proof. The theorem is essentially a formal consequence of Lemma 86.10.8. We give the details but we encourage the reader to think it through for themselves. Let $g : U \to X$ be a surjective étale morphism with $U = \coprod U_ i$ and each $U_ i$ affine. Denote $F_{U, T}$ the functor for $U$ and the inverse image of $T$ in $|U|$.

Since $U = \coprod U_ i$ both the category $\mathcal{C}_{U, T}$ and the category $\mathcal{C}_{U_{/T}}$ decompose as a product of categories, one for each $i$. Since the functors $F_{U_ i, T}$ are equivalences for all $i$ by the lemma we find that the same is true for $F_{U, T}$.

Since $F_{U, T}$ is faithful, it follows that $F_{X, T}$ is faithful too. Namely, if $a, b : Y \to Y'$ are morphisms in $\mathcal{C}_{X, T}$ such that $a_{/T} = b_{/T}$, then we find on pulling back that the base changes $a_ U, b_ U : U \times _ X Y \to U \times _ X Y'$ are equal. Since $U \times _ X Y \to Y$ is surjective étale, this implies that $a = b$.

At this point we know that $F_{X, T}$ is faithful for every situation as in the theorem. Let $R = U \times _ X U$ where $U$ is as above. Let $t, s : R \to U$ be the projections. Since $X$ is Noetherian, so is $R$. Thus the functor $F_{R, T}$ (defined in the obvious manner) is faithful. Let $Y \to X$ and $Y' \to X$ be objects of $\mathcal{C}_{X, T}$. Let $a' : Y_{/T} \to Y'_{/T}$ be a morphism in the category $\mathcal{C}_{X_{/T}}$. Taking the base change to $U$ we obtain a morphism $a'_ U : (U \times _ X Y)_{/T} \to (U \times _ X Y')_{/T}$ in the category $\mathcal{C}_{U_{/T}}$. Since the functor $F_{U, T}$ is fully faithful we obtain a morphism $a_ U : U \times _ X Y \to U \times _ X Y'$ with $F_{U, T}(a_ U) = a'_ U$. Since $s^*(a'_ U) = t^*(a'_ U)$ and since $F_{R, T}$ is faithful, we find that $s^*(a_ U) = t^*(a_ U)$. Since

$\xymatrix{ R \times _ X Y \ar@<1ex>[r] \ar@<-1ex>[r] & U \times _ X Y \ar[r] & Y }$

is an equalizer diagram of sheaves, we find that $a_ U$ descends to a morphism $a : Y \to Y'$. We omit the proof that $F_{X, T}(a) = a'$.

At this point we know that $F_{X, T}$ is faithful for every situation as in the theorem. To finish the proof we show that $F_{X, T}$ is essentially surjective. Let $W \to X_{/T}$ be an object of $\mathcal{C}_{X_{/T}}$. Then $U \times _ X W$ is an object of $\mathcal{C}_{U_{/T}}$. By the affine case we find an object $V \to U$ of $\mathcal{C}_{U, T}$ and an isomorphism $\alpha : F_{U, T}(V) \to U \times _ X W$ in $\mathcal{C}_{U_{/T}}$. By fully faithfulness of $F_{R, T}$ we find a unique morphism $h : s^*V \to t^*V$ in the category $\mathcal{C}_{R, T}$ such that $F_{R, T}(h)$ corresponds, via the isomorphism $\alpha$, to the canonical descent datum on $U \times _ X W$ in the category $\mathcal{C}_{R_{/T}}$. Using faithfulness of our functor on $R \times _{s, U, t} R$ we see that $h$ satisfies the cocycle condition. We conclude, for example by the much more general Bootstrap, Lemma 78.11.3, that there exists an object $Y \to X$ of $\mathcal{C}_{X, T}$ and an isomorphism $\beta : U \times _ X Y \to V$ such that the descent datum $h$ corresponds, via $\beta$, to the canonical descent datum on $U \times _ X Y$. We omit the verification that $F_{X, T}(Y)$ is isomorphic to $W$; hint: in the category of formal algebraic spaces there is descent for morphisms along étale coverings. $\square$

We are often interested as to whether the output of the construction of Theorem 86.10.9 is a separated algebraic space. In the next few lemmas we match properties of $Y \to X$ and the corresponding completion $Y_{/T} \to X_{/T}$.

Lemma 86.10.10. Let $S$ be a scheme. Let $X$ be a locally Noetherian algebraic space over $S$. Let $T \subset |X|$ be a closed subset. Let $W \to X_{/T}$ be an object of the category $\mathcal{C}_{X_{/T}}$ and let $Y \to X$ be the object corresponding to $W$ via Theorem 86.10.9. Then $Y \to X$ is quasi-compact if and only if $W \to X_{/T}$ is so.

Proof. These conditions may be checked after base change to an affine scheme étale over $X$, resp. a formal affine algebraic space étale over $X_{/T}$, see Morphisms of Spaces, Lemma 65.8.8 as well as Formal Spaces, Lemma 85.12.3. If $U \to X$ ranges over étale morphisms with $U$ affine, then the formal completions $U_{/T} \to X_{/T}$ give a family of formal affine coverings as in Formal Spaces, Definition 85.7.1. Thus we may and do assume $X$ is affine.

Let $V \to Y$ be a surjective étale morphism where $V = \coprod _{j \in J} V_ j$ is a disjoint union of affines. Then $V_{/T} \to Y_{/T} = W$ is a surjective étale morphism. Thus if $Y$ is quasi-compact, we can choose $J$ is finite, and we conclude that $W$ is quasi-compact. Conversely, if $W$ is quasi-compact, then we can find a finite subset $J' \subset J$ such that $\coprod _{j \in J'} (V_ j)_{/T} \to W$ is surjective. Then it follows that

$(X \setminus T) \amalg \coprod \nolimits _{j \in J'} V_ j \longrightarrow Y$

is surjective. This either follows from the construction of $Y$ in the proof of Lemma 86.10.8 or it follows since we have

$|Y| = |X \setminus T| \amalg |W_{red}|$

as $Y_{/T} = W$. $\square$

Lemma 86.10.11. Let $S$ be a scheme. Let $X$ be a locally Noetherian algebraic space over $S$. Let $T \subset |X|$ be a closed subset. Let $W \to X_{/T}$ be an object of the category $\mathcal{C}_{X_{/T}}$ and let $Y \to X$ be the object corresponding to $W$ via Theorem 86.10.9. Then $Y \to X$ is quasi-separated if and only if $W \to X_{/T}$ is so.

Proof. These conditions may be checked after base change to an affine scheme étale over $X$, resp. a formal affine algebraic space étale over $X_{/T}$, see Morphisms of Spaces, Lemma 65.4.12 as well as Formal Spaces, Lemma 85.23.5, If $U \to X$ ranges over étale morphisms with $U$ affine, then the formal completions $U_{/T} \to X_{/T}$ give a family of formal affine coverings as in Formal Spaces, Definition 85.7.1. Thus we may and do assume $X$ is affine.

Let $V \to Y$ be a surjective étale morphism where $V = \coprod _{j \in J} V_ j$ is a disjoint union of affines. Then $Y$ is quasi-separated if and only if $V_ j \times _ Y V_{j'}$ is quasi-compact for all $j, j' \in J$. Similarly, $W$ is quasi-separated if and only if $(V_ j \times _ Y V_{j'})_{/T} = (V_ j)_{/T} \times _{Y_{/T}} (V_{j'})_{/T}$ is quasi-compact for all $j, j' \in J$. Since $X$ is Noetherian affine, we see that

$(V_ j \times _ Y V_{j'}) \times _ X (X \setminus T)$

is quasi-compact. Hence we conclude the equivalence holds by the equality

$|V_ j \times _ Y V_{j'}| = |(V_ j \times _ Y V_{j'}) \times _ X (X \setminus T)| \amalg |(V_ j \times _ Y V_{j'})_{/T}|$

and the fact that the second summand is closed in the left hand side. $\square$

Lemma 86.10.12. Let $S$ be a scheme. Let $X$ be a locally Noetherian algebraic space over $S$. Let $T \subset |X|$ be a closed subset. Let $W \to X_{/T}$ be an object of the category $\mathcal{C}_{X_{/T}}$ and let $Y \to X$ be the object corresponding to $W$ via Theorem 86.10.9. Then $Y \to X$ is separated if and only if $W \to X_{/T}$ is separated and $\Delta : W \to W \times _{X_{/T}} W$ is rig-surjective.

Proof. These conditions may be checked after base change to an affine scheme étale over $X$, resp. a formal affine algebraic space étale over $X_{/T}$, see Morphisms of Spaces, Lemma 65.4.12 as well as Formal Spaces, Lemma 85.23.5. If $U \to X$ ranges over étale morphisms with $U$ affine, then the formal completions $U_{/T} \to X_{/T}$ give a family of formal affine coverings as in Formal Spaces, Definition 85.7.1. Thus we may and do assume $X$ is affine. In the proof of both directions we may assume that $Y \to X$ and $W \to X_{/T}$ are quasi-separated by Lemma 86.10.11.

Proof of easy direction. Assume $Y \to X$ is separated. Then $Y \to Y \times _ X Y$ is a closed immersion and it follows that $W \to W \times _{X_{/T}} W$ is a closed immersion too, i.e., we see that $W \to X_{/T}$ is separated. Let

$p : \text{Spf}(R) \longrightarrow W \times _{X_{/T}} W = (Y \times _ X Y)_{/T}$

be an adic morphism where $R$ is a complete discrete valuation ring with fraction field $K$. The composition into $Y \times _ X Y$ corresponds to a morphism $g : \mathop{\mathrm{Spec}}(R) \to Y \times _ X Y$, see Formal Spaces, Lemma 85.26.3. Since $p$ is an adic morphism, so is the composition $\text{Spf}(R) \to X$. Thus we see that $g(\mathop{\mathrm{Spec}}(K))$ is a point of

$(Y \times _ X Y) \times _ X (X \setminus T) \cong X \setminus T \cong Y \times _ X (X \setminus T)$

(small detail omitted). Hence this lifts to a $K$-point of $Y$ and we obtain a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & Y \ar[d] \\ \mathop{\mathrm{Spec}}(R) \ar[r] \ar@{-->}[ru] & Y \times _ X Y }$

Since $Y \to X$ was assumed separated we find the dotted arrow exists (Cohomology of Spaces, Lemma 67.19.1). Applying the functor completion along $T$ we find that $p$ can be lifted to a morphism into $W$, i.e., $W \to W \times _{X_{/T}} W$ is rig-surjective.

Proof of hard direction. Assume $W \to X_{/T}$ separated and $W \to W \times _{X_{/T}} W$ rig-surjective. By Cohomology of Spaces, Lemma 67.19.1 and Remark 67.19.3 it suffices to show that given any commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & Y \ar[d] \\ \mathop{\mathrm{Spec}}(R) \ar[r]^ g \ar@{-->}[ru] & Y \times _ X Y }$

where $R$ is a complete discrete valuation ring with fraction field $K$, there is at most one dotted arrow making the diagram commute. Let $h : \mathop{\mathrm{Spec}}(R) \to X$ be the composition of $g$ with the morphism $Y \times _ X Y \to X$. There are three cases: Case I: $h(\mathop{\mathrm{Spec}}(R)) \subset (X \setminus T)$. This case is trivial because $Y \times _ X (X \setminus T) = X \setminus T$. Case II: $h$ maps $\mathop{\mathrm{Spec}}(R)$ into $T$. This case follows from our assumption that $W \to X_{/T}$ is separated. Namely, if $T$ denotes the reduced induced closed subspace structure on $T$, then $h$ factors through $T$ and

$W \times _{X_{/T}} T = Y \times _ X T \longrightarrow T$

is separated by assumption (and for example Formal Spaces, Lemma 85.23.5) which implies we get the lifting property by Cohomology of Spaces, Lemma 67.19.1 applied to the displayed arrow. Case III: $h(\mathop{\mathrm{Spec}}(K))$ is not in $T$ but $h$ maps the closed point of $\mathop{\mathrm{Spec}}(R)$ into $T$. In this case the corresponding morphism

$g_{/T} : \text{Spf}(R) \longrightarrow (Y \times _ X Y)_{/T} = W \times _{X_{/T}} W$

is an adic morphism (detail omitted). Hence our assumption that $W \to W \times _{X_{/T}} W$ be rig-surjective implies we can lift $g_{/T}$ to a morphism $e : \text{Spf}(R) \to W = Y_{/T}$ (see Lemma 86.9.11 for why we do not need to extend $R$). Algebraizing the composition $\text{Spf}(R) \to Y$ using Formal Spaces, Lemma 85.26.3 we find a morphism $\mathop{\mathrm{Spec}}(R) \to Y$ lifting $g$ as desired. $\square$

Lemma 86.10.13. Let $S$ be a scheme. Let $X$ be a locally Noetherian algebraic space over $S$. Let $T \subset |X|$ be a closed subset. Let $W \to X_{/T}$ be an object of the category $\mathcal{C}_{X_{/T}}$ and let $Y \to X$ be the object corresponding to $W$ via Theorem 86.10.9. Then $Y \to X$ is proper if and only if the following conditions hold

1. $W \to X_{/T}$ is proper,

2. $W \to X_{/T}$ is rig-surjective, and

3. $\Delta : W \to W \times _{X_{/T}} W$ is rig-surjective.

Proof. These conditions may be checked after base change to an affine scheme étale over $X$, resp. a formal affine algebraic space étale over $X_{/T}$, see Morphisms of Spaces, Lemma 65.40.2 as well as Formal Spaces, Lemma 85.24.2. If $U \to X$ ranges over étale morphisms with $U$ affine, then the formal completions $U_{/T} \to X_{/T}$ give a family of formal affine coverings as in Formal Spaces, Definition 85.7.1. Thus we may and do assume $X$ is affine. In the proof of both directions we may assume that $Y \to X$ and $W \to X_{/T}$ are separated and quasi-compact and that $W \to W \times _{X_{/T}} W$ is rig-surjective by Lemmas 86.10.10 and 86.10.12.

Proof of the easy direction. Assume $Y \to X$ is proper. Then $Y_{/T} = Y \times _ X X_{/T} \to X_{/T}$ is proper too. Let

$p : \text{Spf}(R) \longrightarrow X_{/T}$

be an adic morphism where $R$ is a complete discrete valuation ring with fraction field $K$. Then $p$ corresponds to a morphism $g : \mathop{\mathrm{Spec}}(R) \to X$, see Formal Spaces, Lemma 85.26.3. Since $p$ is an adic morphism, we have $p(\mathop{\mathrm{Spec}}(K)) \not\in T$. Since $Y \to X$ is an isomorphism over $X \setminus T$ we can lift to $X$ and obtain a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & Y \ar[d] \\ \mathop{\mathrm{Spec}}(R) \ar[r] \ar@{-->}[ru] & X }$

Since $Y \to X$ was assumed proper we find the dotted arrow exists. (Cohomology of Spaces, Lemma 67.19.2). Applying the functor completion along $T$ we find that $p$ can be lifted to a morphism into $W$, i.e., $W \to X_{/T}$ is rig-surjective.

Proof of hard direction. Assume $W \to X_{/T}$ proper, $W \to W \times _{X_{/T}} W$ rig-surjective, and $W \to X_{/T}$ rig-surjective. By Cohomology of Spaces, Lemma 67.19.2 and Remark 67.19.3 it suffices to show that given any commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & Y \ar[d] \\ \mathop{\mathrm{Spec}}(R) \ar[r]^ g \ar@{-->}[ru] & X }$

where $R$ is a complete discrete valuation ring with fraction field $K$, there is a dotted arrow making the diagram commute. Let $h : \mathop{\mathrm{Spec}}(R) \to X$ be the composition of $g$ with the morphism $Y \times _ X Y \to X$. There are three cases: Case I: $h(\mathop{\mathrm{Spec}}(R)) \subset (X \setminus T)$. This case is trivial because $Y \times _ X (X \setminus T) = X \setminus T$. Case II: $h$ maps $\mathop{\mathrm{Spec}}(R)$ into $T$. This case follows from our assumption that $W \to X_{/T}$ is proper. Namely, if $T$ denotes the reduced induced closed subspace structure on $T$, then $h$ factors through $T$ and

$W \times _{X_{/T}} T = Y \times _ X T \longrightarrow T$

is proper by assumption which implies we get the lifting property by Cohomology of Spaces, Lemma 67.19.2 applied to the displayed arrow. Case III: $h(\mathop{\mathrm{Spec}}(K))$ is not in $T$ but $h$ maps the closed point of $\mathop{\mathrm{Spec}}(R)$ into $T$. In this case the corresponding morphism

$g_{/T} : \text{Spf}(R) \longrightarrow Y_{/T} = W$

is an adic morphism (detail omitted). Hence our assumption that $W \to X_{/T}$ be rig-surjective implies we can lift $g_{/T}$ to a morphism $e : \text{Spf}(R') \to W = Y_{/T}$ for some extension of complete discrete valuation rings $R \subset R'$. Algebraizing the composition $\text{Spf}(R') \to Y$ using Formal Spaces, Lemma 85.26.3 we find a morphism $\mathop{\mathrm{Spec}}(R') \to Y$ lifting $g$. By the discussion in Cohomology of Spaces, Remark 67.19.3 this is sufficient to conclude that $Y \to X$ is proper. $\square$

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