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.

## 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 $X_{/T}$.

**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

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

$X$ is locally Noetherian,

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

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

$f_{/T} : X_{/T} \to Y$ 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

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

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

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

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

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

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

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

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)

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

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

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

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

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

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

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

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

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

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

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

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

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

$W \to X_{/T}$ is rig-surjective, 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 65.40.2 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

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

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

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

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

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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