Lemma 86.25.1. Let $T \subset X$ be a closed subset of a Noetherian affine scheme $X$. Let $W$ be a Noetherian affine formal algebraic space. Let $g : W \to X_{/T}$ be a rig-étale morphism. Then there exists an affine scheme $X'$ and a finite type morphism $f : X' \to X$ étale over $X \setminus T$ such that there is an isomorphism $X'_{/f^{-1}T} \cong W$ compatible with $f_{/T}$ and $g$. Moreover, if $W \to X_{/T}$ is étale, then $X' \to X$ is étale.

## 86.25 Completions and morphisms, I

In this section we put some preliminary results on completions which we will use in the proof of Theorem 86.27.4. Although the lemmas stated and proved here are not trivial (some are based on our work on algebraization of rig-étale algebras), we still suggest the reader skip this section on a first reading.

**Proof.**
The existence of $X'$ is a restatement of Lemma 86.10.3. The final statement follows from More on Morphisms, Lemma 37.12.3.
$\square$

Lemma 86.25.2. Assume we have

Noetherian affine schemes $X$, $X'$, and $Y$,

a closed subset $T \subset |X|$,

a morphism $f : X' \to X$ locally of finite type and étale over $X \setminus T$,

a morphism $h : Y \to X$,

a morphism $\alpha : Y_{/T} \to X'_{/T}$ over $X_{/T}$ (see proof for notation).

Then there exists an étale morphism $b : Y' \to Y$ of affine schemes which induces an isomorphism $b_{/T} : Y'_{/T} \to Y_{/T}$ and a morphism $a : Y' \to X'$ over $X$ such that $\alpha = a_{/T} \circ b_{/T}^{-1}$.

**Proof.**
The notation using the subscript ${}_{/T}$ in the statement refers to the construction which to a morphism of schemes $g : V \to X$ associates the morphism $g_{/T} : V_{/g^{-1}T} \to X_{/T}$ of formal algebraic spaces; it is a functor from the category of schemes over $X$ to the category of formal algebraic spaces over $X_{/T}$, see Section 86.23. Having said this, the lemma is just a reformulation of Lemma 86.8.7.
$\square$

Lemma 86.25.3. 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 85.10. 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 86.25.4. Let $S$ be a scheme. Let $X$, $W$ be algebraic spaces over $S$ with $X$ locally Noetherian. Let $T \subset |X|$ be a closed subset. Let $a, b : X \to W$ be morphisms of algebraic spaces over $S$ such that $a|_{X \setminus T} = b|_{X \setminus T}$ and such that $a_{/T} = b_{/T}$ as morphisms $X_{/T} \to W$. 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 86.25.3 to the two morphisms $f = \text{id} : X \to X$ and $g : E \to X$ and the closed subset $T$ of $|X|$.
$\square$

Lemma 86.25.5. 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.**
Consequence of the more general Lemma 86.25.4.
$\square$

Lemma 86.25.6. 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 (see proof for notation).

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

**Proof.**
The notation using the subscript ${}_{/T}$ in the statement refers to the construction which to a morphism $f : X' \to X$ of algebraic spaces associates the morphism $f_{/T} : X'_{/f^{-1}T} \to X_{/T}$ of formal algebraic spaces, see Section 86.23. 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 86.25.3 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.
$\square$

Lemma 86.25.7. 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 $f : X' \to X$ be a morphism of algebraic spaces which is locally of finite type and étale outside of $T$. There exists a factorization

of $f$ with the following properties: $X'' \to X$ is locally of finite type, $X'' \to X$ is an isomorphism over $X \setminus T$, and $X'_{/T} \to X''_{/T}$ is an isomorphism (see proof for notation).

**Proof.**
The notation using the subscript ${}_{/T}$ in the statement refers to the construction which to a morphism $f : X' \to X$ of algebraic spaces associates the morphism $f_{/T} : X'_{/f^{-1}T} \to X_{/T}$ of formal algebraic spaces, see Section 86.23. We will also use the notion $U \subset X$ and $U' \subset X'$ to denote the open subspaces with $|U| = |X| \setminus T$ and $U' = |X'| \setminus f^{-1}T$ introduced in Section 86.23.

After replacing $X'$ by $X' \amalg U$ we may and do assume the image of $X' \to X$ contains $U$. Let

be the pushout of $U' \to X'$ and the diagonal morphism $U' \to U' \times _ U U' = U' \times _ X U'$. Since $U' \to X$ is étale, this 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 $U' \times _ U U' \to U'$ extend to $R$ and we obtain two étale morphisms $s, t : R \to X'$. Checking on each piece separatedly we find that $R$ is an étale equivalence relation on $X'$. Set $X'' = X'/R$ which is an algebraic space by Bootstrap, Theorem 78.10.1. By construction have the factorization as in the lemma and the morphism $X'' \to X$ is locally of finite type (as this can be checked étale locally, i.e., on $X'$). Since $U' \to U$ is a surjective étale morphism and since $s^{-1}(U') = t^{-1}(U') = U' \times _ U U'$ we see that $U'' = U \times _ X X'' \to U$ is an isomorphism. Finally, we have to show the morphism $X' \to X''$ induces an isomorphism $X'_{/T} \to X''_{/T}$. To see this, note that the formal completion of $R$ along the inverse image of $T$ is equal to the formal completion of $X'$ along the inverse image of $T$ by our choice of $R$! By our construction of the formal completion in Formal Spaces, Section 85.10 we have $X''_{/T} = (X'_{/T}) / (R_{/T})$ as sheaves. Since $X'_{/T} = R_{/T}$ we conclude that $X'_{/T} = X''_{/T}$ and this finishes the proof. $\square$

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