Lemma 80.14.7. Let $B$ be an algebraic space of finite type over $\mathbf{Z}$. Let $U$ be an algebraic space of finite type and separated over $B$. Let $(U_2 \subset U, f : U_1 \to U)$ be an elementary distinguished square. Assume $U_1$ and $U_2$ have compactifications over $B$ and $U_1 \times _ U U_2 \to U$ has dense image. Then $U$ has a compactification over $B$.

Proof. Choose a compactification $U_ i \subset X_ i$ over $B$ for $i = 1, 2$. We may assume $U_ i$ is scheme theoretically dense in $X_ i$. We may assume there is an open $V_ i \subset X_ i$ and a proper morphism $\psi _ i : V_ i \to U$ extending $U_ i \to U$, see Lemma 80.14.6. Picture

$\xymatrix{ U_ i \ar[r] \ar[d] & V_ i \ar[r] \ar[dl]^{\psi _ i} & X_ i \\ U }$

Denote $Z_1 \subset U$ the reduced closed subspace corresponding to the closed subset $|U| \setminus |U_2|$. Recall that $f^{-1}Z_1$ is a closed subspace of $U_1$ mapping isomorphically to $Z_1$. Denote $Z_2 \subset U$ the reduced closed subspace corresponding to the closed subset $|U| \setminus \mathop{\mathrm{Im}}(|f|) = |U_2| \setminus \mathop{\mathrm{Im}}(|U_1 \times _ U U_2| \to |U_2|)$. Thus we have

$U = U_2 \amalg Z_1 = Z_2 \amalg \mathop{\mathrm{Im}}(f) = Z_2 \amalg \mathop{\mathrm{Im}}(U_1 \times _ U U_2 \to U_2) \amalg Z_1$

set theoretically. Denote $Z_{i, i} \subset V_ i$ the inverse image of $Z_ i$ under $\psi _ i$. Observe that $\psi _2$ is an isomorphism over an open neighbourhood of $Z_2$. Observe that $Z_{1, 1} = \psi _1^{-1}Z_1 = f^{-1}Z_1 \amalg T$ for some closed subspace $T \subset V_1$ disjoint from $f^{-1}Z_1$ and furthermore $\psi _1$ is étale along $f^{-1}Z_1$. Denote $Z_{i, j} \subset V_ i$ the inverse image of $Z_ j$ under $\psi _ i$. Observe that $\psi _ i : Z_{i, j} \to Z_ j$ is a proper morphism. Since $Z_ i$ and $Z_ j$ are disjoint closed subspaces of $U$, we see that $Z_{i, i}$ and $Z_{i, j}$ are disjoint closed subspaces of $V_ i$.

Denote $\overline{Z}_{i, i}$ and $\overline{Z}_{i, j}$ the scheme theoretic images of $Z_{i, i}$ and $Z_{i, j}$ in $X_ i$. We recall that $|Z_{i, j}|$ is dense in $|\overline{Z}_{i, j}|$, see Morphisms of Spaces, Lemma 66.17.7. After replacing $X_ i$ by a $V_ i$-admissible blowup we may assume that $\overline{Z}_{i, i}$ and $\overline{Z}_{i, j}$ are disjoint, see Lemma 80.14.2. We assume this holds for both $X_1$ and $X_2$. Observe that this property is preserved if we replace $X_ i$ by a further $V_ i$-admissible blowup. Hence we may replace $X_1$ by another $V_1$-admissible blowup and assume $|\overline{Z}_{1, 1}|$ is the disjoint union of the closures of $|T|$ and $|f^{-1}Z_1|$ in $|X_1|$.

Set $V_{12} = V_1 \times _ U V_2$. We have an immersion $V_{12} \to X_1 \times _ B X_2$ which is the composition of the closed immersion $V_{12} = V_1 \times _ U V_2 \to V_1 \times _ B V_2$ (Morphisms of Spaces, Lemma 66.4.5) and the open immersion $V_1 \times _ B V_2 \to X_1 \times _ B X_2$. Let $X_{12} \subset X_1 \times _ B X_2$ be the scheme theoretic image of $V_{12} \to X_1 \times _ B X_2$. The projection morphisms

$p_1 : X_{12} \to X_1 \quad \text{and}\quad p_2 : X_{12} \to X_2$

are proper as $X_1$ and $X_2$ are proper over $B$. If we replace $X_1$ by a $V_1$-admissible blowing up, then $X_{12}$ is replaced by the strict transform with respect to this blowing up, see Lemma 80.14.5.

Denote $\psi : V_{12} \to U$ the compositions $\psi = \psi _1 \circ p_1|_{V_{12}} = \psi _2 \circ p_2|_{V_{12}}$. Consider the closed subspace

$Z_{12, 2} = (p_1|_{V_{12}})^{-1}Z_{1, 2} = (p_2|_{V_{12}})^{-1}Z_{2, 2} = \psi ^{-1}Z_2 \subset V_{12}$

The morphism $p_1|_{V_{12}} : V_{12} \to V_1$ is an isomorphism over an open neighbourhood of $Z_{1, 2}$ because $\psi _2 : V_2 \to U$ is an isomorphism over an open neighbourhood of $Z_2$ and $V_{12} = V_1 \times _ U V_2$. By Lemma 80.14.3 there exists a $V_1$-admissible blowing up $X_1' \to X_1$ such that the strict tranform $p'_1 : X'_{12} \to X'_1$ of $p_1$ is an isomorphism over an open neighbourhood of the closure of $|Z_{1, 2}|$ in $|X'_1|$. After replacing $X_1$ by $X'_1$ and $X_{12}$ by $X'_{12}$ we may assume that $p_1$ is an isomorphism over an open neighbourhood of $|\overline{Z}_{1, 2}|$.

The result of the previous paragraph tells us that

$X_{12} \cap (\overline{Z}_{1, 2} \times _ B \overline{Z}_{2, 1}) = \emptyset$

where the intersection taken in $X_1 \times _ B X_2$. Namely, the inverse image $p_1^{-1}\overline{Z}_{1, 2}$ in $X_{12}$ maps isomorphically to $\overline{Z}_{1, 2}$. In particular, we see that $|Z_{12, 2}|$ is dense in $|p_1^{-1}\overline{Z}_{1, 2}|$. Thus $p_2$ maps $|p_1^{-1}\overline{Z}_{1, 2}|$ into $|\overline{Z}_{2, 2}|$. Since $|\overline{Z}_{2, 2}| \cap |\overline{Z}_{2, 1}| = \emptyset$ we conclude.

It turns out that we need to do one additional blowing up before we can conclude the argument. Namely, let $V_2 \subset W_2 \subset X_2$ be the open subspace with underlying topological space

$|W_2| = |V_2| \cup (|X_2| \setminus |\overline{Z}_{2, 1}|) = |X_2| \setminus \left(|\overline{Z}_{2, 1}| \setminus |Z_{2, 1}|\right)$

Since $p_2(p_1^{-1}\overline{Z}_{1, 2})$ is contained in $W_2$ (see above) we see that replacing $X_2$ by a $W_2$-admissible blowup and $X_{21}$ by the corresponding strict transform will preserve the property of $p_1$ being an isomorphism over an open neighbourhood of $\overline{Z}_{1, 2}$. Since $\overline{Z}_{2, 1} \cap W_2 = \overline{Z}_{2, 1} \cap V_2 = Z_{2, 1}$ we see that $Z_{2, 1}$ is a closed subspace of $W_2$ and $V_2$. Observe that $V_{12} = V_1 \times _ U V_2 = p_1^{-1}(V_1) = p_2^{-1}(V_2)$ as open subspaces of $X_{12}$ as it is the largest open subspace of $X_{12}$ over which the morphism $\psi : V_{12} \to U$ extends; details omitted1. We have the following equalities of closed subspaces of $V_{12}$:

$p_2^{-1}Z_{2, 1} = p_2^{-1} \psi _2^{-1} Z_1 = p_1^{-1} \psi _1^{-1} Z_1= p_1^{-1}Z_{1, 1} = p_1^{-1}f^{-1}Z_1 \amalg p_1^{-1}T$

Here and below we use the slight abuse of notation of writing $p_2$ in stead of the restriction of $p_2$ to $V_{12}$, etc. Since $p_2^{-1}(Z_{2, 1})$ is a closed subspace of $p_2^{-1}(W_2)$ as $Z_{2, 1}$ is a closed subspace of $W_2$ we conclude that also $p_1^{-1}f^{-1}Z_1$ is a closed subspace of $p_2^{-1}(W_2)$. Finally, the morphism $p_2 : X_{12} \to X_2$ is étale at points of $p_1^{-1}f^{-1}Z_1$ as $\psi _1$ is étale along $f^{-1}Z_1$ and $V_{12} = V_1 \times _ U V_2$. Thus we may apply Lemma 80.14.4 to the morphism $p_2 : X_{12} \to X_2$, the open $W_2$, the closed subspace $Z_{2, 1} \subset W_2$, and the closed subspace $p_1^{-1}f^{-1}Z_1 \subset p_2^{-1}(W_2)$. Hence after replacing $X_2$ by a $W_2$-admissible blowup and $X_{12}$ by the corresponding strict transform, we obtain for every geometric point $\overline{y}$ of the closure of $|p_1^{-1}f^{-1}Z_1|$ a local ring map $\mathcal{O}_{X_{12}, \overline{y}} \to \mathcal{O}$ such that $\mathcal{O}_{X_2, p_2(\overline{y})} \to \mathcal{O}$ is finite flat.

Consider the algebraic space

$W_2 = U \coprod \nolimits _{U_2} (X_2 \setminus \overline{Z}_{2, 1}),$

and with $T \subset V_1$ as in the first paragraph the algebraic space

$W_1 = U \coprod \nolimits _{U_1} (X_1 \setminus \overline{Z}_{1, 2} \cup \overline{T}),$

obtained by pushout, see Lemma 80.9.2. Let us apply Lemma 80.14.1 to see that $W_ i \to B$ is separated. First, $U \to B$ and $X_ i \to B$ are separated. Let us check the quasi-compact immersion $U_ i \to U \times _ B (X_ i \setminus \overline{Z}_{i, j})$ is closed using the valuative criterion, see Morphisms of Spaces, Lemma 66.42.1. Choose a valuation ring $A$ over $B$ with fraction field $K$ and compatible morphisms $(u, x_ i) : \mathop{\mathrm{Spec}}(A) \to U \times _ B X_ i$ and $u_ i : \mathop{\mathrm{Spec}}(K) \to U_ i$. Since $\psi _ i$ is proper, we can find a unique $v_ i : \mathop{\mathrm{Spec}}(A) \to V_ i$ compatible with $u$ and $u_ i$. Since $X_ i$ is proper over $B$ we see that $x_ i = v_ i$. If $v_ i$ does not factor through $U_ i \subset V_ i$, then we conclude that $x_ i$ maps the closed point of $\mathop{\mathrm{Spec}}(A)$ into $Z_{i, j}$ or $T$ when $i = 1$. This finishes the proof because we removed $\overline{Z}_{i, j}$ and $\overline{T}$ in the construction of $W_ i$.

On the other hand, for any valuation ring $A$ over $B$ with fraction field $K$ and any morphism

$\gamma : \mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Im}}(U_1 \times _ U U_2 \to U)$

over $B$, we claim that after replacing $A$ by an extension of valuation rings, there is an $i$ and an extension of $\gamma$ to a morphism $h_ i : \mathop{\mathrm{Spec}}(A) \to W_ i$. Namely, we first extend $\gamma$ to a morphism $g_2 : \mathop{\mathrm{Spec}}(A) \to X_2$ using the valuative criterion of properness. If the image of $g_2$ does not meet $\overline{Z}_{2, 1}$, then we obtain our morphism into $W_2$. Otherwise, denote $\overline{z} \in \overline{Z}_{2, 1}$ a geometric point lying over the image of the closed point under $g_2$. We may lift this to a geometric point $\overline{y}$ of $X_{12}$ in the closure of $|p_1^{-1}f^{-1}Z_1|$ because the map of spaces $|p_1^{-1}f^{-1}Z_1| \to |\overline{Z}_{2, 1}|$ is closed with image containing the dense open $|Z_{2, 1}|$. After replacing $A$ by its strict henselization (More on Algebra, Lemma 15.123.5) we get the following diagram

$\xymatrix{ A \ar@{..>}[rr] & & A' \\ \mathcal{O}_{X_2, \overline{z}} \ar[r] \ar[u] & \mathcal{O}_{X_{12}, \overline{y}} \ar[r] & \mathcal{O} \ar@{..>}[u] }$

where $\mathcal{O}_{X_{12}, \overline{y}} \to \mathcal{O}$ is the map we found in the 5th paragraph of the proof. Since the horizontal composition is finite and flat we can find an extension of valuation rings $A'/A$ and dotted arrow making the diagram commute. After replacing $A$ by $A'$ this means that we obtain a lift $g_{12} : \mathop{\mathrm{Spec}}(A) \to X_{12}$ whose closed point maps into the closure of $|p_1^{-1}f^{-1}Z_1|$. Then $g_1 = p_1 \circ g_{12} : \mathop{\mathrm{Spec}}(A) \to X_1$ is a morphism whose closed point maps into the closure of $|f^{-1}Z_1|$. Since the closure of $|f^{-1}Z_1|$ is disjoint from the closure of $|T|$ and contained in $|\overline{Z}_{1, 1}|$ which is disjoint from $|\overline{Z}_{1, 2}|$ we conclude that $g_1$ defines a morphism $h_1 : \mathop{\mathrm{Spec}}(A) \to W_1$ as desired.

Consider a diagram

$\xymatrix{ W_1' \ar[d] \ar[r] & W & W_2' \ar[l] \ar[d] \\ W_1 & U \ar[l] \ar[lu] \ar[u] \ar[ru] \ar[r] & W_2 }$

as in More on Morphisms of Spaces, Lemma 75.40.1. By the previous paragraph for every solid diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r]_\gamma \ar[d] & W \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar@{..>}[ru] \ar[r] & B }$

where $\mathop{\mathrm{Im}}(\gamma ) \subset \mathop{\mathrm{Im}}(U_1 \times _ U U_2 \to U)$ there is an $i$ and an extension $h_ i : \mathop{\mathrm{Spec}}(A) \to W_ i$ of $\gamma$ after possibly replacing $A$ by an extension of valuation rings. Using the valuative criterion of properness for $W'_ i \to W_ i$, we can then lift $h_ i$ to $h'_ i : \mathop{\mathrm{Spec}}(A) \to W'_ i$. Hence the dotted arrow in the diagram exists after possibly extending $A$. Since $W$ is separated over $B$, we see that the choice of extension isn't needed and the arrow is unique as well, see Morphisms of Spaces, Lemmas 66.41.5 and 66.43.1. Then finally the existence of the dotted arrow implies that $W \to B$ is universally closed by Morphisms of Spaces, Lemma 66.42.5. As $W \to B$ is already of finite type and separated, we win. $\square$

[1] Namely, $V_1 \times _ U V_2$ is proper over $U$ so if $\psi$ extends to a larger open of $X_{12}$, then $V_1 \times _ U V_2$ would be closed in this open by Morphisms of Spaces, Lemma 66.40.6. Then we get equality as $V_{12} \subset X_{12}$ is dense.

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