## 81.13 Coequalizers and glueing

Let $X$ be a Noetherian algebraic space and $Z \to X$ a closed subspace. Let $X' \to X$ be the blowing up in $Z$. In this section we show that $X$ can be recovered from $X'$, $Z_ n$ and glueing data where $Z_ n$ is the $n$th infinitesimal neighbourhood of $Z$ in $X$.

Lemma 81.13.1. Let $S$ be a scheme. Let

\[ g : Y \longrightarrow X \]

be a morphism of algebraic spaces over $S$. Assume $X$ is locally Noetherian, and $g$ is proper. Let $R = Y \times _ X Y$ with projection morphisms $t, s : R \to Y$. There exists a coequalizer $X'$ of $s, t : R \to Y$ in the category of algebraic spaces over $S$. Moreover

The morphism $X' \to X$ is finite.

The morphism $Y \to X'$ is proper.

The morphism $Y \to X'$ is surjective.

The morphism $X' \to X$ is universally injective.

If $g$ is surjective, the morphism $X' \to X$ is a universal homeomorphism.

**Proof.**
Denote $h : R \to X$ denote the composition of either $s$ or $t$ with $g$. Then $h$ is proper by Morphisms of Spaces, Lemmas 67.40.3 and 67.40.4. The sheaves

\[ g_*\mathcal{O}_ Y \quad \text{and}\quad h_*\mathcal{O}_ R \]

are coherent $\mathcal{O}_ X$-algebras by Cohomology of Spaces, Lemma 69.20.2. The $X$-morphisms $s$, $t$ induce $\mathcal{O}_ X$-algebra maps $s^\sharp , t^\sharp $ from the first to the second. Set

\[ \mathcal{A} = \text{Equalizer}\left(s^\sharp , t^\sharp : g_*\mathcal{O}_ Y \longrightarrow h_*\mathcal{O}_ R\right) \]

Then $\mathcal{A}$ is a coherent $\mathcal{O}_ X$-algebra and we can define

\[ X' = \underline{\mathop{\mathrm{Spec}}}_ X(\mathcal{A}) \]

as in Morphisms of Spaces, Definition 67.20.8. By Morphisms of Spaces, Remark 67.20.9 and functoriality of the $\underline{\mathop{\mathrm{Spec}}}$ construction there is a factorization

\[ Y \longrightarrow X' \longrightarrow X \]

and the morphism $g' : Y \to X'$ equalizes $s$ and $t$.

Before we show that $X'$ is the coequalizer of $s$ and $t$, we show that $Y \to X'$ and $X' \to X$ have the desired properties. Since $\mathcal{A}$ is a coherent $\mathcal{O}_ X$-module it is clear that $X' \to X$ is a finite morphism of algebraic spaces. This proves (1). The morphism $Y \to X'$ is proper by Morphisms of Spaces, Lemma 67.40.6. This proves (2). Denote $Y \to Y' \to X$ with $Y' = \underline{\mathop{\mathrm{Spec}}}_ X(g_*\mathcal{O}_ Y)$ the Stein factorization of $g$, see More on Morphisms of Spaces, Theorem 76.36.4. Of course we obtain morphisms $Y \to Y' \to X' \to X$ fitting with the morphisms studied above. Since $\mathcal{O}_{X'} \subset g_*\mathcal{O}_ Y$ is a finite extension we see that $Y' \to X'$ is finite and surjective. Some details omitted; hint: use Algebra, Lemma 10.36.17 and reduce to the affine case by étale localization. Since $Y \to Y'$ is surjective (with geometrically connected fibres) we conclude that $Y \to X'$ is surjective. This proves (3). To show that $X' \to X$ is universally injective, we have to show that $X' \to X' \times _ X X'$ is surjective, see Morphisms of Spaces, Definition 67.19.3 and Lemma 67.19.2. Since $Y \to X'$ is surjective (see above) and since base changes and compositions of surjective morphisms are surjective by Morphisms of Spaces, Lemmas 67.5.5 and 67.5.4 we see that $Y \times _ X Y \to X' \times _ X X'$ is surjective. However, since $Y \to X'$ equalizes $s$ and $t$, we see that $Y \times _ X Y \to X' \times _ X X'$ factors through $X' \to X' \times _ X X'$ and we conclude this latter map is surjective. This proves (4). Finally, if $g$ is surjective, then since $g$ factors through $X' \to X$ we see that $X' \to X$ is surjective. Since a surjective, universally injective, finite morphism is a universal homeomorphism (because it is universally bijective and universally closed), this proves (5).

In the rest of the proof we show that $Y \to X'$ is the coequalizer of $s$ and $t$ in the category of algebraic spaces over $S$. Observe that $X'$ is locally Noetherian (Morphisms of Spaces, Lemma 67.23.5). Moreover, observe that $Y \times _{X'} Y \to Y \times _ X Y$ is an isomorphism as $Y \to X'$ equalizes $s$ and $t$ (this is a categorical statement). Hence in order to prove the statement that $Y \to X'$ is the coequalizer of $s$ and $t$, we may and do assume $X = X'$. In other words, $\mathcal{O}_ X$ is the equalizer of the maps $s^\sharp , t^\sharp : g_*\mathcal{O}_ Y \to h_*\mathcal{O}_ R$.

Let $X_1 \to X$ be a flat morphism of algebraic spaces over $S$ with $X_1$ locally Noetherian. Denote $g_1 : Y_1 \to X_1$, $h_1 : R_1 \to X_1$ and $s_1, t_1 : R_1 \to Y_1$ the base changes of $g, h, s, t$ to $X_1$. Of course $g_1$ is proper and $R_1 = Y_1 \times _{X_1} Y_1$. Since we have flat base change for pushforward of quasi-coherent modules, Cohomology of Spaces, Lemma 69.11.2, we see that $\mathcal{O}_{X_1}$ is the equalizer of the maps $s_1^\sharp , t_1^\sharp : g_{1, *}\mathcal{O}_{Y_1} \to h_{1, *}\mathcal{O}_{R_1}$. Hence all the assumptions we have are preserved by this base change.

At this point we are going to check conditions (1) and (2) of Lemma 81.3.3. Condition (1) follows from Lemma 81.5.1 and the fact that $g$ is proper and surjective (because $X = X'$). To check condition (2), by the remarks on base change above, we reduce to the statement discussed and proved in the next paragraph.

Assume $S = \mathop{\mathrm{Spec}}(A)$ is an affine scheme, $X = X'$ is an affine scheme, and $Z$ is an affine scheme over $S$. We have to show that

\[ \mathop{\mathrm{Mor}}\nolimits _ S(X, Z) \longrightarrow \text{Equalizer}(s, t : \mathop{\mathrm{Mor}}\nolimits _ S(Y, Z) \to \mathop{\mathrm{Mor}}\nolimits _ S(R, Z)) \]

is bijective. However, this is clear from the fact that $X = X'$ which implies $\mathcal{O}_ X$ is the equalizer of the maps $s^\sharp , t^\sharp : g_*\mathcal{O}_ Y \to h_*\mathcal{O}_ R$ which in turn implies

\[ \Gamma (X, \mathcal{O}_ X) = \text{Equalizer}\left( s^\sharp , t^\sharp : \Gamma (Y, \mathcal{O}_ Y) \to \Gamma (R, \mathcal{O}_ R) \right) \]

Namely, we have

\[ \mathop{\mathrm{Mor}}\nolimits _ S(X, Z) = \mathop{\mathrm{Hom}}\nolimits _ A(\Gamma (Z, \mathcal{O}_ Z), \Gamma (X, \mathcal{O}_ X)) \]

and similarly for $Y$ and $R$, see Properties of Spaces, Lemma 66.33.1.
$\square$

We will work in the following situation.

Situation 81.13.2. Let $S$ be a scheme. Let $X$ be a locally Noetherian algebraic space over $S$. Let $Z \to X$ be a closed immersion and let $U \subset X$ be the complementary open subspace. Finally, let $f : X' \to X$ be a proper morphism of algebraic spaces such that $f^{-1}(U) \to U$ is an isomorphism.

Lemma 81.13.3. In Situation 81.13.2 let $Y = X' \amalg Z$ and $R = Y \times _ X Y$ with projections $t, s : R \to Y$. There exists a coequalizer $X_1$ of $s, t : R \to Y$ in the category of algebraic spaces over $S$. The morphism $X_1 \to X$ is a finite universal homeomorphism, an isomorphism over $U$, and $Z \to X$ lifts to $X_1$.

**Proof.**
Existence of $X_1$ and the fact that $X_1 \to X$ is a finite universal homeomorphism is a special case of Lemma 81.13.1. The formation of $X_1$ commutes with étale localization on $X$ (see proof of Lemma 81.13.1). Thus the morphism $X_1 \to X$ is an isomorphism over $U$. It is immediate from the construction that $Z \to X$ lifts to $X_1$.
$\square$

In Situation 81.13.2 for $n \geq 1$ let $Z_ n \subset X$ be the $n$th order infinitesimal neighbourhood of $Z$ in $X$, i.e., the closed subscheme defined by the $n$th power of the sheaf of ideals cutting out $Z$. Consider $Y_ n = X' \amalg Z_ n$ and $R_ n = Y_ n \times _ X Y_ n$ and the coequalizer

\[ \xymatrix{ R_ n \ar@<1ex>[r] \ar@<-1ex>[r] & Y_ n \ar[r] & X_ n \ar[r] & X } \]

as in Lemma 81.13.3. The maps $Y_ n \to Y_{n + 1}$ and $R_ n \to R_{n + 1}$ induce morphisms

81.13.3.1
\begin{equation} \label{spaces-pushouts-equation-system-coequalizers} X_1 \to X_2 \to X_3 \to \ldots \to X \end{equation}

Each of these morphisms is a universal homeomorphism as the morphisms $X_ n \to X$ are universal homeomorphisms.

Lemma 81.13.4. In Situation 81.13.2 assume $X$ quasi-compact. In (81.13.3.1) for all $n$ large enough, there exists an $m$ such that $X_ n \to X_{n + m}$ factors through a closed immersion $X \to X_{n + m}$.

**Proof.**
Let's look a bit more closely at the construction of $X_ n$ and how it changes as we increase $n$. We have $X_ n = \underline{\mathop{\mathrm{Spec}}}(\mathcal{A}_ n)$ where $\mathcal{A}_ n$ is the equalizer of $s_ n^\sharp $ and $t_ n^\sharp $ going from $g_{n , *}\mathcal{O}_{Y_ n}$ to $h_{n, *}\mathcal{O}_{R_ n}$. Here $g_ n : Y_ n = X' \amalg Z_ n \to X$ and $h_ n : R_ n = Y_ n \times _ X Y_ n \to X$ are the given morphisms. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the coherent sheaf of ideals corresponding to $Z$. Then

\[ g_{n, *}\mathcal{O}_{Y_ n} = f_*\mathcal{O}_{X'} \times \mathcal{O}_ X/\mathcal{I}^ n \]

Similarly, we have a decomposition

\[ R_ n = X' \times _ X X' \amalg X' \times _ X Z_ n \amalg Z_ n \times _ X X' \amalg Z_ n \times _ X Z_ n \]

As $Z_ n \to X$ is a monomorphism, we see that $X' \times _ X Z_ n = Z_ n \times _ X X'$ and that this identification is compatible with the two morphisms to $X$, with the two morphisms to $X'$, and with the two morphisms to $Z_ n$. Denote $f_ n : X' \times _ X Z_ n \to X$ the morphism to $X$. Denote

\[ \mathcal{A} = \text{Equalizer}( \xymatrix{ f_*\mathcal{O}_{X'} \ar@<1ex>[r] \ar@<-1ex>[r] & (f \times f)_*\mathcal{O}_{X' \times _ X X'} } ) \]

By the remarks above we find that

\[ \mathcal{A}_ n = \text{Equalizer}( \xymatrix{ \mathcal{A} \times \mathcal{O}_ X/\mathcal{I}^ n \ar@<1ex>[r] \ar@<-1ex>[r] & f_{n, *}\mathcal{O}_{X' \times _ X Z_ n} } ) \]

We have canonical maps

\[ \mathcal{O}_ X \to \ldots \to \mathcal{A}_3 \to \mathcal{A}_2 \to \mathcal{A}_1 \]

of coherent $\mathcal{O}_ X$-algebras. The statement of the lemma means that for $n$ large enough there exists an $m \geq 0$ such that the image of $\mathcal{A}_{n + m} \to \mathcal{A}_ n$ is isomorphic to $\mathcal{O}_ X$. This we may check étale locally on $X$. Hence by Properties of Spaces, Lemma 66.6.3 we may assume $X$ is an affine Noetherian scheme.

Since $X_ n \to X$ is an isomorphism over $U$ we see that the kernel of $\mathcal{O}_ X \to \mathcal{A}_ n$ is supported on $|Z|$. Since $X$ is Noetherian, the sequence of kernels $\mathcal{J}_ n = \mathop{\mathrm{Ker}}(\mathcal{O}_ X \to \mathcal{A}_ n)$ stabilizes (Cohomology of Spaces, Lemma 69.13.1). Say $\mathcal{J}_{n_0} = \mathcal{J}_{n_0 + 1} = \ldots = \mathcal{J}$. By Cohomology of Spaces, Lemma 69.13.2 we find that $\mathcal{I}^ t \mathcal{J} = 0$ for some $t \geq 0$. On the other hand, there is an $\mathcal{O}_ X$-algebra map $\mathcal{A}_ n \to \mathcal{O}_ X/\mathcal{I}^ n$ and hence $\mathcal{J} \subset \mathcal{I}^ n$ for all $n$. By Artin-Rees (Cohomology of Spaces, Lemma 69.13.3) we find that $\mathcal{J} \cap \mathcal{I}^ n \subset \mathcal{I}^{n - c}\mathcal{J}$ for some $c \geq 0$ and all $n \gg 0$. We conclude that $\mathcal{J} = 0$.

Pick $n \geq n_0$ as in the previous paragraph. Then $\mathcal{O}_ X \to \mathcal{A}_ n$ is injective. Hence it now suffices to find $m \geq 0$ such that the image of $\mathcal{A}_{n + m} \to \mathcal{A}_ n$ is equal to the image of $\mathcal{O}_ X$. Observe that $\mathcal{A}_ n$ sits in a short exact sequence

\[ 0 \to \mathop{\mathrm{Ker}}(\mathcal{A} \to f_{n, *}\mathcal{O}_{X' \times _ X Z_ n}) \to \mathcal{A}_ n \to \mathcal{O}_ X/\mathcal{I}^ n \to 0 \]

and similarly for $\mathcal{A}_{n + m}$. Hence it suffices to show

\[ \mathop{\mathrm{Ker}}(\mathcal{A} \to f_{n + m, *}\mathcal{O}_{X' \times _ X Z_{n + m}}) \subset \mathop{\mathrm{Im}}(\mathcal{I}^ n \to \mathcal{A}) \]

for some $m \geq 0$. To do this we may work étale locally on $X$ and since $X$ is Noetherian we may assume that $X$ is a Noetherian affine scheme. Say $X = \mathop{\mathrm{Spec}}(R)$ and $\mathcal{I}$ corresponds to the ideal $I \subset R$. Let $\mathcal{A} = \widetilde{A}$ for a finite $R$-algebra $A$. Let $f_*\mathcal{O}_{X'} = \widetilde{B}$ for a finite $R$-algebra $B$. Then $R \to A \subset B$ and these maps become isomorphisms on inverting any element of $I$.

Note that $f_{n, *}\mathcal{O}_{X' \times _ X Z_ n}$ is equal to $f_*(\mathcal{O}_{X'}/I^ n\mathcal{O}_{X'})$ in the notation used in Cohomology of Spaces, Section 69.22. By Cohomology of Spaces, Lemma 69.22.4 we see that there exists a $c \geq 0$ such that

\[ \mathop{\mathrm{Ker}}(B \to \Gamma (X, f_*(\mathcal{O}_{X'}/I^{n + m + c}\mathcal{O}_{X'})) \]

is contained in $I^{n + m}B$. On the other hand, as $R \to B$ is finite and an isomorphism after inverting any element of $I$ we see that $I^{n + m}B \subset \mathop{\mathrm{Im}}(I^ n \to B)$ for $m$ large enough (can be chosen independent of $n$). This finishes the proof as $A \subset B$.
$\square$

## Comments (3)

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