The Stacks project

80.6 Pushouts along thickenings and affine morphisms

This section is analogue of More on Morphisms, Section 37.14.

Lemma 80.6.1. Let $S$ be a scheme. Let $X \to X'$ be a thickening of schemes over $S$ and let $X \to Y$ be an affine morphism of schemes over $S$. Let $Y' = Y \amalg _ X X'$ be the pushout in the category of schemes (see More on Morphisms, Lemma 37.14.3). Then $Y'$ is also a pushout in the category of algebraic spaces over $S$.

Lemma 80.6.2. Let $S$ be a scheme. Let $X \to X'$ be a thickening of algebraic spaces over $S$ and let $X \to Y$ be an affine morphism of algebraic spaces over $S$. Then there exists a pushout

\[ \xymatrix{ X \ar[r] \ar[d]_ f & X' \ar[d]^{f'} \\ Y \ar[r] & Y \amalg _ X X' } \]

in the category of algebraic spaces over $S$. Moreover $Y' = Y \amalg _ X X'$ is a thickening of $Y$ and

\[ \mathcal{O}_{Y'} = \mathcal{O}_ Y \times _{f_*\mathcal{O}_ X} f'_*\mathcal{O}_{X'} \]

as sheaves on $Y_{\acute{e}tale}= (Y')_{\acute{e}tale}$.

Proof. Choose a scheme $V$ and a surjective étale morphism $V \to Y$. Set $U = V \times _ Y X$. This is a scheme affine over $V$ with a surjective étale morphism $U \to X$. By More on Morphisms of Spaces, Lemma 75.9.6 there exists a $U' \to X'$ surjective étale with $U = U' \times _{X'} X$. In particular the morphism of schemes $U \to U'$ is a thickening too. Apply More on Morphisms, Lemma 37.14.3 to obtain a pushout $V' = V \amalg _ U U'$ in the category of schemes.

We repeat this procedure to construct a pushout

\[ \xymatrix{ U \times _ X U \ar[d] \ar[r] & U' \times _{X'} U' \ar[d] \\ V \times _ Y V \ar[r] & R' } \]

in the category of schemes. Consider the morphisms

\[ U \times _ X U \to U \to V',\quad U' \times _{X'} U' \to U' \to V',\quad V \times _ Y V \to V \to V' \]

where we use the first projection in each case. Clearly these glue to give a morphism $t' : R' \to V'$ which is étale by More on Morphisms, Lemma 37.14.6. Similarly, we obtain $s' : R' \to V'$ étale. The morphism $j' = (t', s') : R' \to V' \times _ S V'$ is unramified (as $t'$ is étale) and a monomorphism when restricted to the closed subscheme $V \times _ Y V \subset R'$. As $V \times _ Y V \subset R'$ is a thickening it follows that $j'$ is a monomorphism too. Finally, $j'$ is an equivalence relation as we can use the functoriality of pushouts of schemes to construct a morphism $c' : R' \times _{s', V', t'} R' \to R'$ (details omitted). At this point we set $Y' = U'/R'$, see Spaces, Theorem 64.10.5.

We have morphisms $X' = U'/U' \times _{X'} U' \to V'/R' = Y'$ and $Y = V/V \times _ Y V \to V'/R' = Y'$. By construction these fit into the commutative diagram

\[ \xymatrix{ X \ar[r] \ar[d]_ f & X' \ar[d]^{f'} \\ Y \ar[r] & Y' } \]

Since $Y \to Y'$ is a thickening we have $Y_{\acute{e}tale}= (Y')_{\acute{e}tale}$, see More on Morphisms of Spaces, Lemma 75.9.6. The commutativity of the diagram gives a map of sheaves

\[ \mathcal{O}_{Y'} \longrightarrow \mathcal{O}_ Y \times _{f_*\mathcal{O}_ X} f'_*\mathcal{O}_{X'} \]

on this set. By More on Morphisms, Lemma 37.14.3 this map is an isomorphism when we restrict to the scheme $V'$, hence it is an isomorphism.

To finish the proof we show that the diagram above is a pushout in the category of algebraic spaces. To see this, let $Z$ be an algebraic space and let $a' : X' \to Z$ and $b : Y \to Z$ be morphisms of algebraic spaces. By Lemma 80.6.1 we obtain a unique morphism $h : V' \to Z$ fitting into the commutative diagrams

\[ \vcenter { \xymatrix{ U' \ar[d] \ar[r] & V' \ar[d]^ h \\ X' \ar[r]^{a'} & Z } } \quad \text{and}\quad \vcenter { \xymatrix{ V \ar[r] \ar[d] & V' \ar[d]^ h \\ Y \ar[r]^ b & Z } } \]

The uniqueness shows that $h \circ t' = h \circ s'$. Hence $h$ factors uniquely as $V' \to Y' \to Z$ and we win. $\square$

In the following lemma we use the fibre product of categories as defined in Categories, Example 4.31.3.

Lemma 80.6.3. Let $S$ be a base scheme. Let $X \to X'$ be a thickening of algebraic spaces over $S$ and let $X \to Y$ be an affine morphism of algebraic spaces over $S$. Let $Y' = Y \amalg _ X X'$ be the pushout (see Lemma 80.6.2). Base change gives a functor

\[ F : (\textit{Spaces}/Y') \longrightarrow (\textit{Spaces}/Y) \times _{(\textit{Spaces}/Y')} (\textit{Spaces}/X') \]

given by $V' \longmapsto (V' \times _{Y'} Y, V' \times _{Y'} X', 1)$ which sends $(\mathit{Sch}/Y')$ into $(\mathit{Sch}/Y) \times _{(\mathit{Sch}/Y')} (\mathit{Sch}/X')$. The functor $F$ has a left adjoint

\[ G : (\textit{Spaces}/Y) \times _{(\textit{Spaces}/Y')} (\textit{Spaces}/X') \longrightarrow (\textit{Spaces}/Y') \]

which sends the triple $(V, U', \varphi )$ to the pushout $V \amalg _{(V \times _ Y X)} U'$ in the category of algebraic spaces over $S$. The functor $G$ sends $(\mathit{Sch}/Y) \times _{(\mathit{Sch}/Y')} (\mathit{Sch}/X')$ into $(\mathit{Sch}/Y')$.

Proof. The proof is completely formal. Since the morphisms $X \to X'$ and $X \to Y$ are representable it is clear that $F$ sends $(\mathit{Sch}/Y')$ into $(\mathit{Sch}/Y) \times _{(\mathit{Sch}/Y')} (\mathit{Sch}/X')$.

Let us construct $G$. Let $(V, U', \varphi )$ be an object of the fibre product category. Set $U = U' \times _{X'} X$. Note that $U \to U'$ is a thickening. Since $\varphi : V \times _ Y X \to U' \times _{X'} X = U$ is an isomorphism we have a morphism $U \to V$ over $X \to Y$ which identifies $U$ with the fibre product $X \times _ Y V$. In particular $U \to V$ is affine, see Morphisms of Spaces, Lemma 66.20.5. Hence we can apply Lemma 80.6.2 to get a pushout $V' = V \amalg _ U U'$. Denote $V' \to Y'$ the morphism we obtain in virtue of the fact that $V'$ is a pushout and because we are given morphisms $V \to Y$ and $U' \to X'$ agreeing on $U$ as morphisms into $Y'$. Setting $G(V, U', \varphi ) = V'$ gives the functor $G$.

If $(V, U', \varphi )$ is an object of $(\mathit{Sch}/Y) \times _{(\mathit{Sch}/Y')} (\mathit{Sch}/X')$ then $U = U' \times _{X'} X$ is a scheme too and we can form the pushout $V' = V \amalg _ U U'$ in the category of schemes by More on Morphisms, Lemma 37.14.3. By Lemma 80.6.1 this is also a pushout in the category of schemes, hence $G$ sends $(\mathit{Sch}/Y) \times _{(\mathit{Sch}/Y')} (\mathit{Sch}/X')$ into $(\mathit{Sch}/Y')$.

Let us prove that $G$ is a left adjoint to $F$. Let $Z$ be an algebraic space over $Y'$. We have to show that

\[ \mathop{\mathrm{Mor}}\nolimits (V', Z) = \mathop{\mathrm{Mor}}\nolimits ((V, U', \varphi ), F(Z)) \]

where the morphism sets are taking in their respective categories. Let $g' : V' \to Z$ be a morphism. Denote $\tilde g$, resp. $\tilde f'$ the composition of $g'$ with the morphism $V \to V'$, resp. $U' \to V'$. Base change $\tilde g$, resp. $\tilde f'$ by $Y \to Y'$, resp. $X' \to Y'$ to get a morphism $g : V \to Z \times _{Y'} Y$, resp. $f' : U' \to Z \times _{Y'} X'$. Then $(g, f')$ is an element of the right hand side of the equation above (details omitted). Conversely, suppose that $(g, f') : (V, U', \varphi ) \to F(Z)$ is an element of the right hand side. We may consider the composition $\tilde g : V \to Z$, resp. $\tilde f' : U' \to Z$ of $g$, resp. $f$ by $Z \times _{Y'} X' \to Z$, resp. $Z \times _{Y'} Y \to Z$. Then $\tilde g$ and $\tilde f'$ agree as morphism from $U$ to $Z$. By the universal property of pushout, we obtain a morphism $g' : V' \to Z$, i.e., an element of the left hand side. We omit the verification that these constructions are mutually inverse. $\square$

Lemma 80.6.4. Let $S$ be a scheme. Let

\[ \xymatrix{ A \ar[r] \ar[d] & C \ar[d] \ar[r] & E \ar[d] \\ B \ar[r] & D \ar[r] & F } \]

be a commutative diagram of algebraic spaces over $S$. Assume that $A, B, C, D$ and $A, B, E, F$ form cartesian squares and that $B \to D$ is surjective étale. Then $C, D, E, F$ is a cartesian square.

Proof. This is formal. $\square$

Lemma 80.6.5. In the situation of Lemma 80.6.3 the functor $F \circ G$ is isomorphic to the identity functor.

Proof. We will prove that $F \circ G$ is isomorphic to the identity by reducing this to the corresponding statement of More on Morphisms, Lemma 37.14.4.

Choose a scheme $Y_1$ and a surjective étale morphism $Y_1 \to Y$. Set $X_1 = Y_1 \times _ Y X$. This is a scheme affine over $Y_1$ with a surjective étale morphism $X_1 \to X$. By More on Morphisms of Spaces, Lemma 75.9.6 there exists a $X'_1 \to X'$ surjective étale with $X_1 = X_1' \times _{X'} X$. In particular the morphism of schemes $X_1 \to X_1'$ is a thickening too. Apply More on Morphisms, Lemma 37.14.3 to obtain a pushout $Y_1' = Y_1 \amalg _{X_1} X_1'$ in the category of schemes. In the proof of Lemma 80.6.2 we constructed $Y'$ as a quotient of an étale equivalence relation on $Y_1'$ such that we get a commutative diagram

80.6.5.1
\begin{equation} \label{spaces-pushouts-equation-cube} \vcenter { \xymatrix{ & X \ar[rr] \ar '[d][dd] & & X' \ar[dd] \\ X_1 \ar[rr] \ar[dd] \ar[ru] & & X_1' \ar[dd] \ar[ru] & \\ & Y \ar '[r][rr] & & Y' \\ Y_1 \ar[rr] \ar[ru] & & Y_1' \ar[ru] } } \end{equation}

where all squares except the front and back squares are cartesian (the front and back squares are pushouts) and the northeast arrows are surjective étale. Denote $F_1$, $G_1$ the functors constructed in More on Morphisms, Lemma 37.14.4 for the front square. Then the diagram of categories

\[ \xymatrix{ (\mathit{Sch}/Y_1') \ar@<-1ex>[r]_-{F_1} \ar[d] & (\mathit{Sch}/Y_1) \times _{(\mathit{Sch}/Y_1')} (\mathit{Sch}/X_1') \ar[d] \ar@<-1ex>[l]_-{G_1} \\ (\textit{Spaces}/Y') \ar@<-1ex>[r]_-F & (\textit{Spaces}/Y) \times _{(\textit{Spaces}/Y')} (\textit{Spaces}/X') \ar@<-1ex>[l]_-G } \]

is commutative by simple considerations regarding base change functors and the agreement of pushouts in schemes with pushouts in spaces of Lemma 80.6.1.

Let $(V, U', \varphi )$ be an object of $(\textit{Spaces}/Y) \times _{(\textit{Spaces}/Y')} (\textit{Spaces}/X')$. Denote $U = U' \times _{X'} X$ so that $G(V, U', \varphi ) = V \amalg _ U U'$. Choose a scheme $V_1$ and a surjective étale morphism $V_1 \to Y_1 \times _ Y V$. Set $U_1 = V_1 \times _ Y X$. Then

\[ U_1 = V_1 \times _ Y X \longrightarrow (Y_1 \times _ Y V) \times _ Y X = X_1 \times _ Y V = X_1 \times _ X X \times _ Y V = X_1 \times _ X U \]

is surjective étale too. By More on Morphisms of Spaces, Lemma 75.9.6 there exists a thickening $U_1 \to U_1'$ and a surjective étale morphism $U_1' \to X_1' \times _{X'} U'$ whose base change to $X_1 \times _ X U$ is the displayed morphism. At this point $(V_1, U'_1, \varphi _1)$ is an object of $(\mathit{Sch}/Y_1) \times _{(\mathit{Sch}/Y_1')} (\mathit{Sch}/X_1')$. In the proof of Lemma 80.6.2 we constructed $G(V, U', \varphi ) = V \amalg _ U U'$ as a quotient of an étale equivalence relation on $G_1(V_1, U_1', \varphi _1) = V_1 \amalg _{U_1} U_1'$ such that we get a commutative diagram

80.6.5.2
\begin{equation} \label{spaces-pushouts-equation-cube-over} \vcenter { \xymatrix{ & U \ar[rr] \ar '[d][dd] & & U' \ar[dd] \\ U_1 \ar[rr] \ar[dd] \ar[ru] & & U_1' \ar[dd] \ar[ru] & \\ & V \ar '[r][rr] & & G(V, U', \varphi ) \\ V_1 \ar[rr] \ar[ru] & & G_1(V_1, U_1', \varphi _1) \ar[ru] } } \end{equation}

where all squares except the front and back squares are cartesian (the front and back squares are pushouts) and the northeast arrows are surjective étale. In particular

\[ G_1(V_1, U_1', \varphi _1) \to G(V, U', \varphi ) \]

is surjective étale.

Finally, we come to the proof of the lemma. We have to show that the adjunction mapping $(V, U', \varphi ) \to F(G(V, U', \varphi ))$ is an isomorphism. We know $(V_1, U_1', \varphi _1) \to F_1(G_1(V_1, U_1', \varphi _1))$ is an isomorphism by More on Morphisms, Lemma 37.14.4. Recall that $F$ and $F_1$ are given by base change. Using the properties of (80.6.5.2) and Lemma 80.6.4 we see that $V \to G(V, U', \varphi ) \times _{Y'} Y$ and $U' \to G(V, U', \varphi ) \times _{Y'} X'$ are isomorphisms, i.e., $(V, U', \varphi ) \to F(G(V, U', \varphi ))$ is an isomorphism. $\square$

Lemma 80.6.6. Let $S$ be a base scheme. Let $X \to X'$ be a thickening of algebraic spaces over $S$ and let $X \to Y$ be an affine morphism of algebraic spaces over $S$. Let $Y' = Y \amalg _ X X'$ be the pushout (see Lemma 80.6.2). Let $V' \to Y'$ be a morphism of algebraic spaces over $S$. Set $V = Y \times _{Y'} V'$, $U' = X' \times _{Y'} V'$, and $U = X \times _{Y'} V'$. There is an equivalence of categories between

  1. quasi-coherent $\mathcal{O}_{V'}$-modules flat over $Y'$, and

  2. the category of triples $(\mathcal{G}, \mathcal{F}', \varphi )$ where

    1. $\mathcal{G}$ is a quasi-coherent $\mathcal{O}_ V$-module flat over $Y$,

    2. $\mathcal{F}'$ is a quasi-coherent $\mathcal{O}_{U'}$-module flat over $X$, and

    3. $\varphi : (U \to V)^*\mathcal{G} \to (U \to U')^*\mathcal{F}'$ is an isomorphism of $\mathcal{O}_ U$-modules.

The equivalence maps $\mathcal{G}'$ to $((V \to V')^*\mathcal{G}', (U' \to V')^*\mathcal{G}', can)$. Suppose $\mathcal{G}'$ corresponds to the triple $(\mathcal{G}, \mathcal{F}', \varphi )$. Then

  1. $\mathcal{G}'$ is a finite type $\mathcal{O}_{V'}$-module if and only if $\mathcal{G}$ and $\mathcal{F}'$ are finite type $\mathcal{O}_ Y$ and $\mathcal{O}_{U'}$-modules.

  2. if $V' \to Y'$ is locally of finite presentation, then $\mathcal{G}'$ is an $\mathcal{O}_{V'}$-module of finite presentation if and only if $\mathcal{G}$ and $\mathcal{F}'$ are $\mathcal{O}_ Y$ and $\mathcal{O}_{U'}$-modules of finite presentation.

Proof. A quasi-inverse functor assigns to the triple $(\mathcal{G}, \mathcal{F}', \varphi )$ the fibre product

\[ (V \to V')_*\mathcal{G} \times _{(U \to V')_*\mathcal{F}} (U' \to V')_*\mathcal{F}' \]

where $\mathcal{F} = (U \to U')^*\mathcal{F}'$. This works, because on affines étale over $V'$ and $Y'$ we recover the equivalence of More on Algebra, Lemma 15.7.5. Details omitted.

Parts (a) and (b) reduce by étale localization (Properties of Spaces, Section 65.30) to the case where $V'$ and $Y'$ are affine in which case the result follows from More on Algebra, Lemmas 15.7.4 and 15.7.6. $\square$

Lemma 80.6.7. In the situation of Lemma 80.6.5. If $V' = G(V, U', \varphi )$ for some triple $(V, U', \varphi )$, then

  1. $V' \to Y'$ is locally of finite type if and only if $V \to Y$ and $U' \to X'$ are locally of finite type,

  2. $V' \to Y'$ is flat if and only if $V \to Y$ and $U' \to X'$ are flat,

  3. $V' \to Y'$ is flat and locally of finite presentation if and only if $V \to Y$ and $U' \to X'$ are flat and locally of finite presentation,

  4. $V' \to Y'$ is smooth if and only if $V \to Y$ and $U' \to X'$ are smooth,

  5. $V' \to Y'$ is étale if and only if $V \to Y$ and $U' \to X'$ are étale, and

  6. add more here as needed.

If $W'$ is flat over $Y'$, then the adjunction mapping $G(F(W')) \to W'$ is an isomorphism. Hence $F$ and $G$ define mutually quasi-inverse functors between the category of spaces flat over $Y'$ and the category of triples $(V, U', \varphi )$ with $V \to Y$ and $U' \to X'$ flat.

Proof. Choose a diagram (80.6.5.1) as in the proof of Lemma 80.6.5.

Proof of (1) – (5). Let $(V, U', \varphi )$ be an object of $(\textit{Spaces}/Y) \times _{(\textit{Spaces}/Y')} (\textit{Spaces}/X')$. Construct a diagram (80.6.5.2) as in the proof of Lemma 80.6.5. Then the base change of $G(V, U', \varphi ) \to Y'$ to $Y'_1$ is $G_1(V_1, U_1', \varphi _1) \to Y_1'$. Hence (1) – (5) follow immediately from the corresponding statements of More on Morphisms, Lemma 37.14.6 for schemes.

Suppose that $W' \to Y'$ is flat. Choose a scheme $W'_1$ and a surjective étale morphism $W'_1 \to Y_1' \times _{Y'} W'$. Observe that $W'_1 \to W'$ is surjective étale as a composition of surjective étale morphisms. We know that $G_1(F_1(W_1')) \to W_1'$ is an isomorphism by More on Morphisms, Lemma 37.14.6 applied to $W'_1$ over $Y'_1$ and the front of the diagram (with functors $G_1$ and $F_1$ as in the proof of Lemma 80.6.5). Then the construction of $G(F(W'))$ (as a pushout, i.e., as constructed in Lemma 80.6.2) shows that $G_1(F_1(W'_1)) \to G(F(W))$ is surjective étale. Whereupon we conclude that $G(F(W)) \to W$ is étale, see for example Properties of Spaces, Lemma 65.16.3. But $G(F(W)) \to W$ is an isomorphism on underlying reduced algebraic spaces (by construction), hence it is an isomorphism. $\square$


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