The Stacks project

75.2 Pushouts in the category of algebraic spaces

This section is analogue of More on Morphisms, Section 36.14. We first prove a general result on colimits and algebraic spaces. To do this we discuss a bit of notation. Let $S$ be a scheme. Let $\mathcal{I} \to (\mathit{Sch}/S)_{fppf}$, $i \mapsto X_ i$ be a diagram (see Categories, Section 4.14). For each $i$ we may consider the small étale site $X_{i, {\acute{e}tale}}$. For each morphism $i \to j$ of $\mathcal{I}$ we have the morphism $X_ i \to X_ j$ and hence a pullback functor $X_{j, {\acute{e}tale}} \to X_{i, {\acute{e}tale}}$. Hence we obtain a pseudo functor from $\mathcal{I}^{opp}$ into the $2$-category of categories. Denote

\[ \mathop{\mathrm{lim}}\nolimits _ i X_{i, {\acute{e}tale}} \]

the $2$-limit (see insert future reference here). What does this mean concretely? An object of this limit is a system of étale morphisms $U_ i \to X_ i$ over $\mathcal{I}$ such that for each $i \to j$ in $\mathcal{I}$ the diagram

\[ \xymatrix{ U_ i \ar[r] \ar[d] & U_ j \ar[d] \\ X_ i \ar[r] & X_ j } \]

is cartesian. Morphisms between objects are defined in the obvious manner. Suppose that $f_ i : X_ i \to T$ is a family of morphisms such that for each $i \to j$ the composition $X_ i \to X_ j \to T$ is equal to $f_ i$. Then we get a functor $T_{\acute{e}tale}\to \mathop{\mathrm{lim}}\nolimits X_{i, {\acute{e}tale}}$. With this notation in hand we can formulate our lemma.

Lemma 75.2.1. Let $S$ be a scheme. Let $\mathcal{I} \to (\mathit{Sch}/S)_{fppf}$, $i \mapsto X_ i$ be a diagram as above. Assume that

  1. $X = \mathop{\mathrm{colim}}\nolimits X_ i$ exists in the category of schemes,

  2. $\coprod X_ i \to X$ is surjective,

  3. if $U \to X$ is étale and $U_ i = X_ i \times _ X U$, then $U = \mathop{\mathrm{colim}}\nolimits U_ i$ in the category of schemes, and

  4. every object $(U_ i \to X_ i)$ of $\mathop{\mathrm{lim}}\nolimits X_{i, {\acute{e}tale}}$ with $U_ i \to X_ i$ separated is in the essential image the functor $X_{\acute{e}tale}\to \mathop{\mathrm{lim}}\nolimits X_{i, {\acute{e}tale}}$.

Then $X = \mathop{\mathrm{colim}}\nolimits X_ i$ in the category of algebraic spaces over $S$ also.

Proof. Let $Z$ be an algebraic space over $S$. Suppose that $f_ i : X_ i \to Z$ is a family of morphisms such that for each $i \to j$ the composition $X_ i \to X_ j \to Z$ is equal to $f_ i$. We have to construct a morphism of algebraic spaces $f : X \to Z$ such that we can recover $f_ i$ as the composition $X_ i \to X \to Z$. Let $W \to Z$ be a surjective étale morphism of a scheme to $Z$. We may assume that $W$ is a disjoint union of affines and in particular we may assume that $W \to Z$ is separated. For each $i$ set $U_ i = W \times _{Z, f_ i} X_ i$ and denote $h_ i : U_ i \to W$ the projection. Then $U_ i \to X_ i$ forms an object of $\mathop{\mathrm{lim}}\nolimits X_{i, {\acute{e}tale}}$ with $U_ i \to X_ i$ separated. By assumption (4) we can find an étale morphism $U \to X$ and (functorial) isomorphisms $U_ i = X_ i \times _ X U$. By assumption (3) there exists a morphism $h : U \to W$ such that the compositions $U_ i \to U \to W$ are $h_ i$. Let $g : U \to Z$ be the composition of $h$ with the map $W \to Z$. To finish the proof we have to show that $g : U \to Z$ descends to a morphism $X \to Z$. To do this, consider the morphism $(h, h) : U \times _ X U \to W \times _ S W$. Composing with $U_ i \times _{X_ i} U_ i \to U \times _ X U$ we obtain $(h_ i, h_ i)$ which factors through $W \times _ Z W$. Since $U \times _ X U$ is the colimit of the schemes $U_ i \times _{X_ i} U_ i$ by (3) we see that $(h, h)$ factors through $W \times _ Z W$. Hence the two compositions $U \times _ X U \to U \to W \to Z$ are equal. Because each $U_ i \to X_ i$ is surjective and assumption (2) we see that $U \to X$ is surjective. As $Z$ is a sheaf for the étale topology, we conclude that $g : U \to Z$ descends to $f : X \to Z$ as desired. $\square$

Lemma 75.2.2. 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 36.14.3). Then $Y'$ is also a pushout in the category of algebraic spaces over $S$.

Lemma 75.2.3. In More on Morphisms, Situation 36.59.1 let $Y \amalg _ Z X$ be the pushout in the category of schemes (More on Morphisms, Proposition 36.59.3). Then $Y \amalg _ Z X$ is also a pushout in the category of algebraic spaces over $S$.

Proof. This is a consequence of Lemma 75.2.1, the proposition mentioned in the lemma and More on Morphisms, Lemmas 36.59.6 and 36.59.7. Conditions (1) and (2) of Lemma 75.2.1 follow immediately. To see (3) and (4) note that an étale morphism is locally quasi-finite and use that the equivalence of categories of More on Morphisms, Lemma 36.59.7 is constructed using the pushout construction of More on Morphisms, Lemmas 36.59.6. Minor details omitted. $\square$

Lemma 75.2.4. 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 70.9.6 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 36.14.3 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 36.14.6. 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 59.10.5.

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 70.9.6. 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 36.14.3 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 75.2.2 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 4.30.3.

Lemma 75.2.5. 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 75.2.4). 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 61.20.5. Hence we can apply Lemma 75.2.4 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 36.14.3. By Lemma 75.2.2 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

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

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 75.2.6. 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 75.2.7. In the situation of Lemma 75.2.5 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 36.14.4.

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 70.9.6 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 36.14.3 to obtain a pushout $Y_1' = Y_1 \amalg _{X_1} X_1'$ in the category of schemes. In the proof of Lemma 75.2.4 we constructed $Y'$ as a quotient of an étale equivalence relation on $Y_1'$ such that we get a commutative diagram

75.2.7.1
\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 36.14.4 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 75.2.2.

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 70.9.6 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 75.2.4 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

75.2.7.2
\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 36.14.4. Recall that $F$ and $F_1$ are given by base change. Using the properties of (75.2.7.2) and Lemma 75.2.6 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 75.2.8. 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 75.2.4). 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 60.30) 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 75.2.9. In the situation of Lemma 75.2.7. 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 (75.2.7.1) as in the proof of Lemma 75.2.7.

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 (75.2.7.2) as in the proof of Lemma 75.2.7. 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 36.14.6 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 36.14.6 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 75.2.7). Then the construction of $G(F(W'))$ (as a pushout, i.e., as constructed in Lemma 75.2.4) 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 60.16.3. 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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