The Stacks project

Proof. This is an immediate consequence of Lemma 81.3.1 and More on Morphisms, Lemmas 37.14.3, 37.14.4, and 37.14.6. $\square$

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 76.9.6 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

in the category of schemes. Consider the morphisms

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 65.10.5.

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

Since $Y \to Y'$ is a thickening we have $Y_{\acute{e}tale}= (Y')_{\acute{e}tale}$, see More on Morphisms of Spaces, Lemma 76.9.6. The commutativity of the diagram gives a map of sheaves

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 81.6.1 we obtain a unique morphism $h : V' \to Z$ fitting into the commutative diagrams

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

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 67.20.5. Hence we can apply Lemma 81.6.2 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 81.6.1 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

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. $\square$

Proof. This is formal. $\square$

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 76.9.6 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 81.6.2 we constructed $Y'$ as a quotient of an étale equivalence relation on $Y_1'$ such that we get a commutative diagram

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

is commutative by simple considerations regarding base change functors and the agreement of pushouts in schemes with pushouts in spaces of Lemma 81.6.1.

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

is surjective étale too. By More on Morphisms of Spaces, Lemma 76.9.6 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 81.6.2 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

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

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 (81.6.5.2) and Lemma 81.6.4 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. $\square$

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

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 66.30) 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. $\square$

Proof. Choose a diagram (81.6.5.1) as in the proof of Lemma 81.6.5.

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 (81.6.5.2) as in the proof of Lemma 81.6.5. 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 81.6.5). Then the construction of $G(F(W'))$ (as a pushout, i.e., as constructed in Lemma 81.6.2) 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 66.16.3. But $G(F(W)) \to W$ is an isomorphism on underlying reduced algebraic spaces (by construction), hence it is an isomorphism. $\square$