Lemma 81.6.5. In the situation of Lemma 81.6.3 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 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
81.6.5.1
\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 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
\[ U_1 = V_1 \times _ Y X \longrightarrow (Y_1 \times _ Y V) \times _ Y X = X_1 \times _ Y V = X_1 \times _ X X \times _ Y V = X_1 \times _ X U \]
is surjective étale too. By More on Morphisms of Spaces, Lemma 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
81.6.5.2
\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 (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. $\square$
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