We will prove that $F \circ G$ is isomorphic to the identity by reducing this to the corresponding statement of More on Morphisms, Lemma \ref{more-morphisms-lemma-equivalence-categories-schemes-over-pushout}. \medskip\noindent Choose a scheme $Y_1$ and a surjective \'etale morphism $Y_1 \to Y$. Set $X_1 = Y_1 \times_Y X$. This is a scheme affine over $Y_1$ with a surjective \'etale morphism $X_1 \to X$. By More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-thickening-equivalence} there exists a $X'_1 \to X'$ surjective \'etale 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 \ref{more-morphisms-lemma-pushout-along-thickening} to obtain a pushout $Y_1' = Y_1 \amalg_{X_1} X_1'$ in the category of schemes. In the proof of Lemma \ref{lemma-pushout-along-thickening} we constructed $Y'$ as a quotient of an \'etale equivalence relation on $Y_1'$ such that we get a commutative diagram \begin{equation} \label{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 \'etale. Denote $F_1$, $G_1$ the functors constructed in More on Morphisms, Lemma \ref{more-morphisms-lemma-equivalence-categories-schemes-over-pushout} for the front square. Then the diagram of categories $$ \xymatrix{ (\Sch/Y_1') \ar@<-1ex>[r]_-{F_1} \ar[d] & (\Sch/Y_1) \times_{(\Sch/Y_1')} (\Sch/X_1') \ar[d] \ar@<-1ex>[l]_-{G_1} \\ (\textit{Spaces}/Y') \ar@<-1ex>[r]_-F & (\textit{Spaces}/Y) \times_{(\textit{Spaces}/Y')} (\textit{Spaces}/X') \ar@<-1ex>[l]_-G } $$ is commutative by simple considerations regarding base change functors and the agreement of pushouts in schemes with pushouts in spaces of Lemma \ref{lemma-pushout-along-thickening-schemes}. \medskip\noindent 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 \'etale 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 \'etale too. By More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-thickening-equivalence} there exists a thickening $U_1 \to U_1'$ and a surjective \'etale 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 $(\Sch/Y_1) \times_{(\Sch/Y_1')} (\Sch/X_1')$. In the proof of Lemma \ref{lemma-pushout-along-thickening} we constructed $G(V, U', \varphi) = V \amalg_U U'$ as a quotient of an \'etale 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 \begin{equation} \label{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 \'etale. In particular $$ G_1(V_1, U_1', \varphi_1) \to G(V, U', \varphi) $$ is surjective \'etale. \medskip\noindent 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 \ref{more-morphisms-lemma-equivalence-categories-schemes-over-pushout}. Recall that $F$ and $F_1$ are given by base change. Using the properties of (\ref{equation-cube-over}) and Lemma \ref{lemma-diagram} 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.