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

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 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

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

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 (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. $\square$

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