Lemma 81.6.7. In the situation of Lemma 81.6.5. If $V' = G(V, U', \varphi )$ for some triple $(V, U', \varphi )$, then
$V' \to Y'$ is locally of finite type if and only if $V \to Y$ and $U' \to X'$ are locally of finite type,
$V' \to Y'$ is flat if and only if $V \to Y$ and $U' \to X'$ are flat,
$V' \to Y'$ 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,
$V' \to Y'$ is smooth if and only if $V \to Y$ and $U' \to X'$ are smooth,
$V' \to Y'$ is étale if and only if $V \to Y$ and $U' \to X'$ are étale, and
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 (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.
$\square$
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