Lemma 40.6.1. Let S be a scheme. Let (U, R, s, t, c, e, i) be a groupoid over S. Let g : U' \to U be a morphism of schemes. Denote h the composition
\xymatrix{ h : U' \times _{g, U, t} R \ar[r]_-{\text{pr}_1} & R \ar[r]_ s & U. }
Let \mathcal{P}, \mathcal{Q}, \mathcal{R} be properties of morphisms of schemes. Assume
\mathcal{R} \Rightarrow \mathcal{Q},
\mathcal{Q} is preserved under base change and composition,
for any morphism f : X \to Y which has \mathcal{Q} there exists a largest open W(\mathcal{P}, f) \subset X such that f|_{W(\mathcal{P}, f)} has \mathcal{P}, and
for any morphism f : X \to Y which has \mathcal{Q}, and any morphism Y' \to Y which has \mathcal{R} we have Y' \times _ Y W(\mathcal{P}, f) = W(\mathcal{P}, f'), where f' : X_{Y'} \to Y' is the base change of f.
If s, t have \mathcal{R} and g has \mathcal{Q}, then there exists an open subscheme W \subset U' such that W \times _{g, U, t} R = W(\mathcal{P}, h).
Proof.
Note that the following diagram is commutative
\xymatrix{ U' \times _{g, U, t} R \times _{t, U, t} R \ar[rr]_-{\text{pr}_{12}} \ar@<1ex>[d]^-{\text{pr}_{02}} \ar@<-1ex>[d]_-{\text{pr}_{01}} & & R \times _{t, U, t} R \ar@<1ex>[d]^-{\text{pr}_1} \ar@<-1ex>[d]_-{\text{pr}_0} \\ U' \times _{g, U, t} R \ar[rr]^{\text{pr}_1} & & R }
with both squares cartesian (this uses that the two maps t \circ \text{pr}_ i : R \times _{t, U, t} R \to U are equal). Combining this with the properties of diagram (40.3.0.2) we get a commutative diagram
\xymatrix{ U' \times _{g, U, t} R \times _{t, U, t} R \ar[rr]_-{c \circ (i, 1)} \ar@<1ex>[d]^-{\text{pr}_{02}} \ar@<-1ex>[d]_-{\text{pr}_{01}} & & R \ar@<1ex>[d]^-{s} \ar@<-1ex>[d]_-{t} \\ U' \times _{g, U, t} R \ar[rr]^ h & & U }
where both squares are cartesian.
Assume s, t have \mathcal{R} and g has \mathcal{Q}. Then h has \mathcal{Q} as a composition of s (which has \mathcal{R} hence \mathcal{Q}) and a base change of g (which has \mathcal{Q}). Thus W(\mathcal{P}, h) \subset U' \times _{g, U, t} R exists. By our assumptions we have \text{pr}_{01}^{-1}(W(\mathcal{P}, h)) = \text{pr}_{02}^{-1}(W(\mathcal{P}, h)) since both are the largest open on which c \circ (i, 1) has \mathcal{P}. Note that the projection U' \times _{g, U, t} R \to U' has a section, namely \sigma : U' \to U' \times _{g, U, t} R, u' \mapsto (u', e(g(u'))). Also via the isomorphism
(U' \times _{g, U, t} R) \times _{U'} (U' \times _{g, U, t} R) = U' \times _{g, U, t} R \times _{t, U, t} R
the two projections of the left hand side to U' \times _{g, U, t} R agree with the morphisms \text{pr}_{01} and \text{pr}_{02} on the right hand side. Since \text{pr}_{01}^{-1}(W(\mathcal{P}, h)) = \text{pr}_{02}^{-1}(W(\mathcal{P}, h)) we conclude that W(\mathcal{P}, h) is the inverse image of a subset of U, which is necessarily the open set W = \sigma ^{-1}(W(\mathcal{P}, h)).
\square
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