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

1. $\mathcal{R} \Rightarrow \mathcal{Q}$,

2. $\mathcal{Q}$ is preserved under base change and composition,

3. 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

4. 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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