Lemma 40.14.2. Let $(U, R, s, t, c)$ be a groupoid scheme over a scheme $S$. Assume $s, t$ are finite. There exists an open subscheme $W \subset U$ and a closed subscheme $W' \subset W$ such that

1. $W$ and $W'$ are $R$-invariant,

2. $U = t(s^{-1}(\overline{W}))$ set theoretically,

3. $W$ is a thickening of $W'$, and

4. the maps $s'$, $t'$ of the restriction $(W', R', s', t', c')$ are finite locally free.

Proof. Consider the stratification $U = Z_0 \supset Z_1 \supset Z_2 \supset \ldots$ of Lemma 40.14.1.

We will construct disjoint unions $W = \coprod _{r \geq 1} W_ r$ and $W' = \coprod _{r \geq 1} W'_ r$ with each $W'_ r \to W_ r$ a thickening of $R$-invariant subschemes of $U$ such that the morphisms $s_ r', t_ r'$ of the restrictions $(W_ r', R_ r', s_ r', t_ r', c_ r')$ are finite locally free of rank $r$. To begin we set $W_1 = W'_1 = U \setminus Z_1$. This is an $R$-invariant open subscheme of $U$, it is true that $W_0$ is a thickening of $W'_0$, and the maps $s_1'$, $t_1'$ of the restriction $(W_1', R_1', s_1', t_1', c_1')$ are isomorphisms, i.e., finite locally free of rank $1$. Moreover, every point of $U \setminus Z_1$ is in $t(s^{-1}(\overline{W_1}))$.

Assume we have found subschemes $W'_ r \subset W_ r \subset U$ for $r \leq n$ such that

1. $W_1, \ldots , W_ n$ are disjoint,

2. $W_ r$ and $W_ r'$ are $R$-invariant,

3. $U \setminus Z_ n \subset \bigcup _{r \leq n} t(s^{-1}(\overline{W_ r}))$ set theoretically,

4. $W_ r$ is a thickening of $W'_ r$,

5. the maps $s_ r'$, $t_ r'$ of the restriction $(W_ r', R_ r', s_ r', t_ r', c_ r')$ are finite locally free of rank $r$.

Then we set

$W_{n + 1} = Z_ n \setminus \left( Z_{n + 1} \cup \bigcup \nolimits _{r \leq n} t(s^{-1}(\overline{W_ r})) \right)$

set theoretically and

$W'_{n + 1} = Z_ n \setminus \left( Z_{n + 1} \cup \bigcup \nolimits _{r \leq n} t(s^{-1}(\overline{W_ r})) \right)$

scheme theoretically. Then $W_{n + 1}$ is an $R$-invariant open subscheme of $U$ because $Z_{n + 1} \setminus \overline{U \setminus Z_{n + 1}}$ is open in $U$ and $\overline{U \setminus Z_{n + 1}}$ is contained in the closed subset $\bigcup \nolimits _{r \leq n} t(s^{-1}(\overline{W_ r}))$ we are removing by property (3) and the fact that $t$ is a closed morphism. It is clear that $W'_{n + 1}$ is a closed subscheme of $W_{n + 1}$ with the same underlying topological space. Finally, properties (1), (2) and (3) are clear and property (5) follows from Lemma 40.14.1.

By Lemma 40.14.1 we have $\bigcap Z_ r = \emptyset$. Hence every point of $U$ is contained in $U \setminus Z_ n$ for some $n$. Thus we see that $U = \bigcup _{r \geq 1} t(s^{-1}(\overline{W_ r}))$ set theoretically and we see that (2) holds. Thus $W' \subset W$ satisfy (1), (2), (3), and (4). $\square$

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