Lemma 77.26.1. Assume $B \to S$ and $(U, R, s, t, c)$ as in Definition 77.20.1 (1). Let $G/U$ be the stabilizer group algebraic space of the groupoid $(U, R, s, t, c, e, i)$, see Definition 77.16.2. Set $R' = R \times _{s, U} G$ and set

1. $s' : R' \to G$, $(r, g) \mapsto g$,

2. $t' : R' \to G$, $(r, g) \mapsto c(r, c(g, i(r)))$,

3. $c' : R' \times _{s', G, t'} R' \to R'$, $((r_1, g_1), (r_2, g_2) \mapsto (c(r_1, r_2), g_1)$.

Then $(G, R', s', t', c')$ is a groupoid in algebraic spaces over $B$ and

$\mathcal{I}_{[U/R]} = [G/ R'].$

i.e., the associated quotient stack is the inertia stack of $[U/R]$.

Proof. By Stacks, Lemma 8.8.5 it suffices to prove that $\mathcal{I}_{[U/_{\! p}R]} = [G/_{\! p} R']$. Let $T$ be a scheme over $S$. Recall that an object of the inertia fibred category of $[U/_{\! p}R]$ over $T$ is given by a pair $(x, g)$ where $x$ is an object of $[U/_{\! \! p}R]$ over $T$ and $g$ is an automorphism of $x$ in its fibre category over $T$. In other words, $x : T \to U$ and $g : T \to R$ such that $x = s \circ g = t \circ g$. This means exactly that $g : T \to G$. A morphism in the inertia fibred category from $(x, g) \to (y, h)$ over $T$ is given by $r : T \to R$ such that $s(r) = x$, $t(r) = y$ and $c(r, g) = c(h, r)$, see the commutative diagram in Categories, Lemma 4.34.1. In a formula

$h = c(r, c(g, i(r))) = c(c(r, g), i(r)).$

The notation $s(r)$, etc is a short hand for $s \circ r$, etc. The composition of $r_1 : (x_2, g_2) \to (x_1, g_1)$ and $r_2 : (x_1, g_1) \to (x_2, g_2)$ is $c(r_1, r_2) : (x_1, g_1) \to (x_3, g_3)$.

Note that in the above we could have written $g$ in stead of $(x, g)$ for an object of $\mathcal{I}_{[U/_{\! p}R]}$ over $T$ as $x$ is the image of $g$ under the structure morphism $G \to U$. Then the morphisms $g \to h$ in $\mathcal{I}_{[U/_{\! p}R]}$ over $T$ correspond exactly to morphisms $r' : T \to R'$ with $s'(r') = g$ and $t'(r') = h$. Moreover, the composition corresponds to the rule explained in (3). Thus the lemma is proved. $\square$

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