## 77.26 Inertia and quotient stacks

The (relative) inertia stack of a stack in groupoids is defined in Stacks, Section 8.7. The actual construction, in the setting of fibred categories, and some of its properties is in Categories, Section 4.34.

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. $\square$

Lemma 77.26.2. 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. There is a canonical $2$-cartesian diagram

$\xymatrix{ \mathcal{S}_ G \ar[r] \ar[d] & \mathcal{S}_ U \ar[d] \\ \mathcal{I}_{[U/R]} \ar[r] & [U/R] }$

of stacks in groupoids of $(\mathit{Sch}/S)_{fppf}$.

Proof. By Lemma 77.25.3 it suffices to prove that the morphism $s' : R' \to G$ of Lemma 77.26.1 isomorphic to the base change of $s$ by the structure morphism $G \to U$. This base change property is clear from the construction of $s'$. $\square$

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