The Stacks project

Proof. We write this out in detail, to make sure that all the definitions work out in exactly the correct manner. Recall that $[X/G]$ is the quotient stack associated to the groupoid in algebraic spaces $(X, G \times _ B X, s, t, c)$, see Groupoids in Spaces, Definition 78.20.1. This means that $[X/G]$ is the stackification of the category fibred in groupoids $[X/_{\! p}G]$ associated to the functor

where $s(g, x) = x$, $t(g, x) = a(g, x)$, and $c((g, x), (g', x')) = (m(g, g'), x')$. By the construction of Categories, Example 4.37.1 an object of $[X/_{\! p}G]$ is a pair $(U, x)$ with $x \in X(U)$ and a morphism $(f, g) : (U, x) \to (U', x')$ of $[X/_{\! p}G]$ is given by a morphism of schemes $f : U \to U'$ and an element $g \in G(U)$ such that $a(g, x) = x' \circ f$. Hence we can define a $1$-morphism of stacks in groupoids

by the following rules: On objects we set

This makes sense because the diagram

commutes, and the two horizontal arrows are $G$-equivariant if we think of the fibre products as trivial $G$-torsors over $U$, resp. $X$. On morphisms $(f, g) : (U, x) \to (U', x')$ we set $F_ p(f, g) = (f, R_{g^{-1}})$ where $R_{g^{-1}}$ denotes right translation by the inverse of $g$. More precisely, the morphism $F_ p(f, g) : F_ p(U, x) \to F_ p(U', x')$ is given by the cartesian diagram

where $R_{g^{-1}}$ on $T$-valued points is given by

To see that this works we have to verify that

which is true because the right hand side applied to the $T$-valued point $(g', u)$ gives the desired equality

because $a(g, x) = x' \circ f$ and hence $a(i(g), x' \circ f) = x$.

By the universal property of stackification from Stacks, Lemma 8.9.2 we obtain a canonical extension $F : [X/G] \to [[X/G]]$ of the $1$-morphism $F_ p$ above. We first prove that $F$ is fully faithful. To do this, since both source and target are stacks in groupoids, it suffices to prove that the $\mathit{Isom}$-sheaves are identified under $F$. Pick a scheme $U$ and objects $\xi , \xi '$ of $[X/G]$ over $U$. We want to show that

is an isomorphism of sheaves. To do this it suffices to work locally on $U$, and hence we may assume that $\xi , \xi '$ come from objects $(U, x)$, $(U, x')$ of $[X/_{\! p}G]$ over $U$; this follows directly from the construction of the stackification, and it is also worked out in detail in Groupoids in Spaces, Section 78.24. Either by directly using the description of morphisms in $[X/_{\! p}G]$ above, or using Groupoids in Spaces, Lemma 78.22.1 we see that in this case

A $T$-valued point of this fibre product corresponds to a pair $(u, g)$ with $u \in U(T)$, and $g \in G(T)$ such that $a(g, x \circ u) = x' \circ u$. (Note that this implies $\pi \circ x \circ u = \pi \circ x' \circ u$.) On the other hand, a $T$-valued point of $\mathit{Isom}_{[[X/G]]}(F(\xi ), F(\xi '))$ by definition corresponds to a morphism $u : T \to U$ such that $\pi \circ x \circ u = \pi \circ x' \circ u : T \to B$ and an isomorphism

of trivial $G_ T$-torsors compatible with the given maps to $X$. Since the torsors are trivial we see that $R = R_{g^{-1}}$ (right multiplication) by some $g \in G(T)$. Compatibility with the maps $a \circ (1_ G, x \circ u), a \circ (1_ G, x' \circ u) : G \times _ B T \to X$ is equivalent to the condition that $a(g, x \circ u) = x' \circ u$. Hence we obtain the desired equality of $\mathit{Isom}$-sheaves.

Now that we know that $F$ is fully faithful we see that Stacks, Lemma 8.4.8 applies. Thus to show that $F$ is an equivalence it suffices to show that objects of $[[X/G]]$ are fppf locally in the essential image of $F$. This is clear as fppf torsors are fppf locally trivial, and hence we win. $\square$