Lemma 77.25.2. Notation and assumption as in Lemma 77.21.1. The morphism of quotient stacks

$[f] : [U/R] \longrightarrow [U'/R']$

is an equivalence if and only if

1. $(U, R, s, t, c)$ is the restriction of $(U', R', s', t', c')$ via $f : U \to U'$, and

2. the map

$\xymatrix{ U \times _{f, U', t'} R' \ar[r]_-{\text{pr}_1} \ar@/^3ex/[rr]^ h & R' \ar[r]_{s'} & U' }$

is a surjection of sheaves.

Part (2) holds for example if $\{ h : U \times _{f, U', t'} R' \to U'\}$ is an fppf covering, or if $f : U \to U'$ is a surjection of sheaves, or if $\{ f : U \to U'\}$ is an fppf covering.

Proof. We already know that part (1) is equivalent to fully faithfulness by Lemma 77.25.1. Hence we may assume that (1) holds and that $[f]$ is fully faithful. Our goal is to show, under these assumptions, that $[f]$ is an equivalence if and only if (2) holds. We may use Stacks, Lemma 8.4.8 which characterizes equivalences.

Assume (2). We will use Stacks, Lemma 8.4.8 to prove $[f]$ is an equivalence. Suppose that $T$ is a scheme and $x' \in \mathop{\mathrm{Ob}}\nolimits ([U'/R']_ T)$. There exists a covering $\{ g_ i : T_ i \to T\}$ such that $g_ i^*x'$ is the image of some element $a'_ i \in U'(T_ i)$, see Lemma 77.24.1. Hence we may assume that $x'$ is the image of $a' \in U'(T)$. By the assumption that $h$ is a surjection of sheaves, we can find an fppf covering $\{ \varphi _ i : T_ i \to T\}$ and morphisms $b_ i : T_ i \to U \times _{g, U', t'} R'$ such that $a' \circ \varphi _ i = h \circ b_ i$. Denote $a_ i = \text{pr}_0 \circ b_ i : T_ i \to U$. Then we see that $a_ i \in U(T_ i)$ maps to $f \circ a_ i \in U'(T_ i)$ and that $f \circ a_ i \cong _{T_ i} h \circ b_ i = a' \circ \varphi _ i$, where $\cong _{T_ i}$ denotes isomorphism in the fibre category $[U'/R']_{T_ i}$. Namely, the element of $R'(T_ i)$ giving the isomorphism is $\text{pr}_1 \circ b_ i$. This means that the restriction of $x$ to $T_ i$ is in the essential image of the functor $[U/R]_{T_ i} \to [U'/R']_{T_ i}$ as desired.

Assume $[f]$ is an equivalence. Let $\xi ' \in [U'/R']_{U'}$ denote the object corresponding to the identity morphism of $U'$. Applying Stacks, Lemma 8.4.8 we see there exists an fppf covering $\mathcal{U}' = \{ g'_ i : U'_ i \to U'\}$ such that $(g'_ i)^*\xi ' \cong [f](\xi _ i)$ for some $\xi _ i$ in $[U/R]_{U'_ i}$. After refining the covering $\mathcal{U}'$ (using Lemma 77.24.1) we may assume $\xi _ i$ comes from a morphism $a_ i : U'_ i \to U$. The fact that $[f](\xi _ i) \cong (g'_ i)^*\xi '$ means that, after possibly refining the covering $\mathcal{U}'$ once more, there exist morphisms $r'_ i : U'_ i \to R'$ with $t' \circ r'_ i = f \circ a_ i$ and $s' \circ r'_ i = \text{id}_{U'} \circ g'_ i$. Picture

$\xymatrix{ U \ar[d]^ f & & U'_ i \ar[ll]^{a_ i} \ar[ld]_{r'_ i} \ar[d]^{g'_ i} \\ U' & R' \ar[l]_{t'} \ar[r]^{s'} & U' }$

Thus $(a_ i, r'_ i) : U'_ i \to U \times _{g, U', t'} R'$ are morphisms such that $h \circ (a_ i, r'_ i) = g'_ i$ and we conclude that $\{ h : U \times _{g, U', t'} R' \to U'\}$ can be refined by the fppf covering $\mathcal{U}'$ which means that $h$ induces a surjection of sheaves, see Topologies on Spaces, Lemma 72.7.5.

If $\{ h\}$ is an fppf covering, then it induces a surjection of sheaves, see Topologies on Spaces, Lemma 72.7.5. If $U' \to U$ is surjective, then also $h$ is surjective as $s$ has a section (namely the neutral element $e$ of the groupoid in algebraic spaces). $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).