The Stacks project

77.25 Restriction and quotient stacks

In this section we study what happens to the quotient stack when taking a restriction.

Lemma 77.25.1. Notation and assumption as in Lemma 77.21.1. The morphism of quotient stacks

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

is fully faithful if and only if $R$ is the restriction of $R'$ via the morphism $f : U \to U'$.

Proof. Let $x, y$ be objects of $[U/R]$ over a scheme $T/S$. Let $x', y'$ be the images of $x, y$ in the category $[U'/R']_ T$. The functor $[f]$ is fully faithful if and only if the map of sheaves

\[ \mathit{Isom}(x, y) \longrightarrow \mathit{Isom}(x', y') \]

is an isomorphism for every $T, x, y$. We may test this locally on $T$ (in the fppf topology). Hence, by Lemma 77.24.1 we may assume that $x, y$ come from $a, b \in U(T)$. In that case we see that $x', y'$ correspond to $f \circ a, f \circ b$. By Lemma 77.22.1 the displayed map of sheaves in this case becomes

\[ T \times _{(a, b), U \times _ B U} R \longrightarrow T \times _{f \circ a, f \circ b, U' \times _ B U'} R'. \]

This is an isomorphism if $R$ is the restriction, because in that case $R = (U \times _ B U) \times _{U' \times _ B U'} R'$, see Lemma 77.17.3 and its proof. Conversely, if the last displayed map is an isomorphism for all $T, a, b$, then it follows that $R = (U \times _ B U) \times _{U' \times _ B U'} R'$, i.e., $R$ is the restriction of $R'$. $\square$

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). $\square$

Lemma 77.25.3. Notation and assumption as in Lemma 77.21.1. Assume that

\[ \xymatrix{ R \ar[d]_ s \ar[r]_ f & R' \ar[d]^{s'} \\ U \ar[r]^ f & U' } \]

is cartesian. Then

\[ \xymatrix{ \mathcal{S}_ U \ar[d] \ar[r] & [U/R] \ar[d]^{[f]} \\ \mathcal{S}_{U'} \ar[r] & [U'/R'] } \]

is a $2$-fibre product square.

Proof. Applying the inverse isomorphisms $i : R \to R$ and $i' : R' \to R'$ to the (first) cartesian diagram of the statement of the lemma we see that

\[ \xymatrix{ R \ar[d]_ t \ar[r]_ f & R' \ar[d]^{t'} \\ U \ar[r]^ f & U' } \]

is cartesian as well. By Lemma 77.21.2 we have a $2$-fibre square

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

where $U'' = U \times _{f, U', t'} R'$ and $R'' = R \times _{f \circ s, U', t'} R'$. By the above we see that $(t, f) : R \to U''$ is an isomorphism, and that

\[ R'' = R \times _{f \circ s, U', t'} R' = R \times _{s, U} U \times _{f, U', t'} R' = R \times _{s, U, t} \times R. \]

Explicitly the isomorphism $R \times _{s, U, t} R \to R''$ is given by the rule $(r_0, r_1) \mapsto (r_0, f(r_1))$. Moreover, $s'', t'', c''$ translate into the maps

\[ R \times _{s, U, t} R \to R, \quad s''(r_0, r_1) = r_1, \quad t''(r_0, r_1) = c(r_0, r_1) \]

and

\[ \begin{matrix} c'' : & (R \times _{s, U, t} R) \times _{s'', R, t''} (R \times _{s, U, t} R) & \longrightarrow & R \times _{s, U, t} R, \\ & ((r_0, r_1), (r_2, r_3)) & \longmapsto & (c(r_0, r_2), r_3). \end{matrix} \]

Precomposing with the isomorphism

\[ R \times _{s, U, s} R \longrightarrow R \times _{s, U, t} R, \quad (r_0, r_1) \longmapsto (c(r_0, i(r_1)), r_1) \]

we see that $t''$ and $s''$ turn into $\text{pr}_0$ and $\text{pr}_1$ and that $c''$ turns into $\text{pr}_{02} : R \times _{s, U, s} R \times _{s, U, s} R \to R \times _{s, U, s} R$. Hence we see that there is an isomorphism $[U''/R''] \cong [R/R \times _{s, U, s} R]$ where as a groupoid in algebraic spaces $(R, R \times _{s, U, s} R, s'', t'', c'')$ is the restriction of the trivial groupoid $(U, U, \text{id}, \text{id}, \text{id})$ via $s : R \to U$. Since $s : R \to U$ is a surjection of fppf sheaves (as it has a right inverse) the morphism

\[ [U''/R''] \cong [R/R \times _{s, U, s} R] \longrightarrow [U/U] = \mathcal{S}_ U \]

is an equivalence by Lemma 77.25.2. This proves the lemma. $\square$


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