The Stacks project

81.3 Invariant morphisms

Definition 81.3.1. Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j = (t, s) : R \to U \times _ B U$ be a pre-relation of algebraic spaces over $B$. We say a morphism $\phi : U \to X$ of algebraic spaces over $B$ is $R$-invariant if the diagram

\[ \xymatrix{ R \ar[r]_ s \ar[d]_ t & U \ar[d]^\phi \\ U \ar[r]^\phi & X } \]

is commutative. If $j : R \to U \times _ B U$ comes from the action of a group algebraic space $G$ on $U$ over $B$ as in Groupoids in Spaces, Lemma 76.14.1, then we say that $\phi $ is $G$-invariant.

In other words, a morphism $U \to X$ is $R$-invariant if it equalizes $s$ and $t$. We can reformulate this in terms of associated quotient sheaves as follows.

Lemma 81.3.2. Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j = (t, s) : R \to U \times _ B U$ be a pre-relation of algebraic spaces over $B$. A morphism of algebraic spaces $\phi : U \to X$ is $R$-invariant if and only if it factors as $U \to U/R \to X$.

Proof. This is clear from the definition of the quotient sheaf in Groupoids in Spaces, Section 76.18. $\square$

Lemma 81.3.3. Let $S$ be a scheme, and let $B$ be an algebraic space over $S$. Let $j = (t, s) : R \to U \times _ B U$ be a pre-relation of algebraic spaces over $B$. Let $U \to X$ be an $R$-invariant morphism of algebraic spaces over $B$. Let $X' \to X$ be any morphism of algebraic spaces.

  1. Setting $U' = X' \times _ X U$, $R' = X' \times _ X R$ we obtain a pre-relation $j' : R' \to U' \times _ B U'$.

  2. If $j$ is a relation, then $j'$ is a relation.

  3. If $j$ is a pre-equivalence relation, then $j'$ is a pre-equivalence relation.

  4. If $j$ is an equivalence relation, then $j'$ is an equivalence relation.

  5. If $j$ comes from a groupoid in algebraic spaces $(U, R, s, t, c)$ over $B$, then

    1. $(U, R, s, t, c)$ is a groupoid in algebraic spaces over $X$, and

    2. $j'$ comes from the base change $(U', R', s', t', c')$ of this groupoid to $X'$, see Groupoids in Spaces, Lemma 76.11.6.

  6. If $j$ comes from the action of a group algebraic space $G/B$ on $U$ as in Groupoids in Spaces, Lemma 76.14.1 then $j'$ comes from the induced action of $G$ on $U'$.

Proof. Omitted. Hint: Functorial point of view combined with the picture:

\[ \xymatrix{ R' = X' \times _ X R \ar[dd] \ar[rr] \ar[rd] & & X' \times _ X U = U' \ar '[d][dd] \ar[rd] \\ & R \ar[dd] \ar[rr] & & U \ar[dd] \\ U' = X' \times _ X U \ar '[r][rr] \ar[rd] & & X' \ar[rd] \\ & U \ar[rr] & & X } \]
$\square$

Definition 81.3.4. In the situation of Lemma 81.3.3 we call $j' : R' \to U' \times _ B U'$ the base change of the pre-relation $j$ to $X'$. We say it is a flat base change if $X' \to X$ is a flat morphism of algebraic spaces.

This kind of base change interacts well with taking quotient sheaves and quotient stacks.

Lemma 81.3.5. In the situation of Lemma 81.3.3 there is an isomorphism of sheaves

\[ U'/R' = X' \times _ X U/R \]

For the construction of quotient sheaves, see Groupoids in Spaces, Section 76.18.

Proof. Since $U \to X$ is $R$-invariant, it is clear that the map $U \to X$ factors through the quotient sheaf $U/R$. Recall that by definition

\[ \xymatrix{ R \ar@<1ex>[r] \ar@<-1ex>[r] & U \ar[r] & U/R } \]

is a coequalizer diagram in the category $\mathop{\mathit{Sh}}\nolimits $ of sheaves of sets on $(\mathit{Sch}/S)_{fppf}$. In fact, this is a coequalizer diagram in the comma category $\mathop{\mathit{Sh}}\nolimits /X$. Since the base change functor $X' \times _ X - : \mathop{\mathit{Sh}}\nolimits /X \to \mathop{\mathit{Sh}}\nolimits /X'$ is exact (true in any topos), we conclude. $\square$

Lemma 81.3.6. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $(U, R, s, t, c)$ be a groupoid in algebraic spaces over $B$. Let $U \to X$ be an $R$-invariant morphism of algebraic spaces over $B$. Let $g : X' \to X$ be a morphism of algebraic spaces over $B$ and let $(U', R', s', t', c')$ be the base change as in Lemma 81.3.3. Then

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

is a $2$-fibre product of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$. For the construction of quotient stacks and the morphisms in this diagram, see Groupoids in Spaces, Section 76.19.

Proof. We will prove this by using the explicit description of the quotient stacks given in Groupoids in Spaces, Lemma 76.23.1. However, we strongly urge the reader to find their own proof. First, we may view $(U, R, s, t, c)$ as a groupoid in algebraic spaces over $X$, hence we obtain a map $f : [U/R] \to \mathcal{S}_ X$, see Groupoids in Spaces, Lemma 76.19.2. Similarly, we have $f' : [U'/R'] \to X'$.

An object of the $2$-fibre product $\mathcal{S}_{X'} \times _{\mathcal{S}_ X} [U/R]$ over a scheme $T$ over $S$ is the same as a morphism $x' : T \to X'$ and an object $y$ of $[U/R]$ over $T$ such that such that the composition $g \circ x'$ is equal to $f(y)$. This makes sense because objects of $\mathcal{S}_ X$ over $T$ are morphisms $T \to X$. By Groupoids in Spaces, Lemma 76.23.1 we may assume $y$ is given by a $[U/R]$-descent datum $(u_ i, r_{ij})$ relative to an fppf covering $\{ T_ i \to T\} $. The agreement of $g \circ x' = f(y)$ means that the diagrams

\[ \vcenter { \xymatrix{ T_ i \ar[rr]_{u_ i} \ar[d] & & U \ar[d] \\ T \ar[r]^{x'} & X' \ar[r]^ g & X } } \quad \text{and}\quad \vcenter { \xymatrix{ T_ i \times _ T T_ j \ar[rr]_{r_{ij}} \ar[d] & & R \ar[d] \\ T \ar[r]^{x'} & X' \ar[r]^ g & X } } \]

are commutative.

On the other hand, an object $y'$ of $[U'/R']$ over a scheme $T$ over $S$ by Groupoids in Spaces, Lemma 76.23.1 is given by a $[U'/R']$-descent datum $(u'_ i, r'_{ij})$ relative to an fppf covering $\{ T_ i \to T\} $. Setting $f'(y') = x' : T \to X'$ we see that the diagrams

\[ \vcenter { \xymatrix{ T_ i \ar[r]_{u'_ i} \ar[d] & U' \ar[d] \\ T \ar[r]^{x'} & X' } } \quad \text{and}\quad \vcenter { \xymatrix{ T_ i \times _ T T_ j \ar[r]_{r'_{ij}} \ar[d] & U' \ar[d] \\ T \ar[r]^{x'} & X' } } \]

are commutative.

With this notation in place, we define a functor

\[ [U'/R'] \longrightarrow \mathcal{S}_{X'} \times _{\mathcal{S}_ X} [U/R] \]

by sending $y' = (u'_ i, r'_{ij})$ as above to the object $(x', (u_ i, r_{ij}))$ where $x' = f'(y')$, where $u_ i$ is the composition $T_ i \to U' \to U$, and where $r_{ij}$ is the composition $T_ i \times _ T T_ j \to R' \to R$. Conversely, given an object $(x', (u_ i, r_{ij})$ of the right hand side we can send this to the object $((x', u_ i), (x', r_{ij}))$ of the left hand side. We omit the discussion of what to do with morphisms (works in exactly the same manner). $\square$


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