Lemma 83.3.6. 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 83.3.3. 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 78.20.
Proof.
We will prove this by using the explicit description of the quotient stacks given in Groupoids in Spaces, Lemma 78.24.1. 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 78.20.2. 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 78.24.1 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 78.24.1 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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