Lemma 100.3.6. Let $g : \mathcal{X}' \to \mathcal{X}$ be a morphism of algebraic stacks which is representable by algebraic spaces. Let $[U/R] \to \mathcal{X}$ be a presentation. Set $U' = U \times _\mathcal {X} \mathcal{X}'$, and $R' = R \times _\mathcal {X} \mathcal{X}'$. Then there exists a groupoid in algebraic spaces of the form $(U', R', s', t', c')$, a presentation $[U'/R'] \to \mathcal{X}'$, and the diagram

$\xymatrix{ [U'/R'] \ar[d]_{[\text{pr}]} \ar[r] & \mathcal{X}' \ar[d]^ g \\ [U/R] \ar[r] & \mathcal{X} }$

is $2$-commutative where the morphism $[\text{pr}]$ comes from a morphism of groupoids $\text{pr} : (U', R', s', t', c') \to (U, R, s, t, c)$.

Proof. Since $U \to \mathcal{Y}$ is surjective and smooth, see Algebraic Stacks, Lemma 94.17.2 the base change $U' \to \mathcal{X}'$ is also surjective and smooth. Hence, by Algebraic Stacks, Lemma 94.16.2 it suffices to show that $R' = U' \times _{\mathcal{X}'} U'$ in order to get a smooth groupoid $(U', R', s', t', c')$ and a presentation $[U'/R'] \to \mathcal{X}'$. Using that $R = V \times _\mathcal {Y} V$ (see Groupoids in Spaces, Lemma 78.22.2) this follows from

$R' = U \times _\mathcal {X} U \times _\mathcal {X} \mathcal{X}' = (U \times _\mathcal {X} \mathcal{X}') \times _{\mathcal{X}'} (U \times _\mathcal {X} \mathcal{X}')$

see Categories, Lemmas 4.31.8 and 4.31.10. Clearly the projection morphisms $U' \to U$ and $R' \to R$ give the desired morphism of groupoids $\text{pr} : (U', R', s', t', c') \to (U, R, s, t, c)$. Hence the morphism $[\text{pr}]$ of quotient stacks by Groupoids in Spaces, Lemma 78.21.1.

We still have to show that the diagram $2$-commutes. It is clear that the diagram

$\xymatrix{ U' \ar[d]_{\text{pr}_ U} \ar[r]_{f'} & \mathcal{X}' \ar[d]^ g \\ U \ar[r]^ f & \mathcal{X} }$

$2$-commutes where $\text{pr}_ U : U' \to U$ is the projection. There is a canonical $2$-arrow $\tau : f \circ t \to f \circ s$ in $\mathop{\mathrm{Mor}}\nolimits (R, \mathcal{X})$ coming from $R = U \times _\mathcal {X} U$, $t = \text{pr}_0$, and $s = \text{pr}_1$. Using the isomorphism $R' \to U' \times _{\mathcal{X}'} U'$ we get similarly an isomorphism $\tau ' : f' \circ t' \to f' \circ s'$. Note that $g \circ f' \circ t' = f \circ t \circ \text{pr}_ R$ and $g \circ f' \circ s' = f \circ s \circ \text{pr}_ R$, where $\text{pr}_ R : R' \to R$ is the projection. Thus it makes sense to ask if

100.3.6.1
$$\label{stacks-properties-equation-verify} \tau \star \text{id}_{\text{pr}_ R} = \text{id}_ g \star \tau '.$$

Now we make two claims: (1) if Equation (100.3.6.1) holds, then the diagram $2$-commutes, and (2) Equation (100.3.6.1) holds. We omit the proof of both claims. Hints: part (1) follows from the construction of $f = f_{can}$ and $f' = f'_{can}$ in Algebraic Stacks, Lemma 94.16.1. Part (2) follows by carefuly working through the definitions. $\square$

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