78.23 The 2-coequalizer property of a quotient stack
On a groupoid we have the composition, which leads to a cocycle condition for the canonical $2$-morphism of the lemma above. To give the precise formulation we will use the notation introduced in Categories, Sections 4.28 and 4.29.
Lemma 78.23.1. Assumptions and notation as in Lemmas 78.20.2 and 78.20.3. The vertical composition of
\[ \xymatrix@C=15pc{ \mathcal{S}_{R \times _{s, U, t} R} \ruppertwocell ^{\pi \circ s \circ \text{pr}_1 = \pi \circ s \circ c}{\ \ \ \ \ \ \alpha \star \text{id}_{\text{pr}_1}} \ar[r]_(.3){\pi \circ t \circ \text{pr}_1 = \pi \circ s \circ \text{pr}_0} \rlowertwocell _{\pi \circ t \circ \text{pr}_0 = \pi \circ t \circ c}{\ \ \ \ \ \ \alpha \star \text{id}_{\text{pr}_0}} & [U/R] } \]
is the $2$-morphism $\alpha \star \text{id}_ c$. In a formula $\alpha \star \text{id}_ c = (\alpha \star \text{id}_{\text{pr}_0}) \circ (\alpha \star \text{id}_{\text{pr}_1}) $.
Proof.
We make two remarks:
The formula $\alpha \star \text{id}_ c = (\alpha \star \text{id}_{\text{pr}_0}) \circ (\alpha \star \text{id}_{\text{pr}_1})$ only makes sense if you realize the equalities $\pi \circ s \circ \text{pr}_1 = \pi \circ s \circ c$, $\pi \circ t \circ \text{pr}_1 = \pi \circ s \circ \text{pr}_0$, and $\pi \circ t \circ \text{pr}_0 = \pi \circ t \circ c$. Namely, the second one implies the vertical composition $\circ $ makes sense, and the other two guarantee the two sides of the formula are $2$-morphisms with the same source and target.
The reason the lemma holds is that composition in the category fibred in groupoids $[U/_{\! p}R]$ associated to the presheaf in groupoids (78.20.0.1) comes from the composition law $c : R \times _{s, U, t} R \to R$.
We omit the proof of the lemma.
$\square$
Note that, in the situation of the lemma, we actually have the equalities $s \circ \text{pr}_1 = s \circ c$, $t \circ \text{pr}_1 = s \circ \text{pr}_0$, and $t \circ \text{pr}_0 = t \circ c$ before composing with $\pi $. Hence the formula in the lemma below makes sense in exactly the same way that the formula in the lemma above makes sense.
Lemma 78.23.2. Assumptions and notation as in Lemmas 78.20.2 and 78.20.3. The $2$-commutative diagram of Lemma 78.20.3 is a $2$-coequalizer in the following sense: Given
a stack in groupoids $\mathcal{X}$ over $(\mathit{Sch}/S)_{fppf}$,
a $1$-morphism $f : \mathcal{S}_ U \to \mathcal{X}$, and
a $2$-arrow $\beta : f \circ s \to f \circ t$
such that
\[ \beta \star \text{id}_ c = (\beta \star \text{id}_{\text{pr}_0}) \circ (\beta \star \text{id}_{\text{pr}_1}) \]
then there exists a $1$-morphism $[U/R] \to \mathcal{X}$ which makes the diagram
\[ \xymatrix{ \mathcal{S}_ R \ar[r]_ s \ar[d]^ t & \mathcal{S}_ U \ar[d] \ar[ddr]^ f \\ \mathcal{S}_ U \ar[r] \ar[rrd]_ f & [U/R] \ar[rd] \\ & & \mathcal{X} } \]
$2$-commute.
Proof.
Suppose given $\mathcal{X}$, $f$ and $\beta $ as in the lemma. By Stacks, Lemma 8.9.2 it suffices to construct a $1$-morphism $g : [U/_{\! p}R] \to \mathcal{X}$. First we note that the $1$-morphism $\mathcal{S}_ U \to [U/_{\! p}R]$ is bijective on objects. Hence on objects we can set $g(x) = f(x)$ for $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U) = \mathop{\mathrm{Ob}}\nolimits ([U/_{\! p}R])$. A morphism $\varphi : x \to y$ of $[U/_{\! p}R]$ arises from a commutative diagram
\[ \xymatrix{ S_2 \ar[dd]_ h \ar[r]_ x \ar[dr]_\varphi & U \\ & R \ar[u]_ s \ar[d]^ t \\ S_1 \ar[r]^ y & U. } \]
Thus we can set $g(\varphi )$ equal to the composition
\[ \xymatrix{ f(x) \ar@{=}[r] \ar[rrrrrd] & f(s \circ \varphi ) \ar@{=}[r] & (f \circ s)(\varphi ) \ar[r]^\beta & (f \circ t)(\varphi ) \ar@{=}[r] & f(t \circ \varphi ) \ar@{=}[r] & f(y \circ h) \ar[d] \\ & & & & & f(y). } \]
The vertical arrow is the result of applying the functor $f$ to the canonical morphism $y \circ h \to y$ in $\mathcal{S}_ U$ (namely, the strongly cartesian morphism lifting $h$ with target $y$). Let us verify that $f$ so defined is compatible with composition, at least on fibre categories. So let $S'$ be a scheme over $S$, and let $a : S' \to R \times _{s, U, t} R$ be a morphism. In this situation we set $x = s \circ \text{pr}_1 \circ a = s \circ c \circ a$, $y = t \circ \text{pr}_1 \circ a = s \circ \text{pr}_0 \circ a$, and $z = t \circ \text{pr}_0 \circ a = t \circ \text{pr}_0 \circ c$ to get a commutative diagram
\[ \xymatrix{ x \ar[rr]_{c \circ a} \ar[rd]_{\text{pr}_1 \circ a} & & z \\ & y \ar[ru]_{\text{pr}_0 \circ a} } \]
in the fibre category $[U/_{\! p}R]_{S'}$. Moreover, any commutative triangle in this fibre category has this form. Then we see by our definitions above that $f$ maps this to a commutative diagram if and only if the diagram
\[ \xymatrix{ & (f \circ s)(c \circ a) \ar[r]_-{\beta } & (f \circ t)(c \circ a) \ar@{=}[rd] & \\ (f \circ s)(\text{pr}_1 \circ a) \ar[rd]^\beta \ar@{=}[ru] & & & (f \circ t)(\text{pr}_0 \circ a) \\ & (f \circ t)(\text{pr}_1 \circ a) \ar@{=}[r] & (f \circ s)(\text{pr}_0 \circ a) \ar[ru]^\beta } \]
is commutative which is exactly the condition expressed by the formula in the lemma. We omit the verification that $f$ maps identities to identities and is compatible with composition for arbitrary morphisms.
$\square$
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