The Stacks project

Lemma 76.22.2. Assumptions and notation as in Lemmas 76.19.2 and 76.19.3. The $2$-commutative diagram of Lemma 76.19.3 is a $2$-coequalizer in the following sense: Given

  1. a stack in groupoids $\mathcal{X}$ over $(\mathit{Sch}/S)_{fppf}$,

  2. a $1$-morphism $f : \mathcal{S}_ U \to \mathcal{X}$, and

  3. 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. $\square$


Comments (2)

Comment #2535 by wpb on

in the second diagram in the proof should be


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