## 91.16 From an algebraic stack to a presentation

Given an algebraic stack over $S$ we obtain a groupoid in algebraic spaces over $S$ whose associated quotient stack is the algebraic stack.

Recall that if $(U, R, s, t, c)$ is a groupoid in algebraic spaces over $S$ then $[U/R]$ denotes the quotient stack associated to this datum, see Groupoids in Spaces, Definition 75.19.1. In general $[U/R]$ is not an algebraic stack. In particular the stack $[U/R]$ occurring in the following lemma is in general not algebraic.

Lemma 91.16.1. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{X}$ be an algebraic stack over $S$. Let $\mathcal{U}$ be an algebraic stack over $S$ which is representable by an algebraic space. Let $f : \mathcal{U} \to \mathcal{X}$ be a 1-morphism. Then

1. the $2$-fibre product $\mathcal{R} = \mathcal{U} \times _{f, \mathcal{X}, f} \mathcal{U}$ is representable by an algebraic space,

2. there is a canonical equivalence

$\mathcal{U} \times _{f, \mathcal{X}, f} \mathcal{U} \times _{f, \mathcal{X}, f} \mathcal{U} = \mathcal{R} \times _{\text{pr}_1, \mathcal{U}, \text{pr}_0} \mathcal{R},$
3. the projection $\text{pr}_{02}$ induces via (2) a $1$-morphism

$\text{pr}_{02} : \mathcal{R} \times _{\text{pr}_1, \mathcal{U}, \text{pr}_0} \mathcal{R} \longrightarrow \mathcal{R}$
4. let $U$, $R$ be the algebraic spaces representing $\mathcal{U}, \mathcal{R}$ and $t, s : R \to U$ and $c : R \times _{s, U, t} R \to R$ are the morphisms corresponding to the $1$-morphisms $\text{pr}_0, \text{pr}_1 : \mathcal{R} \to \mathcal{U}$ and $\text{pr}_{02} : \mathcal{R} \times _{\text{pr}_1, \mathcal{U}, \text{pr}_0} \mathcal{R} \to \mathcal{R}$ above, then the quintuple $(U, R, s, t, c)$ is a groupoid in algebraic spaces over $S$,

5. the morphism $f$ induces a canonical $1$-morphism $f_{can} : [U/R] \to \mathcal{X}$ of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$, and

6. the $1$-morphism $f_{can} : [U/R] \to \mathcal{X}$ is fully faithful.

Proof. Proof of (1). By definition $\Delta _\mathcal {X}$ is representable by algebraic spaces so Lemma 91.10.11 applies to show that $\mathcal{U} \to \mathcal{X}$ is representable by algebraic spaces. Hence the result follows from Lemma 91.9.8.

Let $T$ be a scheme over $S$. By construction of the $2$-fibre product (see Categories, Lemma 4.31.3) we see that the objects of the fibre category $\mathcal{R}_ T$ are triples $(a, b, \alpha )$ where $a, b \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{U}_ T)$ and $\alpha : f(a) \to f(b)$ is a morphism in the fibre category $\mathcal{X}_ T$.

Proof of (2). The equivalence comes from repeatedly applying Categories, Lemmas 4.30.8 and 4.30.10. Let us identify $\mathcal{U} \times _\mathcal {X} \mathcal{U} \times _\mathcal {X} \mathcal{U}$ with $(\mathcal{U} \times _\mathcal {X} \mathcal{U}) \times _\mathcal {X} \mathcal{U}$. If $T$ is a scheme over $S$, then on fibre categories over $T$ this equivalence maps the object $((a, b, \alpha ), c, \beta )$ on the left hand side to the object $((a, b, \alpha ), (b, c, \beta ))$ of the right hand side.

Proof of (3). The $1$-morphism $\text{pr}_{02}$ is constructed in the proof of Categories, Lemma 4.30.9. In terms of the description of objects of the fibre category above we see that $((a, b, \alpha ), (b, c, \beta ))$ maps to $(a, c, \beta \circ \alpha )$.

Unfortunately, this is not compatible with our conventions on groupoids where we always have $j = (t, s) : R \to U$, and we “think” of a $T$-valued point $r$ of $R$ as a morphism $r : s(r) \to t(r)$. However, this does not affect the proof of (4), since the opposite of a groupoid is a groupoid. But in the proof of (5) it is responsible for the inverses in the displayed formula below.

Proof of (4). Recall that the sheaf $U$ is isomorphic to the sheaf $T \mapsto \mathop{\mathrm{Ob}}\nolimits (\mathcal{U}_ T)/\! \cong$, and similarly for $R$, see Lemma 91.8.2. It follows from Categories, Lemma 4.38.8 that this description is compatible with $2$-fibre products so we get a similar matching of $\mathcal{R} \times _{\text{pr}_1, \mathcal{U}, \text{pr}_0} \mathcal{R}$ and $R \times _{s, U, t} R$. The morphisms $t, s : R \to U$ and $c : R \times _{s, U, t} R \to R$ we get from the general equality (91.8.2.1). Explicitly these maps are the transformations of functors that come from letting $\text{pr}_0$, $\text{pr}_0$, $\text{pr}_{02}$ act on isomorphism classes of objects of fibre categories. Hence to show that we obtain a groupoid in algebraic spaces it suffices to show that for every scheme $T$ over $S$ the structure

$(\mathop{\mathrm{Ob}}\nolimits (\mathcal{U}_ T)/\! \cong , \mathop{\mathrm{Ob}}\nolimits (\mathcal{R}_ T)/\! \cong , \text{pr}_1, \text{pr}_0, \text{pr}_{02})$

is a groupoid which is clear from our description of objects of $\mathcal{R}_ T$ above.

Proof of (5). We will eventually apply Groupoids in Spaces, Lemma 75.22.2 to obtain the functor $[U/R] \to \mathcal{X}$. Consider the $1$-morphism $f : \mathcal{U} \to \mathcal{X}$. We have a $2$-arrow $\tau : f \circ \text{pr}_1 \to f \circ \text{pr}_0$ by definition of $\mathcal{R}$ as the $2$-fibre product. Namely, on an object $(a, b, \alpha )$ of $\mathcal{R}$ over $T$ it is the map $\alpha ^{-1} : b \to a$. We claim that

$\tau \circ \text{id}_{\text{pr}_{02}} = (\tau \star \text{id}_{\text{pr}_0}) \circ (\tau \star \text{id}_{\text{pr}_1}).$

This identity says that given an object $((a, b, \alpha ), (b, c, \beta ))$ of $\mathcal{R} \times _{\text{pr}_1, \mathcal{U}, \text{pr}_0} \mathcal{R}$ over $T$, then the composition of

$\xymatrix{ c \ar[r]^{\beta ^{-1}} & b \ar[r]^{\alpha ^{-1}} & a }$

is the same as the arrow $(\beta \circ \alpha )^{-1} : a \to c$. This is clearly true, hence the claim holds. In this way we see that all the assumption of Groupoids in Spaces, Lemma 75.22.2 are satisfied for the structure $(\mathcal{U}, \mathcal{R}, \text{pr}_0, \text{pr}_1, \text{pr}_{02})$ and the $1$-morphism $f$ and the $2$-morphism $\tau$. Except, to apply the lemma we need to prove this holds for the structure $(\mathcal{S}_ U, \mathcal{S}_ R, s, t, c)$ with suitable morphisms.

Now there should be some general abstract nonsense argument which transfer these data between the two, but it seems to be quite long. Instead, we use the following trick. Pick a quasi-inverse $j^{-1} : \mathcal{S}_ U \to \mathcal{U}$ of the canonical equivalence $j : \mathcal{U} \to \mathcal{S}_ U$ which comes from $U(T) = \mathop{\mathrm{Ob}}\nolimits (\mathcal{U}_ T)/\! \! \cong$. This just means that for every scheme $T/S$ and every object $a \in \mathcal{U}_ T$ we have picked out a particular element of its isomorphism class, namely $j^{-1}(j(a))$. Using $j^{-1}$ we may therefore see $\mathcal{S}_ U$ as a subcategory of $\mathcal{U}$. Having chosen this subcategory we can consider those objects $(a, b, \alpha )$ of $\mathcal{R}_ T$ such that $a, b$ are objects of $(\mathcal{S}_ U)_ T$, i.e., such that $j^{-1}(j(a)) = a$ and $j^{-1}(j(b)) = b$. Then it is clear that this forms a subcategory of $\mathcal{R}$ which maps isomorphically to $\mathcal{S}_ R$ via the canonical equivalence $\mathcal{R} \to \mathcal{S}_ R$. Moreover, this is clearly compatible with forming the $2$-fibre product $\mathcal{R} \times _{\text{pr}_1, \mathcal{U}, \text{pr}_0} \mathcal{R}$. Hence we see that we may simply restrict $f$ to $\mathcal{S}_ U$ and restrict $\tau$ to a transformation between functors $\mathcal{S}_ R \to \mathcal{X}$. Hence it is clear that the displayed equality of Groupoids in Spaces, Lemma 75.22.2 holds since it holds even as an equality of transformations of functors $\mathcal{R} \times _{\text{pr}_1, \mathcal{U}, \text{pr}_0} \mathcal{R} \to \mathcal{X}$ before restricting to the subcategory $\mathcal{S}_{R \times _{s, U, t} R}$.

This proves that Groupoids in Spaces, Lemma 75.22.2 applies and we get our desired morphism of stacks $f_{can} : [U/R] \to \mathcal{X}$. We briefly spell out how $f_{can}$ is defined in this special case. On an object $a$ of $\mathcal{S}_ U$ over $T$ we have $f_{can}(a) = f(a)$, where we think of $\mathcal{S}_ U \subset \mathcal{U}$ by the chosen embedding above. If $a, b$ are objects of $\mathcal{S}_ U$ over $T$, then a morphism $\varphi : a \to b$ in $[U/R]$ is by definition an object of the form $\varphi = (b, a, \alpha )$ of $\mathcal{R}$ over $T$. (Note switch.) And the rule in the proof of Groupoids in Spaces, Lemma 75.22.2 is that

91.16.1.1
\begin{equation} \label{algebraic-equation-on-morphisms} f_{can}(\varphi ) = \Big(f(a) \xrightarrow {\alpha ^{-1}} f(b)\Big). \end{equation}

Proof of (6). Both $[U/R]$ and $\mathcal{X}$ are stacks. Hence given a scheme $T/S$ and objects $a, b$ of $[U/R]$ over $T$ we obtain a transformation of fppf sheaves

$\mathit{Isom}(a, b) \longrightarrow \mathit{Isom}(f_{can}(a), f_{can}(b))$

on $(\mathit{Sch}/T)_{fppf}$. We have to show that this is an isomorphism. We may work fppf locally on $T$, hence we may assume that $a, b$ come from morphisms $a, b : T \to U$. By the embedding $\mathcal{S}_ U \subset \mathcal{U}$ above we may also think of $a, b$ as objects of $\mathcal{U}$ over $T$. In Groupoids in Spaces, Lemma 75.21.1 we have seen that the left hand sheaf is represented by the algebraic space

$R \times _{(t, s), U \times _ S U, (b, a)} T$

over $T$. On the other hand, the right hand side is by Stacks, Lemma 8.2.5 equal to the sheaf associated to the following stack in setoids:

$\mathcal{X} \times _{\mathcal{X} \times \mathcal{X}, (f \circ b, f \circ a)} T = \mathcal{X} \times _{\mathcal{X} \times \mathcal{X}, (f, f)} (\mathcal{U} \times \mathcal{U}) \times _{\mathcal{U} \times \mathcal{U}, (b, a)} T = \mathcal{R} \times _{(\text{pr}_0, \text{pr}_1), \mathcal{U} \times \mathcal{U}, (b, a)} T$

which is representable by the fibre product displayed above. At this point we have shown that the two $\mathit{Isom}$-sheaves are isomorphic. Our $1$-morphism $f_{can} : [U/R] \to \mathcal{X}$ induces this isomorphism on $\mathit{Isom}$-sheaves by Equation (91.16.1.1). $\square$

We can use the previous very abstract lemma to produce presentations.

Lemma 91.16.2. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{X}$ be an algebraic stack over $S$. Let $U$ be an algebraic space over $S$. Let $f : \mathcal{S}_ U \to \mathcal{X}$ be a surjective smooth morphism. Let $(U, R, s, t, c)$ be the groupoid in algebraic spaces and $f_{can} : [U/R] \to \mathcal{X}$ be the result of applying Lemma 91.16.1 to $U$ and $f$. Then

1. the morphisms $s$, $t$ are smooth, and

2. the $1$-morphism $f_{can} : [U/R] \to \mathcal{X}$ is an equivalence.

Proof. The morphisms $s, t$ are smooth by Lemmas 91.10.2 and 91.10.3. As the $1$-morphism $f$ is smooth and surjective it is clear that given any scheme $T$ and any object $a \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ T)$ there exists a smooth and surjective morphism $T' \to T$ such that $a|_ T'$ comes from an object of $[U/R]_{T'}$. Since $f_{can} : [U/R] \to \mathcal{X}$ is fully faithful, we deduce that $[U/R] \to \mathcal{X}$ is essentially surjective as descent data on objects are effective on both sides, see Stacks, Lemma 8.4.8. $\square$

Remark 91.16.3. If the morphism $f : \mathcal{S}_ U \to \mathcal{X}$ of Lemma 91.16.2 is only assumed surjective, flat and locally of finite presentation, then it will still be the case that $f_{can} : [U/R] \to \mathcal{X}$ is an equivalence. In this case the morphisms $s$, $t$ will be flat and locally of finite presentation, but of course not smooth in general.

Lemma 91.16.2 suggests the following definitions.

Definition 91.16.4. 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 $B$. We say $(U, R, s, t, c)$ is a smooth groupoid1 if $s, t : R \to U$ are smooth morphisms of algebraic spaces.

Definition 91.16.5. Let $\mathcal{X}$ be an algebraic stack over $S$. A presentation of $\mathcal{X}$ is given by a smooth groupoid $(U, R, s, t, c)$ in algebraic spaces over $S$, and an equivalence $f : [U/R] \to \mathcal{X}$.

We have seen above that every algebraic stack has a presentation. Our next task is to show that every smooth groupoid in algebraic spaces over $S$ gives rise to an algebraic stack.

 This terminology might be a bit confusing: it does not imply that $[U/R]$ is smooth over anything.

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