17

A canonical example of a sheaf of sets on a topological space $X$
is the sheaf that sends an open subset $U$ of $X$ to the set of continuous real-valued functions on $U$.
The gluing property then says that a continuous function on a union of open subsets $U_i$ of $X$ is the same thing as a collection of continuous functions
$f_i: U_i \to \mathbb{R}$ such ...

7

What categories fibered in groupoids over $\mathcal{C}$ corresponds to stacks?
A category fibered in groupoids over $\mathcal{C}$ is given by a functor $p_{\mathcal{F}}:\mathcal{F}\rightarrow \mathcal{C}$ satisfying certain conditions (I am not writing the definition as I assume you already know what is a category fibered in groupoids); look at Definition 4....

5

I'm not entirely sure if this qualifies as an answer, but it is certainly too long for a comment. I hope that someone else will give a better answer.
If you want to define an algebraic stack as something that is locally isomorphic to $[Y/G]$, then you have to say what you mean by this isomorphism. So you need an a priori definition of the objects you're ...

answered Nov 16 '19 at 3:48

R. van Dobben de Bruyn

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5

If all objects of a stack have trivial automorphism groups then it is equivalent to a sheaf, as pointed out by Daniel Litt in the comments. Pick your favourite non-representable sheaf as a counterexample to the claim that the stack is representable. For instance, on the site of manifolds, defined to be Hausdorff, paracompact, locally Euclidean topological ...

3

Pullbacks of stacks coming from Lie groupoids are not always equivalent to Lie groupoids.
Take $G=H=\mathbb{R}$. Define $F(x)=0$ if $x\leq 0$ and $F(x)=exp(−1/x^2)$ if $x>0$.
The pullback is not equivalent to a Lie groupoid in this situation:
the set-theoretical pullback is $(−\infty,0]\times(−\infty,0]\cup \{(x,x)|x\in \mathbb{R}\}$,
which is clearly ...

3

There are no other such gerbes. If $M$ and $N$ are manifolds, and $p\colon \underline{M}\to \underline{N}$ is a gerbe, then the corresponding map of manifolds is a diffeomorphism. The same holds if one merely has representable stacks, rather than something of the form $\underline{M}$ etc.

3

The stack of curves of genus 0 with at most one node is a quotient stack (see The integral Chow ring of the stack of at most 1-nodal rational curves, but Edidin and Fulghesu), so you are fine in this case.
On the other hand, Andrew Kresch proved that the stack of nodal curves of genus 0 with at most two nodes is not a quotient stack (Flattening ...

2

For any algebraic stack $X$, there is a groupoid in schemes whose fppf stackification is equivalent to $X$. You can construct such a groupoid following the stacks project, by choosing a smooth presentation in algebraic spaces (Lemma 04T5), then choosing an étale presentation of the component algebraic spaces (Lemma 0262).
There are stacks associated to ...

2

Think of $BG$ and $BH$ as topological stacks, whereby one can calculate a topological groupoid presenting the stack $BG\times_{BH} BG$, namely the following: the object space is the space underlying $H$ and the morphism space is $G\times H \times G$. The source map $s\colon G\times H \times G \to H$ is the projection on the middle factor; the target map $t\...

answered Jan 20 '19 at 10:16

David Roberts

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2

Take $M = X$ and $f=p$, so that $\underline{P} = \underline{X} \times_\mathcal{D} \underline{X}$. The Lie groupoid $P\rightrightarrows X$ you get should be proper, in the sense the source-target map $(s,t)\colon P\to X\times X$ is a proper map, and $(s,t)$ should be injective, so that there are no nontrivial automorphism groups. Then your original stack is ...

1

If R→X is an epimorphism, then each fiber F_x is nonempty. Then if R→R ×_X R is an epimorphism, this means that π_0(F_x) is a point (otherwise π_0(F_x)→π_0(F_x) × π_0(F_x) cannot be an epimorphism), i.e., F_x is (noncanonically) equivalent to the stack BG for some smooth group G. But a gerbe over X is precisely a fiber bundle whose typical fiber is ...

answered Jul 29 '18 at 20:54

Dmitri Pavlov

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