...apart from stating properties of $(s,t):X_1 \to X_0\times X_0$ for the a presenting algebraic groupoid $X_1 \rightrightarrows X_0$?

Once we know that given a stack $\mathcal{X}$ we have a smooth representable $X_0 \to \mathcal{X}$ where $X_0$ is a scheme, then we can talk about the algebraic groupoid $X_1 :=X_0\times_\mathcal{X} X_0 \rightrightarrows X_0$, which has source and target smooth maps. We thus have the map $(s,t)$, and can talk about its properties, such as being separated or whatever. We can specify its properties (such as having 'property P') by demanding that the diagonal $\Delta:\mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is representable and has 'property P'. But where else is representability and property $P$ of $\Delta$ used, in a way that couldn't be otherwise derived from property $P$ of $(s,t)$? 'Everybody knows' that algebraic stacks and algebraic groupoids form the objects of two equivalent bicategories (there is a 1996 article by Dorette Pronk that makes this precise, and recent work by myself - available on the arXiv if you care - expands hers to be applied in more general situations). Thus I wonder what properties of the diagonal $\Delta$ are used that couldn't be instead derived from $(s,t)$ of a presenting groupoid. (Edit: or can all (stable under pullback) properties of the diagonal be so described - and also used?)

Pointers (in comments) to any relevant questions where example situations are discussed in detail would be appreciated.

(NB This is a spin-off from comments at this question.)

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    $\begingroup$ I've always thought that the condition that the diagonal is representable is useful because it is equivalent to requiring that any map from a scheme to your stack is representable. In particular, it implies that the map from an atlas of the stack to the stack is representable, which is certainly important. $\endgroup$ Commented Mar 4, 2011 at 2:14
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    $\begingroup$ @Mike - but we can require representability of the atlas map separately, which is essential, and if pressed, require that any map from a scheme to the stack is representable. If it is this latter property which is used, then that is the sort of answer I am looking for. And any other reasons too, of course. $\endgroup$
    – David Roberts
    Commented Mar 4, 2011 at 3:00
  • $\begingroup$ Since you can reconstruct the diagonal up to 2-isomorphism from a presenting groupoid, it seems tautological that there isn't any (isomorphism-invariant) property of the diagonal that can't be derived from properties of a presenting groupoid. Is that really the question you mean to ask? Also, I protest at your use of the word "the" in front of "presenting". $\endgroup$
    – S. Carnahan
    Commented Mar 4, 2011 at 5:41
  • $\begingroup$ @Scott - regarding 'the' - whoops! I didn't mean that. And yes, I think you've hit the nail on the head: I really want to know if one can avoid using representability of the diagonal explicitly in favour of properties arising from a presenting groupoid. $\endgroup$
    – David Roberts
    Commented Mar 4, 2011 at 5:54
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    $\begingroup$ One point (which doesn't answer your question, I know) is that representability of the diagonal follows from the existence of an atlas (i.e. a smooth surjective and representable morphism from an algebraic space $X\to \mathcal{X}$). See for example prop (4.3.2) of Champs Algébriques. $\endgroup$ Commented Mar 4, 2011 at 11:54

1 Answer 1


If $X_1 \rightrightarrows X_0$ is a groupoid and $\mathcal{X}$ is the associated stack, consider the atlas $p:X_0 \to \mathcal{X}$. If you form the weak $2$-pullback of $p \times p:X_0 \times X_0 \to \mathcal{X} \times \mathcal{X}$ against the diagonal $\Delta:\mathcal{X} \to \mathcal{X} \times \mathcal{X}$ you get by first projection $\left(s,t\right):X_1 \to X_0 \times X_0$. But $p \times p$ is an atlas for $\mathcal{X} \times \mathcal{X}$ so it follows that $\Delta$ is representable, and it has "property $P$" if and only if $\Delta$ does.

  • $\begingroup$ Of course, by "property P", we need a property which is invariant under restriction and local on the target. $\endgroup$ Commented Mar 4, 2011 at 14:40
  • $\begingroup$ Hi David - local on the target might be too strong an assumption for me... :) $\endgroup$
    – David Roberts
    Commented May 24, 2011 at 4:26
  • $\begingroup$ What kind of property $P$ do you have in mind exactly? Maybe then I can give a better answer. $\endgroup$ Commented May 24, 2011 at 11:36

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