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I once heard a rumour that various nice categories of stacks were co-complete. Gepner and Henriques, working from the groupoids point of view, give a construction [link] of 2-colimits of topological groupoids, but I haven't seen any definitive statement on whether there is a construction that works for, eg., differential or holomorphic stacks (let alone a proof). Personally I don't expect it, but I just want a definitive answer, so that I know that certain constructions I'm working on aren't a waste of time.

So what I'm asking is: are differential stacks co-complete? How about holomorphic stacks? If so, are there easy constructions? (Groupoid point of view preferred, but I'll take what I can get.) If the answer's no, are there any counterexamples? If none of these questions can be answered, can an expert come on here and say "it's not known"?

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Just because there are 2-colimits of topological groupoids does not mean that there are 2-colimits of topological stacks. The Yoneda embedding generally fails to preserve colimits! –  David Carchedi May 30 '13 at 11:26
    
P.S. Exactly what types of colimits are you wanting to exist/compute? –  David Carchedi May 30 '13 at 11:35
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1 Answer

up vote 3 down vote accepted

The answer is no.

Claim: If $\mathfrak{DiffSt}$ were cocomplete, it would be reflective in $St\left(Mfd\right).$

Proof:

Let $i:Mfd \hookrightarrow \mathfrak{DiffSt}$ be the full and faithful inclusion of manifolds into differentiable stacks. Take the $(2,1)$-categorical left Kan extension of $i$ along the Yoneda embedding into (weak) presheaves of groupoids $$Lan_y i:Fun(Mfd,Gpd) \to \mathfrak{DiffSt},$$ which would exist, since we could use the pointwise formula, if $\mathfrak{DiffSt}$ were cocomplete. It comes with a right adjoint $R,$ which by the Yoneda lemma, is identified with the canonical full and faithful inclusion $$R:\mathfrak{DiffSt} \hookrightarrow Fun(Mfd,Gpd).$$ Notice, the essential image of $R$ lands in stacks, $St\left(Mfd\right),$ so the adjunction restricts. The result now follows.

Ok: How does this help us?

If $\mathfrak{DiffSt}$ were reflective in $St\left(Mfd\right),$ it would have to also be complete. But now look at say, an infinite product of manifolds, or non-tranvsersal pullbacks. It is easy to cook up examples like this that are not represented by differentiable stacks, since the limits must agree with point-wise limits, since $R$ would be a right-adjoint. Hence, $\mathfrak{DiffSt}$ is not co-complete.

Remark: Incidentally, certain non-full subcategories of $\mathfrak{DiffSt}$ are cocomplete. E.g., if you restrict to etale differentiable stacks, and look at only local diffeomorphisms between them, this actually forms a $2$-topos: http://arxiv.org/abs/1212.2282 - and moreover, these colimits are preserved by the inclusion into $\mathfrak{DiffSt}$ (I will be updating this preprint in a few days)

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