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This question is related to a previous question of mine, which has so far gone unanswered.

For a fixed site $\mathcal C$, the fibred categories over $\mathcal C$ form a (strict) $2$-category, see here. The category $\operatorname{Stack}_{\mathcal C}$ of stacks in groupoids over $\mathcal C$ is a subcategory. By a result of Street, $\operatorname{Stack}_{\mathcal C}$ has all lax colimits.

Let $F: A \to \operatorname{Stack}_{\mathcal C}$ be a lax functor from a small 1-category (considered as a 2-category with trivial 2-cells).

We can take the lax colimit of $F$. If instead of $\operatorname{Stack}_{\mathcal C}$ the codomain of $F$ was the $2$-category of $1$-categories, this lax colimit would be the usual Grothendieck construction $\int^A F$. It has a concrete characterisation as a category in terms of objects and arrows. For example, the objects of $\int^A F$ are pairs $(a, x)$ where $a$ is an object of $A$ and $x$ an object of $F(a)$, see here.

Can we give a similar description of the lax colimit of $F: A \to \operatorname{Stack}_{\mathcal C}$?

$F(a)$ will not be a category but a family of categories, so it is already unclear to me what "object of $F(a)$" should be replaced with.

Any stack over $\mathcal C$ can already be viewed as a Grothendieck construction, so I'm really asking about an iterated Grothendieck construction. And Grothendieck constructions satisfy a Fubini theorem,, so the object I'm interested in might be viewed as a category fibred over $A \times \mathcal C$. To define a stack in this setting, one would need to induce a Grothendieck topology on $A \times \mathcal C$.

I am most interested in the case where $A$ has only one object, or furthermore is a one-object groupoid. Then $A \times \mathcal C$ has objects the same as $\mathcal C$, but I'm not sure what the associated Grothendieck topology should be. Anyway, I would like to view the Grothendieck construction as a stack over $\mathcal C$.

Any hints are welcome, thank you!

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If your codomain is a (2, 1)-category then lax colimits are the same as pseudocolimits, which are a strict kind of homotopy colimit. For the very special case of diagrams over a one-object groupoid, this is basically a generalised semidirect product.

I find it conceptually easier to work with presheaves than fibrations. In this picture, what you knew about colimits of sheaves of sets transfers easily: the (lax, pseudo, etc.) colimit of a diagram of stacks is the stackification of the colimit of the (lax, pseudo, etc.) diagram of underlying presheaves. As such, I do not think it is feasible to give an understandable general description – you would first need to have an explicit description of stackification.

Even in the special case of the diagram over a one-object groupoid $A$ with constant value $1$ (!!!), the problem is at least as hard as classifying principal $G$-bundles on $\mathcal{C}$, where $G$ is the automorphism group of the unique object in the groupoid. Indeed, the presheaf lax colimit of such a diagram is just the constant presheaf on $\mathcal{C}$ with value $A$, but the stackification of this is precisely the stack of principal $G$-bundles on $\mathcal{C}$. In particular, the set of connected components (i.e. isomorphism classes) of the global sections of the stack is $H^1 (\mathcal{C}, G)$.

Incidentally, the above also shows that it is inappropriate to think of the colimit as a stack on $A \times \mathcal{C}$. What is instead true is that you get a stack on the Grothendieck construction of the stack of principal $G$-bundles on $\mathcal{C}$.

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  • $\begingroup$ Regarding your last point, are you saying that the stackification of the Grothendieck construction on $A \times \mathcal C$ does not agree with the stackification of the Grothendieck construction on $\mathcal C$ of the stackification of the Grothendieck construction on $A$? $\endgroup$ Commented Nov 12 at 23:15
  • $\begingroup$ Yes. Take the case of the constant diagram with value $1$, as mentioned. $\endgroup$
    – Zhen Lin
    Commented Nov 12 at 23:29
  • $\begingroup$ I don't see why the stackification of Grothendieck construction on $A \times \mathcal C$ wouldn't agree with the double stackification you outline. Could it not be as complicated? $\endgroup$ Commented Nov 12 at 23:39
  • $\begingroup$ The constant presheaf with value $1$ is already a stack. $\endgroup$
    – Zhen Lin
    Commented Nov 12 at 23:41

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