A stack is usually given in terms of:

-A category $F$ fibered over another $C$ such that the functor $Hom(x,y), x,y \in F(\alpha), \alpha \in C$ is a sheaf

-The descent data are effective.

There is an equivalent definition, using the Grothendieck construction, which is a correspondence between fibered categories and pseudofunctors in $Cat$. Given this correspondence a stack becomes a (contravariant) pseudofunctor such that descent is effective.

Now, with this definition in mind it seems obvious that a stack is a generalised sheaf. But here is my doubt:

When defining a sheaf $T$ over a space $X$, is not unusual to see the following diagram (naively)

$ T(X) \rightarrow T(U) \stackrel{\longrightarrow}{\longrightarrow} T(U \cap U)$

While, saying that descent data are effective is like asking a similar diagram, but there is the difference that here descent satisfies cocycle condition, something which is naively in $U \cap U \cap U$.

Recalling that (for example) locally constant sheaves automatically satisfy the cocycle condition because of their correspondent covering spaces do.

So my question is, at the light of my interpretation, the cocycle condition seems to me the only obstruction to the fact that a stack is a generalised sheaf. Am i wrong? Every sheaf trivially satisfies cocycle condition?

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    $\begingroup$ This diagram is not naive. It's actually correct once you use homotopy colimits and add all the intersections (not only the double intersections). In other words, in a sheaf you require that sections coinciding in a intersection must glue to a global section, while in a stack, sections connected by an arrow in each double intersection (that is coherent with other arrows) must glue to a global section. More generally, you could relax these commutative triangles to be only commutative up to a 2-morphism with coherences between these 2-morphisms and so on (this would be an $\infty$-stack) $\endgroup$
    – user40276
    Jun 12, 2017 at 14:50
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    $\begingroup$ If I understand correctly the comment by @user40276, it is about maximally general formulation: sheaf/stack/... condition is that certain kind of colimits are carried to the same kind of limits. In every case, coproducts must go to products; the rest is about some kind of coequalizers go to the same kind of equalizers. For sheaves it is that if $U\times_XU\rightrightarrows U\to X$ is a coequalizer then $F(X)\to F(U)\rightrightarrows F(U\times_XU)$ must be an equalizer. Higher and higher, you get $n$-truncated simplicial objects, which, if exact, must go to exact $n$-truncated cosimplicials,... $\endgroup$ Jun 13, 2017 at 8:54
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    $\begingroup$ (where exact truncated simplicial, resp. cosimplicial object means that $X$, resp. $F(X)$, is the homotopy colimit, resp. homotopy limit, of this truncated simplicial, resp. cosimplicial object considered as a diagram) $\endgroup$ Jun 13, 2017 at 8:58
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    $\begingroup$ For the $n$-categorial case you can truncate this $\infty$-groupoid to an n-groupoid and you will get the same result up to homotopy. Furthermore, as it was already noticed in the comments, this is equivalent to the preservation of a homotopy limit. This process of requiring things to be isomorphic instead of equal is the usual procedure for vertical categorification. However when requiring things to be isomorphic, you must require these isomorphisms to be compatible and so on. $\endgroup$
    – user40276
    Jun 13, 2017 at 19:18
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    $\begingroup$ @HaroldF Ah! Ok. But there are no problems with Vistoli's presentation. Most of the cases that deals with moduli do not require 2-stacks (or 3-sheaves). Ah! And I forgot to write this. That cocycle condition guarantees that every possible way of restricting a local section gives you the same result. It's analogous to that theorem of Maclane which says that when the Maclane pentagon commutes then all the possible ways of associating an n-ary monoidal product gives you isomorphic objects. $\endgroup$
    – user40276
    Jun 14, 2017 at 21:06

1 Answer 1


Let us start with what we know about sheaves, i.e. the "1-level". A sheaf on a (Grothendieck) site $\mathcal{C}$ is a contravariant functor $F : \mathcal{C}^\text{op} \to \textbf{Set}$ such that for any cover $\{ X\ \to Y\}$ , the diagram $$F(Y) \to F(X) \stackrel{\longrightarrow}{\longrightarrow} F (X \times_Y X)$$ is an equalizer in the category of sets. For the sake of exposition, I will only consider the case where the cover consists of a single element.

Now suppose we want to move to the "2-level" and talk about stacks. Then we need to throw in the cocycle condition, so we can naively define a stack $F$ to be a contravariant functor $\mathcal{C}^{\text{op}} \to \textbf{Set}$ such that

$$F(Y) \to F(X) \stackrel{\longrightarrow}{\longrightarrow} F (X \times_Y X)\stackrel{\stackrel{\longrightarrow}{\longrightarrow}}{\longrightarrow} F(X \times_Y X \times_Y X)$$

is an equalizer in the category of sets. The problem now is the following:

If $F$ is some kind of moduli stack, e.g. $F = \mathcal{M}_{1,1}$ then to make $F$ set valued often involves quotiening out isomorphisms. However, this is very bad as the presence of quadratic twists of elliptic curves means $F$ is not injective.

So now what do we do? Well, we can try to not quotient out isomorphisms, and think of $F$ as a groupoid valued functor. But now we have a new problem:

If $F$ is valued in groupoids, what does it mean to say that $F(Y) \to F(X)$ is "injective"?

The solution is the following. Let's go back to situation where $F$ is a plain old sheaf, and let us think of the set $F(X)$ as a category where the only arrow $x \to x'$ is when $x = x'$, otherwise $\operatorname{Hom}(x,x') = \emptyset$. Then now to say that $F(Y) \to F(X)$ is injective is exactly equivalent to the statement that the functor $F(Y) \to F(X)$ is fully faithful.

The upshot is that to make the right definition (of a prestack), we now know that:

  1. $F(X)$ should be a groupoid.
  2. $F(Y) \to F(X)$ should be fully faithful.

We're not there yet, and we need one last modification (at least for $F$ to be a prestack. We need to replace $F(X)$ with $F(X \to Y)$, namely the category of covering data. The objects of this category are pairs $(y, \phi)$ where $y \in F(Y)$ and $\phi : \text{pr}_1^\ast y \to \text{pr}_2^\ast y$ is an isomorphism. A morphism of covering data $ (y, \phi) \to (y', \phi')$ is a map $f : y\to y'$ such that an appropriate diagram commutes (see chapter 8 of the book "Neron Models" by BLR for the exact definition). We can now define:

A groupoid valued functor $F$ (or pseudofunctor in Vistoli's language) is a prestack if the natural pullback functor $F(Y) \to F(X \to Y)$ is fully faithful.

If you unravel what the morphisms are in the category $F(X \to Y)$, you will see this is exactly the condition that the set-valued functor $\underline{\operatorname{Isom}}$ is a plain old sheaf!

So we can finally get to your question. In my view, a stack is a generalized sheaf if you replace sets with groupoids, and if you introduce the category of covering data.


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