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A site C with pullbacks is subcanonical (all representable presheaves are sheaves) if and only if its codomain fibration $Arr(C) \to C$ is a prestack (all hom-presheaves are sheaves). Is there a common name for a site whose codomain fibration is a stack? The canonical topology on a Grothendieck topos has this property, as does the coherent topology on a pretopos, the regular topology on a Barr-exact category, the extensive topology on a lextensive category, etc.

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1 Answer 1

I don't have an answer to your question, but I'm going to post whatever thoughts I had about it. Maybe something here will help someone answer the question, or at least help more people understand what's involved. I'm sorry that it's come out so long.

(skip this unless you suspect we mean different things by "(pre)stack")

A functor $F\to C$ is a fibered category if for every arrow $f:U\to X$ in $C$ and every object $Y$ in $F$ lying over $X$, there is a cartesian arrow $V\to Y$ in $F$ lying over $f$ (see Definition 3.1 of Vistoli's notes). This arrow is determined up to unique isomorphism (by the cartesian property), so I'll call $V$ "the" pullback of $Y$ along $f$ and maybe denote it $f^*Y$. A fibered category is roughly a "category-valued presheaf (contravariant functor) on $C$".

Given an object $X$ in $C$, let $F(X)$ be the subcategory of objects in $F$ lying over $X$, with morphisms being those morphisms in $F$ which lie over the identity morphism of $X$. I'll call $F(X)$ the "fiber over $X$." Given a morphism $f:U\to X$ in $C$, let $F(U\to X)$ be "the category of descent data along $f$," whose objects consist of an element $Z$ of $F(U)$ and an isomorphism $\sigma:p_2^*Z\to p_1^*Z$ (where $p_1,p_2:U\times_XU\to U$ are the projections) satisfying the usual cocycle condition over $U\times_XU\times_XU$ (see Definition 4.2 of Vistoli's notes). A morphism in $F(U\to X)$ is a morphism $Z\to Z'$ in $F(U)$ such that the following square commutes:
$\begin{matrix} p_2^*Z & \xrightarrow{\sigma} & p_1^*Z \\ \downarrow & & \downarrow\\ p_2^*Z' & \xrightarrow{\sigma'} & p_1^*Z' \end{matrix}$

Suppose $C$ has the structure of a site. Then we say that $F$ is a prestack (resp. stack) over $C$ if for any cover $U\to X$ in $C$, the functor $F(X)\to F(U\to X)$ given by pullback is fully faithful (resp. an equivalence). Roughly, a prestack is a "separated presheaf of categories" and a stack is a "sheaf of categories" over $C$.

The domain fibration (not your question, but related)

Consider the domain functor $Arr(C)\to C$ given by $(X\to Y)\mapsto X$. You can check that a cartesian arrow over $f:U\to X$ is a commutative square
$\begin{matrix} U & \xrightarrow{f} & X \\ \downarrow & & \downarrow\\ Y & = & Y \end{matrix}$
If I haven't made a mistake,

  • This fibered category is a prestack iff every cover $U\to X$ is an epimorphism.
  • It is a stack if furthermore every cover $U\to X$ is the coequalizer of the projection maps $p_1,p_2:U\times_XU\to U$. This last condition is equivalent to saying that every object $Y$ of $C$ satisfies the sheaf axiom with respect to the morphism $U\to X$. In particular, the domain fibration is a stack if and only if the topology is subcanonical.

The codomain fibration (your question)

Consider the codomain functor $Arr(C)\to C$ given by $(U\to X)\mapsto X$. You can check that a cartesian arrow over a morphism $f:U\to X$ is a cartesian square
$\begin{matrix} V & \to & U \\ \downarrow & & \downarrow\\ Y & \xrightarrow{f} & X \end{matrix}$
There is a general result that says that the fibered category of sheaves on a site is itself a stack (I usually call this result "descent for sheaves on a site"). If you're working with the canonical topology on a topos (where every sheaf is representable), it follows that the codomain fibration is a stack. If the topology is subcanonical, then objects are sheaves, so descent for sheaves tells you that the pullback functor is fully faithful (i.e. the codomain fibration is a prestack), but when you "descend" a representable sheaf, it may no longer be representable, so the codomain fibration may not be a stack. In your question you say that being a prestack is actually equivalent to the topology being subcanonical, but I can't see the other implication (prestack⇒subcanonical).

Supposing the codomain fibration is a prestack, saying that it is a stack roughly says that when you glue representable sheaves along a "cover relation," you get a representable sheaf, but with the strange condition that the "cover relation" you started with came from a relation where you could glue to get a representable sheaf. That is, given this diagram, where the squares on the left are cartesian ($\Rightarrow$ is meant to be two right arrows), can you fill in the "?" so that the square on the right is cartesian?
$\begin{matrix} Z' & \Rightarrow & Z & \to & ?\\ \downarrow & & \downarrow & & \downarrow\\ U\times_XU & \Rightarrow & U & \to & X \end{matrix}$

A more natural (to me) condition is to ask that the only sheaves you can glue together from representable sheaves are already representable. That is, if $R\Rightarrow U$ is a "covering relation" (i.e. each of the maps $R\to U$ is a covering and $R\to U\times U$ is an equivalence relation), then the quotient sheaf $U/R$ is representable. I would call such a site "closed under gluing."

For example, the category of schemes with the Zariski topology is closed under gluing (it's the "Zariski gluing closure" of the category affine schemes). The category of algebraic spaces with the etale topology is closed under gluing (it's the "etale gluing closure" of the category of affine schemes). In fact, I think that a standard structure theorem for smooth morphisms and a theorem of Artin (∃ fppf cover ⇒ ∃ smooth cover) imply that the category of algebraic spaces with the fppf topology is closed under gluing.

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Wait. Aren't algebraic spaces usually defined to be quasi-seperated? Doesn't this mean that you also need a condition on the diagonal map $U \to U \times_{U/R} U$? or on the map $U \times_{U/R} U \to U \times U$? This seems to suggest that algebraic spaces (viewed as sheaves) are not closed in the way you describe. See also this MO question: mathoverflow.net/questions/9043/… –  Chris Schommer-Pries Dec 24 '09 at 13:14
Though quasi-separatedness is necessary for many results, it just doesn't make sense to build it into the definition. After all, not all schemes are quasi-separated, and all schemes should be algebraic spaces. But there is a separation hypothesis that I swept under the rug: when I take the "gluing closure" of a site, I only want to throw in sheaves with representable diagonal (otherwise I wouldn't be able to extend the site structure to the enlarged category). I seem to recall that this isn't completely automatic for algebraic spaces, but very close to free. –  Anton Geraschenko Dec 24 '09 at 18:03
Incidently, you may need to apply the "gluing closure construction" several times (possibly countably many?) to get a site closed under gluing. For example, the applying it to affine schemes with the Zariski topology gives you the category of separated schemes (because of the representability condition on the diagonal). You have to apply it again to get the category of all schemes. –  Anton Geraschenko Dec 24 '09 at 18:05
Thanks! My definition of "prestack" is the same as yours, except that I would consider arbitrary covering families, not just single covers (see mathoverflow.net/questions/9705/… ). Then regarding prestack ⇒ subcanonical, $C$ is subcanonical if for any covering family $(U_i\to X)$, maps $X\to Y$ are the same as compatible families of maps $(U_i\to Y)$. But maps $X\to Y$ are the same as maps $X\to Y\times X$ in the fiber $C/X$ of the codomain fibration, and similarly for maps $U_i\to Y$, which the prestack condition lets you glue. –  Mike Shulman Dec 24 '09 at 22:01
Also, I wonder if our definitions of "covering" are different? The projections $R\to U$ of any equivalence relation are split epic, and hence coverings in any Grothendieck topology, so it seems to me like any equivalence relation is a "covering relation" in your sense. I think your notion of "closed under gluing" would then just reduce to saying that the category is Barr-exact and its topology includes the regular topology (a "superexact site"). –  Mike Shulman Dec 24 '09 at 22:07

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