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.
Definitions
(skip this unless you suspect we mean different things by "(pre)stack")
A functor $Fto C$ is a fibered category if for every arrow $f:Uto X$ in $C$ and every object $Y$ in $F$ lying over $X$, there is a cartesian arrow $Vto Y$ 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 $f:Uto X$ in $C$, let $F(Uto X)$ be "the category of descent data along $f$," whose objects consist of an element $Z$ of $F(U)$ and an isomorphism $sigma:p_2^*Zto p_1^*Z$ (where $p_1,p_2:Utimes_XUto U$ are the projections) satisfying the usual cocycle condition over $Utimes_XUtimes_XU$ (see Definition 4.2 of Vistoli's notes). A morphism in $F(Uto X)$ is a morphism $Zto Z'$ 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 $Uto X$ in $C$, the functor $F(X)to F(Uto X)$ 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 $(Xto Y)mapsto X$. You can check that a cartesian arrow over $f:Uto X$ 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 $Uto X$ is an epimorphism.
- It is a stack if furthermore every cover $Uto X$ is the coequalizer of the projection maps $p_1,p_2:Utimes_XUto U$. This last condition is equivalent to saying that every object $Y$ of $C$ satisfies the sheaf axiom with respect to the morphism $Uto X$. 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 $(Uto X)mapsto X$. You can check that a cartesian arrow over a morphism $f:Uto X$ 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?
$begin{matrix}
Z' & Rightarrow & Z & to & ?\
downarrow & & downarrow & & downarrow\
Utimes_XU & Rightarrow & U & to & X
end{matrix}$
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 $RRightarrow U$ is a "covering relation" (i.e. each of the maps $Rto U$ is a covering and $Rto Utimes U$ 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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