Prestack
In algebraic geometry,[] a prestack F over a category C equipped with some Grothendieck topology is a category together with a functor p: F → C satisfying a certain lifting condition and such that locally isomorphic objects are isomorphic. A stack is a prestack with effective descents, meaning local objects may be patched together to become a global object.
Prestacks that appear in nature are typically stacks but some naively constructed prestacks may not be stacks. Prestacks may be studied on their own or [|passed to stacks].
Since a stack is a prestack, all the results on prestacks are valid for stacks as well. Throughout the article, we work with a fixed base category C; for example, C can be the category of all schemes over some fixed scheme equipped with some Grothendieck topology.
Definition
Let F be a category and suppose it is fibered over C through the functor ; this means that one can construct pullbacks along morphisms in C, up to canonical isomorphisms.Given an object U in C and objects x, y in, for each morphism in C, after fixing pullbacks, we let
be the set of all morphisms from to ; here, the bracket means we canonically identify different Hom sets resulting from different choices of pullbacks. For each over U, define the restriction map from f to g:
to be the composition
where a canonical isomorphism is used to get the = on the right. Then is a presheaf on the slice category, the category of all morphisms in C with target U.
By definition, F is a prestack if, for each pair x, y, is a sheaf of sets with respect to the induced Grothendieck topology on.
This definition can be equivalently phrased as follows. First, for each covering family, we "define" the category as a category where: writing, etc.,
- an object is a set of pairs consisting of objects in and isomorphisms that satisfy the cocycle condition:
- a morphism consists of in such that
There is an obvious functor that sends an object to the descent datum that it defines. One can then say: F is a prestack if and only if, for each covering family, the functor is fully faithful. A statement like this is independent of choices of canonical identifications mentioned early.
The essential image of consists precisely of effective descent data. Thus, F is a stack if and only if, for each covering family, is an equivalence of categories.
These reformulations of the definitions of prestacks and stacks make intuitive meanings of those concepts very explicit: "fibered category" means one can construct a pullback "prestack in groupoids" additionally means "locally isomorphic" implies "isomorphic" "stack in groupoids" means, in addition to the previous properties, a global object can be constructed from local data subject to cocycle conditions. All these work up to canonical isomorphisms.
Morphisms
Definitions
Given prestacks over the fixed base category C, a morphism is a functor such that and it maps cartesian morphisms to cartesian morphisms. Note is automatic if G is fibered in groupoids; e.g., an algebraic stackIf is the stack associated to a scheme S in the base category C, then the fiber is, by construction, the set of all morphisms from U to S in C. Analogously, given a scheme U in C viewed as a stack and a category F fibered in groupoids over C, the 2-Yoneda lemma says: there is a natural equivalence of categories
where refers to the relative functor category; the objects are the functors from U to F over C and the morphisms are the base-preserving natural transformations.
Fiber product
Let be morphisms of prestacks. Then, by definition, the fiber product is the category where- an object is a triple consisting of an object x in F, an object y in G, both over the same object in C, and an isomorphism in G over the identity morphism in C, and
- a morphism consists of in F, in G, both over the same morphism in C, such that.
This fiber product behaves like a usual fiber product but up to natural isomorphisms. The meaning of this is the following. Firstly, the obvious square does not commute; instead, for each object in :
That is, there is an invertible natural transformation
Secondly, it satisfies the strict universal property: given a prestack H, morphisms,, a natural isomorphism, there exists a together with natural isomorphisms and such that is. In general, a fiber product of F and G over B is a prestack canonically isomorphic to above.
When B is the base category C, B is dropped and one simply writes. Note, in this case, in objects are all identities.
Example: For each prestack, there is the diagonal morphism given by.
Example: Given,.
Example: Given and the diagonal morphism,
this isomorphism is constructed simply by hand.
Representable morphisms
A morphism of prestacks is said to be strongly representable if, for every morphism from a scheme S in C viewed as a prestack, the fiber product of prestacks is a scheme in C.In particular, the definition applies to the structure map . Then p is strongly representable if and only if is a scheme in C.
The definition applies also to the diagonal morphism. If is strongly representable, then every morphism from a scheme U is strongly representable since is strongly representable for any T → X.
If is a strongly representable morphism, for any, S a scheme viewed as a prestack, the projection is a morphism of schemes; this allows one to transfer many notions of properties on morphisms of schemes to the stack context. Namely, let P be a property on morphisms in the base category C that is stable under base changes and that is local on the topology of C. Then a strongly representable morphism of prestacks is said to have the property P if, for every morphism, T a scheme viewed as a prestack, the induced projection has the property P.
Example: the prestack given by an action of an algebraic group
Let G be an algebraic group acting from the right on a scheme X of finite type over a field k. Then the group action of G on X determines a prestack over the category C of k-schemes, as follows. Let F be the category where- an object is a pair consisting of a scheme U in C and x in the set,
- a morphism consists of an in C and an element such that xg = y where we wrote.
When X is a point and G is affine, the quotient is the classifying prestack of G and its stackification is the classifying stack of G.
One viewing X as a prestack, there is the obvious canonical map
over C; explicitly, each object in the prestack X goes to itself, and each morphism, satisfying x equals by definition, goes to the identity group element of G.
Then the above canonical map fits into a 2-coequalizer :
where t: → xg is the given group action and s a projection. It is not 1-coequalizer since, instead of the equality, one has given by
The prestack of equivalence classes
Let X be a scheme in the base category C. By definition, an equivalence pre-relation is a morphism in C such that, for each scheme T in C, the function has the image that is an equivalence relation. The prefix "pre-" is because we do not require to be an injective function.Example: Let an algebraic group G act on a scheme X of finite type over a field k. Take and then for any scheme T over k let
By Yoneda's lemma, this determines a morphism f, which is clearly an equivalence pre-relation.
To each given equivalence pre-relation , there is an associated prestack F defined as follows. Firstly, F is a category where: with the notations,
Via a forgetful functor, the category F is fibered in groupoids. Finally, we check F is a prestack; for that, notice: for objects x, y in F and an object in,
Now, this means that is the fiber product of and. Since the fiber product of sheaves is a sheaf, it follows that is a sheaf.
The prestack F above may be written as and the stackification of it is written as.
Note, when X is viewed as a stack, both X and have the same set of objects. On the morphism-level, while X has only identity morphisms as morphisms, the prestack have additional morphisms specified by the equivalence pre-relation f.
One importance of this construction is that it provides an atlas for an algebraic space: every algebraic space is of the form for some schemes U, R and an étale equivalence pre-relation such that, for each T, is an injective function
Starting from a Deligne–Mumford stack, one can find an equivalence pre-relation for some schemes R, U so that is the stackification of the prestack associated to it:. This is done as follows. By definition, there is an étale surjective morphism from some scheme U. Since the diagonal is strongly representable, the fiber product is a scheme and then let
be the first and second projections. Taking, we see is an equivalence pre-relation. We finish, roughly, as follows.
- Extend to
- By the universal property of stackification, factors through.
- Check the last map is an isomorphism.
Stacks associated to prestacks
As it turns out, it is a stack and comes with a natural morphism such that F is a stack if and only if θ is an isomorphism.
In some special cases, the stackification can be described in terms of torsors for affine group schemes or the generalizations. In fact, according to this point of view, a stack in groupoids is nothing but a category of torsors, and a prestack a category of trivial torsors, which are local models of torsors.