3.1. 基础定义和米田引理

The next topic is fibered category. At this point, we talked about descents. Then, descents plus fibered category gives categorical stacks, then plus geometry then we get algebraic stacks. In particular, categorical stacks are special kind of fibered categories which satisfy descent.

Frequently, given a diagramwe say take the fibered product and we get(where we use to indicate Cartesian diagram) However, is only defined up to (canonical) isomorphism.

Thus, any two objects that claim to be the fibered product are isomorphic up to unique isomorphism. Usually it is good enough to make a choice between and and the choice does not make a difference.

However, since stacks are about automorphisms, we need to keep track of the choices we made, which is bad. Thus, we can speak of whenis Cartesian, rather than saying is “the” fibered product.

定义 3.1.0.1. Let be a category, then a category over is a category and a functor .

A morphism in is Cartesian if for the following diagram and for all and factorization, there exists unique withsuch that and .

The point of this is that, with this, the diagramlooks like a pullback.

定义 3.1.0.2. Let be category over and be Cartesian, then we say is a pullback of along .

If and are both pullbacks along the samething , then we haveand hence which implies is an isomorphism (with ).

注 3.1.0.3. Now given , for any , let be a category defined as follows: the objects are such that and morphisms are such that .

The idea above is that we are supposed to think of as a map from to categories, where we input and output a “category” .

定义 3.1.0.4. Let be a category over , then we say is a fibered category if “pullbacks exist”, i.e. given diagramthen there exists Cartesian arrow such that , i.e.is a pullback.

例 3.1.0.5. Let be the category of schemes, and let be the category of genus curves. In other words, objects of are where is smooth and geometric fibers of are genus curves(recall geometric fiber means ). The morphisms are diagrams

Then, our is going to be . Since pullback exists because they exists in , we see is a fibered category. Indeed, just take the fibered product in , saythen we getwhere since is smooth, then is smooth, and has the same geometric fibers as .

Thus, we just constructed the moduli space of genus curves.

例 3.1.0.6. If is any category, , then consider as the category with objects and morphismsThen with gives the structure of a fibered category. This is because every arrow in is Cartesian. Indeed, take an arrow in , and let be any arrow in with a factorization , then we see we indeed have dashed arrow in the following diagramThat is, we just take the dashed arrow be , which is indeed unique and it exists.

例 3.1.0.7. Let’s consider the category over . Here objects of are pairs where is quasi-coherent sheaf on , and morphisms between are given by and .

In the above examples, as in Remark 3.1.0.3, we see is exactly the category of quasi-coherent sheaves on , and is exactly the category of genus curves on .

Indeed, is by definition the category with objects being such that and morphisms being such that . Well, means we need to have is the identity, i.e. and is just a morphism between quasi-coherent sheaves on .

Similarly is category of genus curves because the projection forces any object to be live over .

定义 3.1.0.8. If and be two fibered categories, then a morphism of fibered categories is a functor such thatand sends Cartesian arrows to Cartesian arrows.

We note, for all , we get . Indeed, since are categories, is a functor between categories. It is just the same as , i.e. . We can check this is indeed a functor. Indeed, let be two arrows, we need to show we get a diagramHowever, note we get the following diagramwhere the two triples of vertical arrows (i.e. ) commutes, and the outer square and inner square both commutes, which forces the upper square to commute as desired.

定义 3.1.0.9. If are two morphisms of fibered categories, then a base-preserving natural transformation is a natural transformation of functors such that for all , the map satisfies , i.e. if we have the following diagramthen we must have . In other word, we want to be a morphism in .

We note this gives the category with objects being morphisms of fibered category and morphisms being base-preserving natural transformations.

Now, suppose we have , then we get for all . This is kind of looks like a map of presheaves.

Last time we defined a fibered category, which is a functor such that pullbacks exist, i.e. for all and for all , there exists unique so the following commutesThe morphisms of fibered categories are given by functors such that sends Cartesian arrows to Cartesian arrows.

Then, for fibered category, for all , we let be the category with objects being such that , and morphisms being such that .

Thus, say be maps of fibered categories over , then we get for all . At the end of last class, we mentioned this implies fibered categories look like presheaves but instead of being sets, we have is category.

例 3.1.0.10. Consider be the fibered category of genus curves with marked points. In this category, objects of are with smooth and on geometric fibers of , is genus curves. Next, we need to explain what marked points are. Those are given by sections of , say , which are distinct points on the geometric fibers. Then the morphisms are diagramssuch that and . Then, the projection is given by .

One should check that pullbacks exist.

In particular, is the moduli space of genus curves with one marked point, i.e. they are exactly elliptic curves.

Next, note we have the fibered category and we get map of fibered category, for . The map is given bywhere we take the pullback of relative differential via . We also need to define what does on morphisms. Well, suppose we have morphismWe need a map between . Well, we do have a canonical morphism (and in fact its isomorphism), as we will show next. First, note is equal to as the diagram commutes, and we also have . Hence we get the desired canonical (iso)morphism as desired.

引理 3.1.0.11. Suppose is a map of fibered category. Then is fully faithful as a map of categories (not fibered category) if and only if , is fully faithful.

This result should remind you of: if is a map of presheaves then it is injective if and only if for all , is injective.

证明. Recall fully faithful means we have a bijection between Hom sets (full means surjection between hom sets and faithful means injection between hom sets). Thus, let , we getThen is fully faithful if and only if for all in , induces a bijectionThis is sort of like we show bijection on each of the fibers.

Thus, we fix downstairsand let be a fibered product/pullbackThen we see for all , we get unique arrow , i.e. we getso, we have a bijectionThis is because when we actually using definition of pullback, what we get is a diagramNow, since is Cartesian arrow as is Cartesian (by def of morphism of fibered cat), we see we getWe also have the floating around, and we getHence we get bijectionwith the similar reasoning as above. Hence,is a bijection if and only if is fully faithful. This concludes the proof.

定义 3.1.0.12. We say a map of fibered categories is an equivalence if there exists map of fibered categories and exists a base preserving isomorphism and .

命题 3.1.0.13. For a map of fibered categories , is equivalence iff , is an equivalence (in category theory sense) iff , is fully faithful and essentially surjective.

证明. We already showed is fully faithful if and only if all are. Thus we just need to show the claims about essentially surjective (recall essentially surjective for means each object is isomorphic to an object of the form where ).

if is equivalence, then we want to be essentially surjective. Given , we have and is a morphism of fibered cats1, so .

: now assume is essentially surjective for all . We need to construct an equivalence of fibered categories . Given , since is equivalence, we know there exists such that . Given any in , there exists unique in such thatbecause is fully faithful.

This gives functor and also . We need sends Cartesian arrows to Cartesian arrows. If is Cartesian in , then we getand suppose we are given arbitrary with the following diagramwhere we want to show there exists unique arrow . Since is fully faithful, there exists dotted arrow if and only if it holds after we apply . Thus we want to show there exists unique dotted arrow in the following:However, note we can complete the diagram withIn other word, we getbut is Cartesian, hence we indeed have the dotted arrow as desired.

Lastly, we need . If , we want . By full faithfulness of , we just need to show . But we do have an isomorphism, . Then, we left as an exercise that is a natural transformation.

Now we have seen fibered categories are analogous to presheaves (over ), we ask a natural question: what is analogue of sheaf? The answer is stacks.

Next, we ask if fibered categories are analogous to presheaves, is there a type of Yoneda lemma? Well, there is, and its called -Yoneda lemma.

Before we do this, let’s recall if , we have fibered category maps with morphism . The analogy is that, should correspond to .

定理 3.1.0.14 (-Yoneda Lemma). For any fibered category , and all , we have a category with morphisms being base-preserving natural transformations. Then, we have a equivalence of categories

We will prove the -Yoneda lemma next time, and we first prove an corollary of this.

推论 3.1.0.15. For , we haveis an equivalence of categories.

Note in the above, is a set viewed as a category with objects equal set elements, and morphisms being only identity maps.
证明. Well, apply -Yoneda lemma to , we getwhere the objects are and morphisms areliving over , i.e. . Thus, the only morphism we have in are identity maps, i.e. is exactly the category .

Thus, we will introduce some notations: we will frequently write in place of if is fibered category. This is justified by -Yoneda because is the same as . So it is just a convenience to write in place of . The corollary shows , so in place of is unambiguous.

Last time, we have the -Yoneda lemma

定理 3.1.0.16 (-Yoneda Lemma). For any fibered category , and all , we have a category with morphisms being base-preserving natural transformations. Then, we have a equivalence of categories

Just like the proof of Yoneda lemma, this is not involved, but also not a short proof.

证明. We need to construct that maps to a fibered category .

First, we define on objects: an object in is a map . In particular, we have diagramWell, the natural thing to do is just take a pullback , i.e. we make a choice and get the following diagramand define .

Next, we define on morphisms. Suppose we are given morphismin . We want to know what is. Well, we get the following diagrambut then we get a dotted arrow between to as the squares are Cartesian. Hence we haveThis unique dotted arrow gives the desired map on morphisms (i.e. ).

Now we know as a functor , why is a morphism of fibered categories. We have to check two things: the first thing is that it respect fibers. However, we checked that already, i.e. .

The second thing we need to show is that takes Cartesian arrows to Cartesian. First, in all arrows are Cartesian, so we need . However, note by basic category theory, since the inner square and outer squares are both Cartesian, the dotted arrow must also be Cartesian.

So, we now know on objects. What about morphisms?

Given in , we want