# 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.

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.

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

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

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.

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 .

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.

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.

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.

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

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.

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 .

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

Note in the above, is a set viewed as a category with objects equal set elements, and morphisms being only identity maps.

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

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

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