It might be a stupid question.
Suppose There is a category of categories,denoted by CAT,where objects are categories, morpshims are functors between categories
Take multiplicative system S={category equivalences}. Then we take localization at S. Then we get localized category S^(-1)CAT
Another Category, denoted by |CAT|,where objects are categories, morphism are isomorphism classes of functors between categories.
I want to prove this two categories are equivalent. I want to prove this problem in two ways:
-
Naive prove
Because either |CAT| or S^(-1)CAT has same objects as CAT. So one can prove the morphism class of these two category has same equivalent relationship. For the S^(-1)Cat, suppose F and G are two functors in the same equivalent class, we have sF=sG,for s belongs to S. Then I can prove F and G belongs to the same equivalent class in |CAT|. But on the other hand, if F and G are isomorphic functor. I can not prove sF=sG,for some s belongs to S. What I can only prove is sF is isomorphic to sG. -
Using adjoint functors
Because I want to prove the projection functor CAT—>|CAT| is a localization functor at S. So I want to construct an adjoint functor |CAT|—>CAT which is fully faithful. But what is this functor. It seems I can not make it well defined.
This is an exercise in Toen's lecture: Lectures on DG-categories
I attach the lecture notes on DG-categories here Toen notes
page 5,exercise 2
Best Answer
There are set theoretic issues that will hinder any proof of this statement. In particular I don't believe you can construct the adjunction in part (2) without applying the axiom of choice to CAT, and it's not clear the localization S^(-1)CAT makes sense either.
That said, let's ignore these issues and pretend that CAT is a small category of (small) categories.
I was originally confused by this. I believe that there are some problems in the formulation of this question. In particular if S consists of those functors which are equivalences of categories, then it does not form a multiplicative system. Specifically, the right Ore condition fails. I'll give an example later. What this means though, is that the property your are trying to check in part (1) actually fails. To see this pick a group G and view it as a category with one object. Then the functors from H to H are the homomorphisms, and the equivalence classes of this functors are the orbits of Hom(H,H) under the conjugation action. Choose H such that there are distinct automorphisms of H which are equivalent under conjugation.
Claim: If F,G: H --> H are two such automorphisms, then there is no equivalence s:H --> C such that sF = sG. I think this is easy to check, so I'll leave it at that.
So the second part of (1) is impossible to prove.
I first thought that we can answer this question be going to skeletal categories. This is not correct, but let's look at why it is not correct. Let SKEL be the full subcategory of CAT consisting of the Skeletal categories, i.e. those categories in which $a \cong b$ implies $a = b$, i.e. those categories which have a single object in each isomorphism class.
Every object in CAT is equivalent to one in SKEL. This is an easy exercise you should work out for yourself. By applying the axiom of choice to all of CAT we can construct an equivalence between CAT and SKEL, and in particular we have L: CAT ---> SKEL with an equivalence $C \cong L(C)$ for every category C.
Notice that a functor between skeletal categories is an equivalence if and only if it is an isomorphism on the nose. So we are part way there. This made me think that SKEL was the localizing subcategory we were after.
However two functors into a skeletal category can be naturally isomorphic without being equal. The group example above is actually an example of this. The functors F and G are naturally isomorphic, but not equal.
Let J denote the "Joyal interval", the category with two objects and an isomorphism between them. J can be used to say when two functors are naturally isomorphic. F,G: C --> D are nat. isomorphic if they extend to a functor $C \times J \to D$. There is a quotient of J where we identify the two objects. This is in fact the category Z (the group of integers viewed as a category). Let W be the set of morphisms consisting of the single morphism J --> Z. Notice that SKEL consists of exactly the "W-local" objects. But this is not quite what we want.
Okay, now for the rest. First, I promised an example where the right Ore condition fails. This is given by considering the two inclusions of pt into J. Both of these are equivalences, but this cannot be completed to an appropriate square. So the class of equivalences in not a multiplicative system.
Next, Toen doesn't assume that S is a multiplicative system. He (correctly) defines the localization in terms of a (weak) universal property. Namely, there exists a locization functor:
$ L: CAT \to S^{-1}CAT$
such that fro any other category D, $L^* : Fun( S^{-1}CAT, D) \to Fun(CAT, D)$ is fully-faithfull and the essential image consists of those functors which send elements of S to isomorphisms in D. In particular the quotient functor from CAT to |CAT| does this, so this gives us functor from S^(-1)CAT to |CAT| defined up to unique natural isomorphism.
On the other hand, |CAT| is also defined by a universal property. This gives us maps between |CAT| and S^(-1)CAT and by the usual sort of general nonsense this is an equivalence. The details aren't too hard, so I'll leave them to you.
Now the second part of your question asks about whether we can realize |CAT| as a full subcategory of CAT such that the quotient $ q:CAT \to |CAT|$ is left adjoint to the inclusion. The answer is no. Let's first look at whether q can have a right adjoint at all and what this would mean.
Suppose that R: |CAT| --> CAT is right adjoint to the quotient q. Be abuse, we will identify the objects of CAT and |CAT|. This means that for all categories X and Y we have:
$CAT(X, R(Y)) = |CAT|(X, Y) = |CAT|(X \times J, Y) = CAT( X \times J, R(Y))$
So $R(Y)$ must be cartesian local with respect to the map J --> pt. In particular we see, taking X=pt, that R(Y) has no non=identity automorphisms, i.e. R(Y) is rigid.
Now let's look at an example. Consider the group Z viewed as a category with one object. Then the endomorphisms of Z in |CAT| is the monoid Z (under multiplication). Now if R is fully-faithful then we need:
$ \mathbb{Z} = |CAT| (\mathbb{Z}, \mathbb{Z}) = CAT (R(\mathbb{Z}), R(\mathbb{Z}))$
but this last is equal to the automorphism of some set R(Z), hence we have a contradiction. So there is no such fully-faithful right adjoint.