Sorry about the delay, I’ve been really busy with other things. Most (probably all) people have completely forgotten what I was talking about. Luckily you don’t need to in order to follow this post!
Today we’ll look at what it means for a category to be fibered in groupoids over another one. Suppose we have a (covariant) functor . I’ll refer to as the “base” category and as lying over . We’ll say is fibered in groupoids over (by ) if the functor satisfies two conditions.
First, suppose we have objects in the base with an arrow between them and an object lying over (i.e. some object such that ). The condition is that whenever we have this situation we can “complete the square”. This means that we can find an object and an arrow such that the arrow maps to the base arrow. So and .
It might be good to visualize this in the following way: Given you can always complete to
The second condition is that whenever you have objects and arrows in the base category and you have lying over respectively with the two arrows lying over and lying over the composite , there is a unique arrow so that everying lies over .
There is a nice way to visualize this as well, but I am still awful at making nice things in wordpress, so we’ll do it as follows, you have the following two pieces of information
the whole thing can be completed to .
To get a feel for this, let’s look at a motivating example for the terminology. Consider any functor , then if we pick an object in the base we can form the over , which we’ll denote . This is just a category whose objects lie over i.e. all so that , and the morphisms are the ones the functor maps to the identity morphsim .
Claim: Any fiber category that is fibered in groupoids (via the functor that we are taking the fiber of) is a groupoid. By groupoid here we just mean that every morphism is an isomorphism.
Suppose there is a map between and in , say . Then by definition of the category . So we can build the situation of the second condition. The base is just all the identity and hence the composition is the identity. We have lying over the second identity map, and we have the identity lying over the composition, so we get a unique map such that the composite is the identity. We can now repeat this process with this unique map to get which is the identity on and by uniqueness the completed map must be the original and hence is an isomorphism. Thus the fiber category is a groupoid since any morphism is an isomorphism.