A Mind for Madness

Musings on art, philosophy, mathematics, and physics

Categories Fibered in Groupoids

1 Comment


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 {F:\mathcal{C}\rightarrow\mathcal{D}}. I’ll refer to {\mathcal{D}} as the “base” category and {\mathcal{C}} as lying over {\mathcal{D}}. We’ll say {\mathcal{C}} is fibered in groupoids over {\mathcal{D}} (by {F}) if the functor satisfies two conditions.

First, suppose we have objects in the base with an arrow between them {A\rightarrow B} and an object lying over {B} (i.e. some object {Y\in \mathcal{C}} such that {F(Y)=B}). 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 {X\rightarrow Y} such that the arrow maps to the base arrow. So {F(X)=A} and {F(X\rightarrow Y)= A\rightarrow B}.

It might be good to visualize this in the following way: Given {\begin{matrix} & & Y \\ & & \\ A & \rightarrow & B \end{matrix}} you can always complete to {\begin{matrix} X & \rightarrow & Y \\ & & \\ A & \rightarrow & B\end{matrix}}

The second condition is that whenever you have objects and arrows {A\rightarrow B\rightarrow C} in the base category and you have {X, Y, Z} lying over {A, B, C} respectively with the two arrows {Y\rightarrow Z} lying over {B\rightarrow C} and {X\rightarrow Z} lying over the composite {A\rightarrow C}, there is a unique arrow {X\rightarrow Y} so that everying lies over {A\rightarrow B\rightarrow C}.

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

{\begin{matrix} Y & \rightarrow & Z \\ & & \\ B & \rightarrow & C \end{matrix}} and {\begin{matrix} X & \rightarrow & \rightarrow & \rightarrow & Z \\ \\ A & \rightarrow & B & \rightarrow & C \end{matrix}}

the whole thing can be completed {\mathit{uniquely}} to {\begin{matrix} X & \rightarrow & Y & \rightarrow & Z \\ \\ A & \rightarrow & B & \rightarrow & C \end{matrix}}.

To get a feel for this, let’s look at a motivating example for the terminology. Consider any functor {F: \mathcal{C}\rightarrow \mathcal{D}}, then if we pick an object {D} in the base we can form the {\mathit{fiber \ category}} over {D}, which we’ll denote {\mathcal{C}_D}. This is just a category whose objects lie over {D} i.e. all {X} so that {F(X)=D}, and the morphisms are the ones the functor maps to the identity morphsim {id_D: D\rightarrow D}.

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 {Y} and {Z} in {\mathcal{C}_D}, say {Y\stackrel{f}{\rightarrow} Z}. Then by definition of the category {F(f)=id_D}. So we can build the situation of the second condition. The base is just {D\rightarrow D \rightarrow D} all the identity and hence the composition is the identity. We have {Y\rightarrow Z} lying over the second identity map, and we have the identity {Z\rightarrow Z} lying over the composition, so we get a unique map {Z\rightarrow Y} such that the composite {Z\rightarrow Y\rightarrow Z} is the identity. We can now repeat this process with this unique map {Z\rightarrow Y} to get {Y \rightarrow Z \rightarrow Y} which is the identity on {Y} and by uniqueness the completed map must be the original {f} and hence {f} is an isomorphism. Thus the fiber category is a groupoid since any morphism is an isomorphism.

About these ads

Author: hilbertthm90

I am a mathematics graduate student fascinated in how all my interests fit together.

One thought on “Categories Fibered in Groupoids

  1. based on the stacks for stacks phase….
    my most simplistic understanding of my personal relativity began with the Pythagorean theorem, I cant remember what philosopher’s theory I was learning about that afternoon.
    Why should I go more in to specificity about this and how would it even benefit what I think Ive already theorized???????
    I am not a mathematician and never was, I majored in political science and understand philosophy and art to be interrelated. I am going to law school. WHY would studying math in my spare time benefit me??????
    PLEASE EMAIL ME

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 163 other followers