# Categories Fibered in Groupoids

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.