Dose the word "fiber" in "fibered category" and "fiber bundle" mean something similar? For a fiber bundle, a point in the base space relates to a fiber in the total space, in some sense the total space is more complex and has more information than the base space. Recently I'm studying fibered category, it's definition is quite different from fiber bundle's and I'm still puzzled.
Does the word "fiber" in fibered category also mean it somehow has more information than the category it fibers over?
 A: It's not too surprising that the definitions look different, because while they're related, a fibered category is a special case of a generalization of a fiber bundle. The generalization from fiber bundles is, first, to fibrations. A (Serre) fibration $E\to B$ of spaces can be characterized by the property that given any map from a "horn" $\Lambda^n_k$ (which is the $n$-simplex minus its interior and $k$th face) into $E$, any extension of the induced map $\Lambda^n_k\to B$ to $\Delta^n$ can be lifted into $E$ in a way consistent with the given $\Lambda^n_k\to E$. 
Now we generalize further. If we only allow horns $\Lambda^n_k$ with $k>0$, then we have a "right fibration." This is equivalent to the Serre fibration for spaces, but we can interpret it to mean something more interesting for categories. A map $\Delta^n\to B$, for a category $B$, is just a sequence of $n$ composable arrows in $B$, while a map from $\Lambda^n_k$ is defined similarly by sticking together a bunch of maps from $\Delta^{n-1}$. So, if we say a right fibration of categories is a functor $p:E\to B$ such that every extension of a map $\Lambda^n_k\to E$ to a map $\Delta^n\to B$ lifts into $E$, then these turn out to be exactly the functors such that:


*

*Given any map $g:b\to p(e)$  in $B$ there exists a map $\tilde g:\tilde b\to e$ in $E$ with $p(\tilde g)=g$

*Given maps $f:b'\to b,f':b''\to b,$ and $g:p(b'')\to p(b')$ such that $p(f)\circ g=p(f')$, there exists a unique $\tilde g:b''\to b'$ with $p(\tilde g)=g$ and $f\circ \tilde g=f'$. 


These conditions correspond to the lifting properties against the inclusions of $\Lambda^1_1,\Lambda^2_2,$ and $\Lambda^3_3$ in their respective $\Delta^n$s. All the other lifting properties are automatic for functors of categories. Now we almost have the definition of fibered category! To turn the definition of a right fibration (often called a "fibration in groupoids") of categories into the definition of a fibered category, we just have to assume in point 1. that $\tilde g$ is Cartesian, where Cartesian maps are those that satisfy condition 2. when assigned the role of $f$. 
To make the relation with fiber bundles more concrete, we can note that the fibers $p^{-1}(\{b\})$ and $p^{-1}(\{b'\})$ over isomorphic objects $b,b'$ are equivalent categories. This is a bit like the condition of local triviality for a fiber bundle, although again it's more closely related to the behavior of Serre fibrations, in which the fibers over points connected by a path are always homotopy equivalent.
Finally, certainly a fibered category has much more information than is in the base. For instance, a category fibered over the point is simply an arbitrary category!
A: Yes.  In the case of a fibre or vector bundle we have a projection $\pi: T\to B$ and the fibre above $p\in B$ is $\pi^{-1}(p)\subset T$.  Fibred categories concerns inverse images or pull-backs of objects like vector bundles, when we have two such and a map from one total space to the other. 
