Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I found the following exercise in Vistoli's notes. He proves a theorem stating that any category $\mathcal{F}$ fibered over $\mathcal{C}$ is equivalent, as a fibered category, to a split one. Namely $\mathcal{F}$ is equivalent to the category $\mathcal{F}' = Hom_\mathcal{C}(\cdot, \mathcal{F})$, which is the fibered category associated to the following functor $F : \mathcal{C}^{op} \rightarrow Cat$.

For every $U \in \mathcal{C}$ we set $F(U) = Hom_\mathcal{C}(\mathcal{C}/U, \mathcal{F})$, where $\mathcal{C}/U$ is the comma category and $Hom_\mathcal{C}$ denotes the category of morphism of fibered categories. (In particular an arrow in this category is a morphism of functors over the identity of $\mathcal{C}$). The action of $F$ on arrows is the obvious one: an arrow $U \rightarrow V$ in $\mathcal{C}$ gives a functor $\mathcal{C}/U \rightarrow \mathcal{C}/V$, and $F$ acts by composition with this functor.

The exercise requires to carry out the construction explicitly for the following situation. A group $G$ can be seen as a category with a single object. If $G \rightarrow H$ is a surjective homomorphism, then we can see $G$ as a category fibered over $H$, and the exercise is to work out what $\mathcal{F}'$ is in this case.

I am able to do this exercise, but I think I am missing something. Vistoli says that this is a nice exercise, so I guess I should obtain as a result something which I can recognize, but I don't. If needed I can post here my answer, but it is not very enlightening.

I was tempted to write here the relevant terminology, but it is pointless, as everything is clearly defined in chapter 3 of the above mentioned notes. If you need any clarification, I'll be happy to provide more details.

share|improve this question
    
I deleted my answer because, as Mike Shulman pointed out, it's wrong. –  Tom Leinster Feb 3 '10 at 2:49
add comment

1 Answer

In Anton Geraschenko's notes from Martin Olsson's course on stacks, you can find the following quote:

The upshot is that if you choose a splitting, you really have no idea what’s going on.

That's at the end of example 23.8, page 94, precisely after calculating a splitting for $\mathbb{Z}/4 \to \mathbb{Z}/2$ as a fibered category.

share|improve this answer
    
Uhm... that kinda explains why the computation in general seemed so cumbersome :-D If this is the case, I may retire the question. I'll just wait a few days to see if someone finds out a nice interpretation. –  Andrea Ferretti Feb 2 '10 at 23:08
    
I don't think you should retire the question: a negative answer can still be useful! –  Alberto García-Raboso Feb 3 '10 at 4:13
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.