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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.

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I deleted my answer because, as Mike Shulman pointed out, it's wrong. –  Tom Leinster Feb 3 '10 at 2:49

1 Answer 1

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.

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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

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