I found the following exercise in Vistoli's notes. He proves a theorem (Theorem $3.45$, page number $64$) 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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    $\begingroup$ I deleted my answer because, as Mike Shulman pointed out, it's wrong. $\endgroup$ Commented Feb 3, 2010 at 2:49

2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$ Commented Feb 2, 2010 at 23:08
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    $\begingroup$ I don't think you should retire the question: a negative answer can still be useful! $\endgroup$ Commented Feb 3, 2010 at 4:13

This construction may not be the most natural (or general) one, but I find it reasonably enlightening.

Let $\mathcal F$ and $\mathcal C$ denote the categories with one object associated to $G$ and $H$, respectively. Notice that if $\mathcal F'$ is any category equivalent to $\mathcal F$, then in particular it admits a fully faithful functor to $\mathcal F$. Since $\mathcal F$ has only one object, with isomorphisms in bijection with $G$, this implies that every hom-set in $\mathcal F'$ must be in bijection with $G$ as well. It's easy to check that every morphism in $\mathcal F'$ must be an isomorphism, so this proves that $\mathcal F'$ is a groupoid, with exactly $|G|$ isomorphisms between any two objects.

I claim that we can choose $\mathcal F'$ to have objects indexed by $H$. To be explicit, let's say that the morphisms between any two objects $h_1, h_2$ are identified with $G$, and that the composition of $g_1: h_1 \to h_2$ and $g_2: h_2 \to h_3$ is $g_1 g_2: h_1 \to h_3$. Then this admits a natural "projection" functor to $\mathcal F$, by sending every object to the unique object $*$ of $\mathcal F$ and sending each morphism to the morphism of the same name. We get a functor in the other direction by sending $*$ to the object labeled by the identity of $H$, and preserving names of morphisms. The composition $\mathcal F \to \mathcal F' \to \mathcal F$ is literally the identity functor, and $\mathcal F' \to \mathcal F \to \mathcal F'$ is easily seen to be naturally isomorphic to the identity functor via a base-preserving natural transformation. So $\mathcal F'$ and $\mathcal F$ are equivalent fibered categories over $\mathcal C$.

Now let's construct a splitting of $\mathcal F' \to \mathcal C$. Fix a representative $\widetilde h \in G$ for each element $h \in H$. Consider the subcategory of $\mathcal F'$ that includes all objects, but only the morphisms of the form $\widetilde h_1 \widetilde h_2^{-1}: h_1 \to h_2$. (Note that this does contain identities and compositions.) For any given morphism $h$ in $\mathcal C$ and any object $h_2 \in \mathcal F'$, our chosen subcategory contains a unique pullback $h_1 \to h_2$ of $h$, namely the morphism $\widetilde h_1 \widetilde h_2^{-1}: h_1 \to h_2$ with $h_1$ chosen so that $h_1 h_2^{-1} = h$.

To make this more concrete, let's look at the simplest possible non-split group extension: $\mathbb Z/4\mathbb Z \twoheadrightarrow \mathbb Z/2\mathbb Z$. Here, the category $\mathcal F'$ has two objects, with four morphisms between any pair, all of them isomorphisms. This category deserves to be equivalent to $\mathcal F$: it has two objects, which both look exactly like the object of $\mathcal F$ and are isomorphic to each other. Make $\mathcal F'$ into a fibered category over $\mathcal C$ by composing the "projection" functor to $\mathcal F$ with the given functor $\mathcal F \to \mathcal C$. Recall that we can't construct a splitting of the original fibered category $\mathcal F \to \mathcal C$ precisely because we would need to choose a lift of the morphism $1 \in \mathbb Z/2\mathbb Z$ to $\mathbb Z/4\mathbb Z$, and neither of the two choices gives something that respects composition. But in our new fibered category $\mathcal F'$, we need to choose a lift of $1 \in \mathbb Z/2\mathbb Z$ to some morphism between the two objects of $\mathcal F'$, instead of an automorphism of one of the objects. So we don't need to worry about composing the lift with itself, and the problem is avoided.


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