Given a fibred category $\mathcal{F} \to \mathcal{C}$, we can choose a cleavage, which is a class of cartesian arrows $K$ in $\mathcal{F}$ s.t. for each arrow $f:U\to V$ in $\mathcal{C}$ and each object $\eta$ in $\mathcal{F}(V)$ there exists a unique arrow in $K$ with target $\eta$ mapping to $f$ in $C$. A split fibred category is a fibred category admit a splitting, i.e. there is a splitting cleavage which contains all identities and is closed under composition of arrows. Such a splitting doesn’t always exist.

For example, if we consider a group $G$ as a category with only one object, and arrows are multiplied by group elements, then a surjective group homomorphism $G\to H$ can be seen as a fibred category. A cleavage is a subset $K$ of $G$ that maps bijectively onto $H$. Such a cleavage splits iff $K$ is a subgroup of $G$. Thus we have a homomorphism $H\to G$ s.t. the composition $H\to G\to H$ is identity. Surely, such a homomorphism doesn’t always exist.

There is a statement that a fibred category is equivalent to a split fibred category by using 2-Yoneda lemma. If $U$ is an object of $\mathcal{C}$, we identify the functor $h_U=Hom(-,U)$ with comma category ($\mathcal{C}/U$). We can have a functor $Hom(-,\mathcal{F})$ sending $U$ into the category $Hom(h_U,\mathcal{F})$. Now we denote by $\mathcal{F}’$ the fibred category associated with this functor. Then we have a split fibred category equivalent to $\mathcal{F}$ by 2-Yoneda lemma.

My first question is how can we see the fibred category constructed above is split? Second, what does this construction yields in the case of surjective group homomorphism $G\to H$ as the example above? Is it $Hom(H,G)$? Thank in advance!


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