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!
The best thing I can say on the splitness is that "it is by the associativity of composition in the base category". You have to either meditate on this, or read some detailed but inevitably unintelligible proof.
For the latter question, the objects of the new category $G'$ are the subsets of $G$ that maps bijectively to $H$, containing the identity $e$. The morphisms in this category $S \to T$ are all just elements in $G$, and composition is group multiplication. There is an obvious functor mapping $G'$ to $H$, and obviously as categories $G' \cong G$.
This new functor is a split fibration, because for each $u \in H$ and an object $S$ in $G'$, we choose the domain of the lift to be $S \cdot (u^*)^{-1}$, where $u^* \in S \subseteq G$ is in the preimage of $u$ under the given group homomorphism, and the multiplication is element-wise. And the lift is just $u^* \in G$.