 Solution

For an arbitrary functor $F:\mathcal{C}\to \mathcal{D}$, define a category $\mathcal{E}$ as follows. Its objects are pairs $(B,c)$, where $B\in \mathcal{D}$ and $c$ is a connected component of $(B\downarrow F)$. Precomposition by an arrow $g:B\to B^{\prime }$ definesThere is one morphism $(B,c)\to (B^{\prime },c^{\prime })$ over $g$ exactly when $g^{\ast }(c^{\prime })=c$. Functoriality of precomposition makes this a category, and projectionis a discrete fibration: given $g:B\to B^{\prime }=G(B^{\prime },c^{\prime })$, its unique lift with codomain $(B^{\prime },c^{\prime })$ has domain $(B,g^{\ast }{c}^{\prime })$.

Define $H:\mathcal{C}\to \mathcal{E}$ bywhere $[1_{FA}]$ is the component of the identity object in $(FA\downarrow F)$. For $u:A\to A^{\prime }$, use the unique arrow over $Fu$. It exists because $(Fu{)}^{\ast }[1_{F{A}^{\prime }}]$ contains the object $Fu:FA\to F{A}^{\prime }$, and this object is joined to $1_{FA}$ by the morphism $u$ in $(FA\downarrow F)$. Plainly $GH=F$.

For $(B,c)\in \mathcal{E}$, an object of $((B,c)\downarrow H)$ is exactly an arrow $\beta :B\to FA$ lying in the component $c$: the condition for an arrow $(B,c)\to H(A)$ is precisely ${\beta }^{\ast }[1_{FA}]=c$. Morphisms agree with those in $(B\downarrow F)$. Hencewhich is nonempty and connected by definition. Therefore $H$ is final. We have factored $F=GH$ as a final functor followed by a discrete fibration, giving the final-discrete-fibration factorization.