 Solution

Consider the commutative squarewhere $F$ is a final functor and $G$ is a discrete fibration. For $B\in \mathcal{B}$, choose an object $(A,\beta :B\to FA)$ of the nonempty comma category $(B\downarrow F)$. Commutativity gives an arrowLift it uniquely through $G$ with codomain $HA$, and define $LB$ to be the domain of this lift.

This definition does not depend on the choice of $(A,\beta )$. A morphism $u:(A,\beta )\to (A^{\prime },{\beta }^{\prime })$ in the comma category satisfies $Fu\beta ={\beta }^{\prime }$. The composite of the lift of $K\beta$ with $Hu$ is then a lift of $K{\beta }^{\prime }$ with codomain $H{A}^{\prime }$, so uniqueness of discrete-fibration lifts says that it is the chosen lift and has the same domain. Since $(B\downarrow F)$ is connected, all choices give the same object $LB$.

For $v:B\to B^{\prime }$, define $Lv$ as the unique lift through $G$ of $Kv:KB\to K{B}^{\prime }=G(L{B}^{\prime })$ with codomain $L{B}^{\prime }$. Its domain is $LB$: choose ${\beta }^{\prime }:B^{\prime }\to FA$, and observe that composing this lift with the lift of $K{\beta }^{\prime }$ yields the lift associated with ${\beta }^{\prime }v:B\to FA$. Uniqueness of lifting also proves preservation of identities and composition, so $L$ is a functor and $GL=K$.

For $B=FA$, choose $(A,1_{FA})$ in $(FA\downarrow F)$. The lift of $1_{GHA}$ is $1_{HA}$, hence $LFA=HA$; the same lifting argument on arrows gives $LF=H$. Finally, if $L^{\prime }$ is another filler, then for every $\beta :B\to FA$, the arrow $L^{\prime }\beta :L^{\prime }B\to HA$ is a lift of $K\beta$. Unique lifting forces $L^{\prime }B=LB$ and then forces equality on arrows. Thus $L$ is unique, proving orthogonality of final functors and discrete fibrations.