 Solution

Write the square as $GH=KF$, with $F:\mathcal{A}\to \mathcal{B}$ final and $G:\mathcal{C}\to \mathcal{D}$ a discrete fibration. For $b\in \mathcal{B}$, choose an object $(a,h:b\to Fa)$ of the nonempty comma category $(b\downarrow F)$. Lift $K(h):Kb\to GHa$ uniquely to an arrowover $K(h)$, and tentatively put $L(b)=c_h$.

A morphism $(a,h)\to (a^{\prime },h^{\prime })$ in this comma category is an arrow $j:a\to a^{\prime }$ with $F(j)h=h^{\prime }$. The composite $H(j){\ell }_h$ is a lift of $K(h^{\prime })$ with codomain $H{a}^{\prime }$. Uniqueness of lifts gives $H(j){\ell }_h={\ell }_{h^{\prime }}$, including equality of their source objects. Since the comma category is connected by zigzags, $c_h$ is independent of the choice. Thus $L(b)$ is well defined and $GL(b)=Kb$.

For $f:b\to b^{\prime }$, take $(a^{\prime },h^{\prime }:b^{\prime }\to F{a}^{\prime })$. Lift $K(f)$ into $L(b^{\prime })$, obtaining $t:c\to L(b^{\prime })$. Composing with ${\ell }_{h^{\prime }}$ gives a lift of $K(h^{\prime }f)$ into $H{a}^{\prime }$, whose source must be $L(b)$. Hence $t:L(b)\to L(b^{\prime })$. Set $L(f)=t$. Uniqueness of this lift makes the definition independent of the chosen comma object and proves $L(1_b)=1_{L(b)}$ and $L(gf)=L(g)L(f)$.

At $b=Fa$, choose $(a,1_{Fa})$. Its lift is $1_{Ha}$, so $L(Fa)=Ha$. For $j:a\to a^{\prime }$, $Hj$ is the unique lift of $K(Fj)$ into $H{a}^{\prime }$, so $L(Fj)=Hj$. We have constructed the required functor with $LF=H$ and $GL=K$.

If $L^{\prime }$ is another filler, $L^{\prime }(h):L^{\prime }b\to Ha$ must equal ${\ell }_h$ by unique lifting. Thus $L^{\prime }b=L(b)$ for every object. Its arrows are then forced by unique lifting into those objects. ThereforeThis proves the orthogonality of final functors and discrete fibrations.