 Solution

Use the square's notation $HF=KG$. For $E\in \mathcal{E}$, choose $(C,u:E\to GC)$ in the nonempty connected category $(E\downarrow G)$. Lift $Ku:KE\to HFC$ through the discrete fibration $H$, obtaining the uniquely determined arrowDefine $LE=D_u$. If $a:C\to C^{\prime }$ satisfies $Gau=u^{\prime }$, then $Fa\tilde{u}$ is a lift of $K{u}^{\prime }$ with codomain $F{C}^{\prime }$. Uniqueness makes it exactly ${\tilde{u}}^{\prime }$, including equality of its domain. Connectedness of the comma category therefore makes $LE$ independent of the chosen object, with strict equality rather than merely an isomorphism.

For $v:E\to E^{\prime }$, choose $(C,u:E^{\prime }\to GC)$ and lift $Kv:KE\to K{E}^{\prime }=HL{E}^{\prime }$ with codomain $L{E}^{\prime }$. Write this lift as $w:D\to L{E}^{\prime }$. Its composite with $\tilde{u}$ is the lift of $K(uv)$ to $FC$, so its domain is $LE$. Set $Lv=w$. The lift of an identity is an identity, and the composite of two lifts is the lift of the composite with the specified codomain. Thus $L$ is a functor, and $HL=K$.

At $E=GC$, choose $(C,1_{GC})$; its lift is $1_{FC}$, giving $LGC=FC$. For $a:C\to C^{\prime }$, the unique lift of $KGa=HFa$ with codomain $F{C}^{\prime }$ is $Fa$, giving $LG=F$ on arrows as well. Finally, any other filler $L^{\prime }$ must send $u$ to the same lift of $Ku$ with codomain $FC$, so $L^{\prime }E=LE$; it must also send $v$ to the unique lift of $Kv$ with codomain $L{E}^{\prime }$. This proves orthogonality of final functors and discrete fibrations: