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Suppose $F=GJ=G^{\prime }{J}^{\prime }$ are two final-discrete-fibration factorizations. Apply the orthogonality of final functors and discrete fibrations to the square with left side $J$, top $J^{\prime }$, bottom $G$ and right side $G^{\prime }$. It gives a unique $S:\mathcal{E}\to {\mathcal{E}}^{\prime }$ satisfying $SJ=J^{\prime }$ and $G^{\prime }S=G$. Reversing the roles gives $T:{\mathcal{E}}^{\prime }\to \mathcal{E}$.

Both $TS$ and $1_{\mathcal{E}}$ fill the same square for $J,G$, so uniqueness gives $TS=1_{\mathcal{E}}$; similarly $ST=1_{{\mathcal{E}}^{\prime }}$. This proves the uniqueness of a final-discrete-fibration factorization: the connecting functor is a canonical isomorphism, not just a noncanonical equivalence.

We prove the orthogonality of final functors and discrete fibrations. For $b\in \mathcal{B}$, finality supplies an object $(a,u:b\to Fa)$ of $(b\downarrow F)$. Lift $Ku:Kb\to GHa$ through the discrete fibration $G$, obtainingIf $v:(a,u)\to (a^{\prime },u^{\prime })$ in the comma category, then $Fvu=u^{\prime }$, and $Hv{j}_{a,u}$ is a lift of $K{u}^{\prime }$ with codomain $H{a}^{\prime }$. Uniqueness of that lift makes its domain equal to $c_{a^{\prime },u^{\prime }}$. Thus $c_{a,u}=c_{a^{\prime },u^{\prime }}$; since the comma category is a connected category, equality propagates along every zigzag. Define $Lb$ to be this common object.

For $w:b\to b^{\prime }$, choose $(a,u:b^{\prime }\to Fa)$. Lift $Kw$ into $L{b}^{\prime }$, obtaining $\ell :c\to L{b}^{\prime }$. Then $j_{a,u}\ell$ lifts $K(uw)$ into $Ha$, so its domain is $Lb$ by the preceding construction. Define $Lw=\ell$. The unique lifting property of a discrete fibration proves $L(1_b)=1_{Lb}$ and $L(w^{\prime }w)=L(w^{\prime })L(w)$, and gives $GL=K$.

At $b=Fa$, use $(a,1_{Fa})$; its lift is $1_{Ha}$, so $LFa=Ha$. Both $L(Fv)$ and $Hv$ are lifts of $GHv$ with the same codomain, hence agree. Therefore $LF=H$.

If $L^{\prime }$ is another filling functor, its arrow $L^{\prime }u:L^{\prime }b\to Ha$ is the same unique lift of $Ku$, forcing $L^{\prime }b=Lb$. Its arrows are likewise forced by their codomains and images under $G$. The diagonal functor exists and is unique.

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.

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:

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.

Two factorizations of one functor have a unique connecting functor in each direction, by the orthogonality of final functors and discrete fibrations. Their composites and the identity fill the same lifting square, so uniqueness makes both composites identities. The intermediate categories are therefore canonically isomorphic.