Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 23 / 3

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.

Use the connected-component presheaf of a functor. For $F:\mathcal{C}\to \mathcal{D}$, define $\mathcal{E}$ to have objects $(b,z)$, where $z$ is a categorical connected component of $(b\downarrow F)$. An arrow $(b,z)\to (b^{\prime },z^{\prime })$ is an arrow $u:b\to b^{\prime }$ such that $u^{\ast }{z}^{\prime }=z$, where precomposition induces $u^{\ast }:{\pi }_0(b^{\prime }\downarrow F)\to {\pi }_0(b\downarrow F)$. Identity and composition follow from precomposition, so this is a category.

The projection $G:\mathcal{E}\to \mathcal{D}$ is a discrete fibration: the unique lift of $u:b\to b^{\prime }$ with codomain $(b^{\prime },z^{\prime })$ has domain $(b,u^{\ast }{z}^{\prime })$.

Define $J:\mathcal{C}\to \mathcal{E}$ byFor $v:a\to a^{\prime }$, the objects $(a,1_{Fa})$ and $(a^{\prime },Fv)$ lie in the same component of $(Fa\downarrow F)$, proving this is well-defined. Clearly $GJ=F$.

For $(b,z)\in \mathcal{E}$, the objects and arrows of $((b,z)\downarrow J)$ are exactly the objects and arrows of the component $z$ of $(b\downarrow F)$. Indeed, an arrow $(b,z)\to Ja$ is $u:b\to Fa$ with $[(a,u)]=z$, and its commuting triangles are precisely those in the original comma category. This category is nonempty and connected. Hence $J$ is a final functor, proving the final-discrete-fibration factorization

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.