Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 25 / 3

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.

For each $d\in \mathcal{D}$, let $P(d)={\pi }_0(d\downarrow F)$, the collection of connected components, where connectedness means connection by a zigzag of arrows. Precomposition with $u:d\to d^{\prime }$ gives a function $P(u):P(d^{\prime })\to P(d)$. Identity and composition are inherited from precomposition.

Construct a category $\mathcal{E}$ whose objects are pairs $(d,c)$ with $c\in P(d)$. An arrow $(d,c)\to (d^{\prime },c^{\prime })$ is an arrow $u:d\to d^{\prime }$ satisfying $c=P(u)c^{\prime }$. Composition is that of $\mathcal{D}$, and the component equation is preserved by composition. The projection $p:\mathcal{E}\to \mathcal{D}$ is a discrete fibration: an arrow $u:d\to d^{\prime }$ has precisely one lift into $(d^{\prime },c^{\prime })$, namely the arrow from $(d,P(u)c^{\prime })$ represented by $u$.

Define $J:\mathcal{C}\to \mathcal{E}$ byThis is well defined because $(a,1_{Fa})$ and $(a^{\prime },Fj)$ lie in the same component of $(Fa\downarrow F)$ whenever $j:a\to a^{\prime }$. Clearly $pJ=F$.

For an object $(d,c)$ of $\mathcal{E}$, an object of $((d,c)\downarrow J)$ is exactly a pair $(a,u:d\to Fa)$ representing the component $c$. Its morphisms are exactly the arrows in $(d\downarrow F)$ between such objects. Thus this comma category is the component $c$ itself, which is nonempty and connected. This proves that $J$ is a final functor, and henceThis is the final-discrete-fibration factorization. For small categories the collections $P(d)$ are sets and this construction is the category of elements of the presheaf $P$. For categories of a larger size, the construction is interpreted in the same chosen ambient size framework.

Suppose $F=pJ=p^{\prime }{J}^{\prime }$ are two such factorizations, with $J,J^{\prime }$ final and $p,p^{\prime }$ discrete fibrations. Apply the unique diagonal construction to the square with left side $J$, top $J^{\prime }$, bottom $p$ and right side $p^{\prime }$. It gives a unique functor $L:\mathcal{E}\to {\mathcal{E}}^{\prime }$ such that $LJ=J^{\prime }$ and $p^{\prime }L=p$. Reversing the two factorizations gives $M:{\mathcal{E}}^{\prime }\to \mathcal{E}$ with $M{J}^{\prime }=J$ and $pM=p^{\prime }$.

Both $ML$ and $1_{\mathcal{E}}$ fill the square with left side $J$, top $J$, bottom $p$ and right side $p$. Uniqueness from part (i) gives $ML=1_{\mathcal{E}}$. Similarly $LM=1_{{\mathcal{E}}^{\prime }}$. ThusIn particular the conclusion is a canonical isomorphism of intermediate categories, not just an unspecified equivalence.