Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 119 / 3

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.

For an arbitrary functor $F:\mathcal{C}\to \mathcal{D}$, define a category $\mathcal{E}$ as follows. Its objects are pairs $(B,c)$, where $B\in \mathcal{D}$ and $c$ is a connected component of $(B\downarrow F)$. Precomposition by an arrow $g:B\to B^{\prime }$ definesThere is one morphism $(B,c)\to (B^{\prime },c^{\prime })$ over $g$ exactly when $g^{\ast }(c^{\prime })=c$. Functoriality of precomposition makes this a category, and projectionis a discrete fibration: given $g:B\to B^{\prime }=G(B^{\prime },c^{\prime })$, its unique lift with codomain $(B^{\prime },c^{\prime })$ has domain $(B,g^{\ast }{c}^{\prime })$.

Define $H:\mathcal{C}\to \mathcal{E}$ bywhere $[1_{FA}]$ is the component of the identity object in $(FA\downarrow F)$. For $u:A\to A^{\prime }$, use the unique arrow over $Fu$. It exists because $(Fu{)}^{\ast }[1_{F{A}^{\prime }}]$ contains the object $Fu:FA\to F{A}^{\prime }$, and this object is joined to $1_{FA}$ by the morphism $u$ in $(FA\downarrow F)$. Plainly $GH=F$.

For $(B,c)\in \mathcal{E}$, an object of $((B,c)\downarrow H)$ is exactly an arrow $\beta :B\to FA$ lying in the component $c$: the condition for an arrow $(B,c)\to H(A)$ is precisely ${\beta }^{\ast }[1_{FA}]=c$. Morphisms agree with those in $(B\downarrow F)$. Hencewhich is nonempty and connected by definition. Therefore $H$ is final. We have factored $F=GH$ as a final functor followed by a discrete fibration, giving the final-discrete-fibration factorization.