Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-119/1/solution

A category is well-powered when the isomorphism classes of monomorphisms into each object form a set. For , every subobject is represented by a subfunctor with . Since is small, all choices lie in the set , and naturality merely cuts out a subset. Thus the functor category is well-powered. Quotients are similarly represented by compatible equivalence relations on the sets , so they form a subset of ; hence it is well-copowered.
A cocone under the identity diagram consists of maps satisfying for every . A terminal object supplies the unique such cocone and has the required universal property. Conversely, if is a colimit of the identity, both and mediate its cocone to itself, so uniqueness gives . For any , cocone compatibility gives ; thus there is exactly one arrow , and is terminal.
For , let be the set of isomorphism classes of quotients of the representable . This is a set by well-copoweredness. A map sends a quotient of to the image quotient of the composite , making a functor. For any functor and , Yoneda gives ; factor it as an epimorphism followed by a monomorphism and send to the resulting quotient class. These maps agree along every monomorphism. Conversely, a cone with apex assigns to a quotient the element obtained by applying its leg at to . Yoneda and epi-mono factorization show that this is well-defined and is the unique map . Therefore is a local state classifier.
An object of is a finite set with a permutation. For each , let with trivial action and map a finite -set to by sending every point to the length of its orbit modulo . Equivariant injections preserve orbit lengths, so these maps form a cocone under the monomorphism subcategory; varying cyclic orbits shows that its legs are collectively surjective. If a local state classifier existed, its universal map onto every would be surjective because the universal legs are jointly epic. This would force the finite set to have at least elements for every , a contradiction.
Solved by gpt-5.6-sol high.

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