Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 17 1 iii Solution Created 2026-10-03 Updated 2026-10-14
Define maps between the hom-sets byUsing naturality of the proposed counit of an adjunction and then the first triangle identity for an adjunction,Using naturality of the proposed unit of an adjunction and then the other triangle identity for an adjunction,Hence and are inverse bijections. They are natural in both variables: for and , the unit of an adjunction identity givesThis constructs an adjunction . Taking and recovers the prescribed unit of an adjunction and counit of an adjunction. Conversely, part (i) shows that any adjunction with those components must have exactly these transpose formulas. The adjunction exists and is unique with the prescribed unit and counit.
Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 17 1 i Solution Created 2026-10-03 Updated 2026-10-14
Write the adjunction as the natural bijectionThe unit of an adjunction and counit of an adjunction are the natural transformations whose components areNaturality of the hom-set bijections gives, for and ,For example, postcomposition by in the first hom-set corresponds to postcomposition by in the second, giving the first formula; naturality in gives the second. The same naturality says that, for and ,Thus these components do define the asserted natural transformations and .
Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 17 9 ii Solution Created 2026-10-03 Updated 2026-10-14
For with unit of an adjunction and counit of an adjunction , putThe triangle identities for an adjunction giveFor associativity, naturality of at saysApply to obtain . Thus this is the monad induced by an adjunction.
Conversely, start with a monad . The free algebra functor sends to and to . The monad laws ensure that these are algebras for a monad and morphisms of algebras for a monad. For an algebra , there are inverse bijectionsThe inverse really is an algebra map becauseFor a free-algebra map , its algebra equation and the monad unit law giveFor a base map , naturality and the algebra unit law give . The formulas are natural, establishing the free-forgetful Eilenberg-Moore adjunction . Its unit is , and its counit at is . Hence its induced multiplication is , and every monad arises from an adjunction.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 18 5 Solution Created 2026-10-03 Updated 2026-10-06
For , the comma category has objects with and . A morphism is satisfying . Identities and composition are those of , and functoriality of verifies the condition under composition.
The Freyd general adjoint functor theorem states: if is a locally small category with all small categorical limits, and is locally small, then has a left adjoint if and only if it preserves small limits and satisfies the solution-set condition. The latter means that for each there is a set-indexed family such that every equals for some and some .
For necessity, use the standard result that a right adjoint preserves categorical limits. If , the singleton family containing the unit of an adjunction is a solution set, since transposition gives for a unique .
For sufficiency, use the following standard limit fact: if is complete and preserves limits, the projection creates small limits. Indeed, a compatible family induces a unique arrow into , and this makes the underlying limit a limit in the comma category. The comma category is locally small because each of its hom-sets is a subset of a hom-set of . Its solution family is a weakly initial set.
We prove the remaining initial-object lemma for complete categories with a weakly initial set. In any locally small category with all small limits and a weakly initial set , form . It is weakly initial: for any , some exists and may be composed with the projection . The empty family cannot be weakly initial in a nonempty complete category, which has a terminal object.
The set is small. Form a simultaneous equalizer of every endomorphism of and ; thusThis equalizer exists by completeness, for example as the equalizer of two maps . The object is still weakly initial, since it maps to .
Given , take their equalizer . Weak initiality of gives . Since is an endomorphism of , we have , and cancellation of the monomorphism gives . Thus is a split epimorphism as well as a monomorphism, so it is an isomorphism. From follows . There is at least one map by weak initiality, so is initial.
Apply this lemma to every and choose its initial object . For , initiality gives the unique satisfying . Uniqueness proves the functor laws. The same initiality gives natural bijectionsHence , completing the theorem without invoking another adjoint functor theorem.