Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 119 1 a Solution Created 2026-10-03 Updated 2026-10-05
Write . The covariant Yoneda lemma gives the natural bijectionwhose inverse sends to the natural transformation with component . This also identifies the two variables' naturality.
To prove that is a projective object in a category, let be an epimorphism in the functor category and . By pointwise epimorphism in a functor category, lift through the surjective function . The Yoneda lemma turns this lift into with .
For a small category, the set of pairs with indexes the canonical mapIt is natural, and every is the image of . Thus it is an epimorphism, establishing the projective cover of a set-valued functor by representables construction. Here cover means a projective presentation; no minimality is asserted.
For the equivalence's first step, if is a monomorphism, then , the map , is injective for every . Hence every representable functor is a monofunctor. Conversely, injectivity of every is exactly left cancellation against every , so it implies that is a monomorphism. Therefore
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 119 1 b Solution Created 2026-10-03 Updated 2026-10-05
Assume (b). A coproduct in a category of monofunctors in is again a monofunctor: at an object it is the disjoint union of the values, and every arrow acts within the same summand by an injective function. Distinct summands cannot become identified, and injectivity within a summand follows from its monofunctor property.
Apply this to the coproduct of representable functors constructed in part (a). Its canonical map to is a pointwise epimorphism in a functor category. Thus every is an epimorphic image of a monofunctor, proving
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 119 1 c Solution Created 2026-10-03 Updated 2026-10-05
Assume (c), and take an epimorphism with a monofunctor. Since is a projective object in a category, there is a natural transformation with . A retract in a category of a monofunctor is a monofunctor: if , then naturality of gives , injectivity of gives , and applying gives . Thus (c) implies (b), completing
The stronger assertion that every set-valued functor is a monofunctor holds exactly when is a groupoid. Sufficiency follows because a functor preserves inverses, so sends every arrow to a bijection.
For necessity, fix , and form the pointwise pushout in a categoryusing twice the natural transformation given by precomposition with . At an object , consists of two copies of , with the two copies of identified for every . No different underlying elements are identified: each generating relation simply joins the two copies of one element. In , the two copies of coincide, since . Hence sends the two copies of in to the same element. If is a monofunctor, those copies of coincide. The description of the pushout implies for some . Thus every morphism is a split monomorphism.
Apply the same conclusion to : there is with . Then , and so as well. Every arrow is invertible, proving
Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 119 1 Solution 2026-09-28
For a locally small category , an object , and a functor , the covariant Yoneda lemma is the natural bijectionIts inverse sends to the natural transformation whose component at maps to .
Suppose now that is a small category. For , form the coproduct in a categoryThe Yoneda lemma associates to every summand the natural transformation determined by , and these transformations combine to a map . At an object , the element is the image of in the summand indexed by , so is a pointwise epimorphism in a functor category. Each representable functor is a projective object in a category, sinceand evaluation preserves pointwise epimorphisms. A coproduct of projectives is projective, so is the required projective object. This is the projective cover of a set-valued functor by representables.
We next prove the three equivalent conditions. If every morphism of is a monomorphism, then for and every , postcompositionis injective. Thus every covariant representable functor is a monofunctor. Conversely, taking shows that injectivity for every representable implies that forces , so every is monic.
If all representables are monofunctors, the object above is a monofunctor because a coproduct of injective functions is injective. Hence every is an epimorphic image of a monofunctor. Conversely, suppose every functor is an epimorphic image of a monofunctor and apply this to a representable . Choose an epimorphism with a monofunctor. Since is projective, lifts to with . Thus is a retract in a category of . Every retract of a monofunctor is a monofunctor: if , then injectivity of applied to and gives , and applying gives . This completes the equivalence.
Finally, every functor is a monofunctor exactly when every morphism of is a split monomorphism. The forward implication is immediate because every functor preserves a left inverse. For the converse, fix and form a quotient of by identifying the distinguished point with every arrow of the form , where . In the resulting functor , the two elements and of have equal images under because . If every functor is a monofunctor, is injective, so . By construction this means for some . Thus is split monic. Equivalently, every morphism of must be an absolute monomorphism.