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Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 18 8 b Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-07
In an abelian category, the image factorization in an abelian category of iswhere the abelian-category axiom identifies coimage with image. Thus is epic and is monic. Any other epi-mono factorization has and . Since epimorphisms are cokernels of their kernels, its middle object is canonically isomorphic to , uniquely compatibly with the two factors.
For a square , define byThe first arrow exists because factors through , so annihilates . Its composite with equals after the epimorphism , proving the second equation. Uniqueness after proves preservation of identities and composition. This gives the functoriality of abelian image factorization as a functor from the arrow category.
For pullback stability of abelian image factorization, state the standard facts that pullbacks preserve monomorphisms, epimorphisms in an abelian category are stable under pullback, and two adjoining pullback squares have pullback outer rectangle. In the given diagram, is therefore monic and is epic, while the composite is the pullback of . Its epi-mono factorization is an image factorization by the uniqueness just proved. Thus the top row is the image factorization of the pulled-back arrow, with its middle object canonically the pullback of the original image subobject.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 18 8 a Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-07
Use the kernel squares in an abelian category argument. If is monic and , satisfy , then , so . The kernel in a category property gives a unique with . The equation and monicity of give . This proves the left square is a pullback in a category.
Now suppose the right square is a pullback, without imposing the earlier monicity hypothesis on . The pair gives a unique with and . Factor through the kernel. Then , so . Also and have the same two pullback projections, hence . Thus and . Therefore is an isomorphism. These arguments use only the relevant kernels, zero arrows and pullback properties; the abelian hypothesis supplies them.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 18 6 b Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-07
We prove full faithfulness from counit coequalizers. Let and let be an monad algebra morphism. ThusSet . Its composites with the two arrows of the printed presentation are equal: naturality of givesIn the last equality we used naturality at . The coequalizer property therefore gives a unique satisfying .
Apply to that identity. The algebra-morphism equation gives . The triangle identity makes a split epimorphism with section , so . This proves fullness of .
If have , naturality gives . The counit is epic since it is a coequalizer, so . Thus is a faithful functor. Both parallel arrows matter: the converted TeX loses the second one, , which is visible in the original PDF.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 18 6 a Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-07
For the monad , an algebra for a monad is with and . A morphism of algebras for a monad satisfies . These objects and arrows form the Eilenberg-Moore category .
The free algebra functor sends to and to . The monad identities verify the algebra laws. For the forgetful functor , the free adjunction iswith inverse . The algebra law makes the latter an monad algebra morphism. The identities and prove the bijection.
In the category of adjunctions inducing a fixed monad, objects are adjunctions with their induced monad identified with . A morphism to is a functor between the right-hand categories satisfying , and compatibility with units and counits. With these strict identifications, defineThe triangle identities give the unit algebra law; naturality of at gives the multiplication law. Naturality at makes an monad algebra morphism. Moreover , because , and is the free-adjunction counit at . Hence is a morphism into the Eilenberg-Moore adjunction.
For any other such , its underlying object at must be . Counit compatibility forces its algebra action to be , and the forgetful functor, which is a faithful functor, forces . Thus . This proves terminality of the Eilenberg-Moore adjunction. If adjunctions are specified only up to coherent isomorphisms, the same argument gives uniqueness up to the corresponding compatible natural isomorphism.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 18 5 d Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-07
If is representable, it is a limit-preserving functor, and its universal element gives an initial object of , hence a weakly initial singleton.
Conversely, a limit-preserving makes a complete category. For a small diagram , take in . Its distinguished elements form a compatible family in . Categorical limit preservation gives a unique with . The underlying categorical limit factorization of a categorical cone preserves this element, proving the comma-category universal property. For the empty diagram, this uses . This is the construction of limits in a comma category of a limit-preserving functor.
The comma category is locally small since its arrows are subsets of the hom-sets in . By the previous part, its weakly initial set therefore yields an initial object. Part (b) then yields a representation of . Thus representability from a solution set follows with all small categorical limits, rather than finite categorical limits alone.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 18 5 c Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-07
An initial object by itself is a weakly initial set. For the converse, let be a small weakly initial family and form its product in a category . Given , choose an arrow ; its composite with the projection shows that is weakly initial.
Local smallness makes a set, so completeness supplies the simultaneous equalizer of all endomorphisms of with . Thus for every . The object is weakly initial because it maps to .
For parallel arrows , take their equalizer . Weak initiality of supplies . The endomorphism of satisfies , and monicity of gives . Therefore is both a monomorphism and a split epimorphism, hence an isomorphism. Since , we obtain . There is already at least one arrow from to every , so is initial. This proves the initial-object lemma for complete categories with a weakly initial set, with smallness used exactly where the endomorphisms are equalized.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 18 5 b Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-07
A representation of a functor for says precisely that for every object of there is a unique arrow with . This is exactly the initial object property for in that comma category. Conversely, an initial object supplies these unique arrows, hence the representing bijections .
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 18 5 a Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-07
The comma category has objects with and a function . A morphism is an arrow such that . Identities and composition come from . When , a map selects an element of , so this is the covariant category of elements of .
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 18 4 c Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-07
The currying adjunction for small categories uses the bijectionIt sends to the functor , with an arrow inducing the natural transformation whose -component is . Conversely, for , define its uncurried functor by andNaturality of allows the two factors to be interchanged in the appropriate order, giving functoriality. The constructions are inverse and natural in and . Hence is a left adjoint to on the category of small categories.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 18 4 b Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-07
Define the correspondence from the given natural transformations by , with candidate inverse . For , naturality of and a triangle identity for an adjunction giveFor , naturality of and the other triangle identity giveThese are inverse bijections. Naturality of , , and makes the bijections natural in and , so they define an adjunction with the required unit and counit. Uniqueness follows from the formulas in the previous part: any adjunction with that unit and counit must have exactly these transposition maps.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 18 4 a Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-07
An adjunction is a family of bijectionsnatural in both variables. Its adjunction unit and adjunction counit are and . Naturality of the correspondence givesFor , naturality in both variables evaluates in two ways, giving . Thus is a natural transformation; the dual calculation gives naturality of .
Applying the inverse correspondence to and the correspondence to yields the triangle identities for an adjunction:They express that transposing an identity morphism and transposing back returns that identity.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 18 3 b Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-07
Construct pointwise limits in a functor category. For , choose at each object a categorical limit of , with projections . For , the family is compatible; define uniquely byThe universal property gives and , since those equalities hold after every projection. Thus is a functor, and each is a natural transformation.
For any categorical cone , the pointwise categorical limits give unique maps . To check naturality, compose and with every ; both become . The projections distinguish arrows into their categorical limit, so the two maps agree. Componentwise uniqueness gives uniqueness of the natural transformation . This proves that really is the required categorical limit, rather than merely a family of objectwise candidates.
A specified categorical limit categorical cone in fixes these objectwise vertices and projections. The displayed equation forces every arrow , and the argument forces every categorical cone factorization. Hence the forgetful functor uniquely lifts that categorical cone and is a limit-creating functor. The construction only takes small categorical limits in ; it does not require to be small. As usual, the functor categories are understood in a universe where their collections of transformations are meaningful.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 18 3 a Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-07
For a diagram , a limit-preserving functor carries every limiting cone over a diagram to a limiting categorical cone. A limit-reflecting functor has the converse property: a categorical cone is limiting whenever its image is limiting. A limit-creating functor uniquely lifts every specified limiting categorical cone over the image diagram to a categorical cone over , and the lift is limiting. These definitions concern diagrams of the stipulated shape; creation includes the lifting requirement, not merely reflection.
Let be a categorical limit of . The mapis a bijection: the right side consists exactly of compatible families of arrows from , and the categorical limit's universal property gives their unique factorization through . The bijection is induced by the categorical limit projections, so it proves that covariant representables preserve limits, including the empty diagram and its terminal object.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 18 2 b Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-07
A split coequalizer consists of , , and with , , and . If , then . Thus factors through , and the factor is unique because has the right inverse . This proves the coequalizer property directly. Every functor preserves the diagram, since all these equations are preserved.
For idempotent splitting through a coequalizer, first suppose with . Then . Any satisfying factors as , uniquely since is a split epimorphism. Hence coequalizes .
Conversely, let coequalize . Since , the arrow itself equalizes the pair, so there is a unique with . Then , and every coequalizer is an epimorphism, so . Thus is a splitting of an idempotent morphism.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 18 2 a Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-07
A monomorphism satisfies , and an epimorphism satisfies . A strong monomorphism is a monomorphism with the right lifting property against all epimorphisms: every square has a diagonal satisfying and . The diagonal is unique by monicity. A regular monomorphism is an equalizer of some pair . It is monic because two equalizer factorizations of the same arrow must agree.
The converted TeX omits the remainder of this subpart. For the printed strict monomorphism condition, an arrow is admissible when, for every pair out of , the implication holds; strictness says each such factors uniquely through . If equalizes , every admissible satisfies , and the equalizer property supplies its unique factorization. Hence every regular monomorphism is strict.
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