Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 25 8 a Solution Created 2026-10-03 Updated 2026-10-07
Take any morphism and let be its categorical cokernel. Let be the categorical kernel of . Since , kernel universality gives with . The map is a monomorphism. We must establish the pseudo-epimorphism property of , rather than merely name an image.
Suppose satisfies . Let be its categorical kernel. Then for some . The composite is a monomorphism, so the hypothesis that every monomorphism is normal provides a map of which is a categorical kernel, up to its unique compatible isomorphism. Now . The categorical cokernel property of gives , and therefore .
Since is the categorical kernel of , the map factors through it: for some . Cancel the monomorphism to get . As is also a monomorphism, , so is an isomorphism. Finally , proving that is a pseudo-epimorphism.
ThusThe pseudo-epimorphism factorization through the kernel of a cokernel proof uses a pointed category, existence of categorical kernels and categorical cokernels, and normality of every monic map. It does not silently replace pseudo-epic cancellation against zero by the stronger epimorphism condition.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 25 8 b Solution Created 2026-10-03 Updated 2026-10-07
In an abelian category, for set and . The canonical map from coimage to image is an isomorphism. Thus the image factorization in an abelian category isunique up to a unique compatible isomorphism.
The short five lemma states that in a commuting diagram of short exact sequences and in an abelian category, if the vertical maps and are isomorphisms, then so is .
For the Five lemma, consider commuting exact rowswith vertical maps . We prove the usual stronger version: epic, and invertible, and monic imply that is invertible. In particular the conclusion holds if all four outer vertical maps are isomorphisms.
Put and , and define similarly. Exactness gives short exact sequencesThe functoriality of abelian image factorization supplies their vertical outer maps. We show that both are isomorphisms.
First, and : exactness at identifies with , and the image-coimage isomorphism for gives the claimed cokernel. Write and for these categorical cokernels. To construct an inverse to the induced map , observeCancellation of the epimorphism gives , so factors uniquely as with . From , composing with the epic gives and .
Second, and by exactness at the fourth objects. Write their inclusions as and . The equality gives the induced map . Sinceand is a monomorphism, . Kernel universality gives with . Cancelling the monic proves and .
Apply the short five lemma to the two short exact sequences: their outer maps are invertible, so is an isomorphism. This five lemma via image factorization argument uses only universal properties and therefore works in any abelian category, without treating its objects as literal sets of elements.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 18 2 a Solution 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.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 18 5 c Solution 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 8 b Solution 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 2014 iii Paper 18 1 a Solution Created 2026-10-03 Updated 2026-10-06
A monomorphism is a morphism such that implies for every pair . Dually, an epimorphism satisfies . A regular epimorphism is a coequalizer of some parallel pair. An isomorphism has a two-sided inverse.
For a product in a category, write for its categorical diagonal. Since , equality gives . Thus the diagonal is a monomorphism, indeed a split monomorphism.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 18 1 b Solution Created 2026-10-03 Updated 2026-10-06
Here use the left lifting property against monomorphisms as the definition of “strong”; epimorphicity will be established separately. In a lifting square write , , and , with and a monomorphism. Any two lifts agree because .
First let be the coequalizer of . Every coequalizer is an epimorphism: if , the uniqueness clause for the coequalizer applied to this common composite gives . Moreover,so by monomorphism cancellation. The coequalizer therefore supplies with . Then , and epimorphism cancellation gives . This proves that regular epimorphisms are strong epimorphisms.
If is also a monomorphism, take , and . Its lift satisfies and . Hence monic lifting-only strong morphisms are invertible.
Next suppose has the left lifting property against monomorphisms. Given a lifting square for , with , precompose its top arrow with . A lift for gives with and . Crucially, one does not cancel : instead , and the monomorphism gives . Thus the right factor of a strong composite is strong, proving right-factor cancellation for lifting-only strong morphisms.
Finally, in with strong and monic, the preceding result makes strong. The monic-strong argument then makes an isomorphism. None of these arguments assumed that a lifting-only strong morphism was already epic.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 18 1 c Solution Created 2026-10-03 Updated 2026-10-06
Let have the left lifting property against monomorphisms, and suppose satisfy . Use the categorical diagonal , which is a monomorphism by part (a). The square with top arrow , bottom arrow , left arrow and right arrow commutes, since both product components are .
Its lift satisfies . Applying the two product in a category projections gives and . Thus is an epimorphism. This binary-product criterion for lifting-only strong epimorphisms requires binary products, rather than any assumption about equalizers.
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.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 20 1 i Solution Created 2026-10-03 Updated 2026-10-06
Write the comonad as and its category of coalgebras for a comonad as . A coalgebra for a comonad is a map with and . A morphism satisfies . Let be the forgetful functor and let be the cofree coalgebra. The adjunction has the explicit correspondenceWe construct the three pieces of the elementary topos structure.
Because preserves finite limits, each underlying finite limiting cone has a unique coalgebra structure induced by the structures on its vertices. The counit and coassociativity equations can be checked after its jointly monic projections. Thus creates finite limits and reflects isomorphisms. A morphism of coalgebras is monic exactly when its underlying morphism is monic, by the diagonal criterion using the created pullback.
For exponentials in a coalgebra topos, fix coalgebras and and put in . On the cofree coalgebra there is an underlying evaluationThe two mapsuse in the second expression. Transpose them in to maps , and then transpose across to coalgebra morphisms . Take their equalizer in .
An underlying map corresponds to a coalgebra map . The equation saying that the original map is a coalgebra morphism is precisely , since for the structure . By the cofree adjunction, this is equivalent to , hence to unique factorization through . Thereforenaturally in . This constructs the required exponential object.
For the subobject classifier of a coalgebra topos, let be the underlying subobject classifier, and let classify the mono . Its cofree transpose is the coalgebra endomorphismDefine as the equalizer of and . The transpose of factors through this equalizer and gives .
Indeed, for a subobject classified by , the pullback has characteristic map . It is always contained in , by naturality of the counit. Equality holds exactly when restricts to a coalgebra structure on ; its axioms then follow by composing with the monomorphisms and . Under , the classifying map becomes . The equality of subobjects is . Transposing this equality gives , so exactly the coalgebra subobjects correspond to maps . Their pullback of is the desired subcoalgebra, and uniqueness follows from uniqueness of .
Thus we have finite limits, exponentials and a subobject classifier:The construction does not assume that preserves the underlying exponentials or underlying subobject classifier.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 22 4 b Solution Created 2026-10-03 Updated 2026-10-06
The Special adjoint functor theorem says: if is a locally small category, a complete category, a well-powered category, and has a small cogenerating family, then a functor into any locally small category has a left adjoint if and only if it preserves small categorical limits.
A small cogenerating family , with a set, distinguishes unequal parallel morphisms by postcomposition: for , some has . Being well-powered means that the subobjects of each object form a set up to the usual equivalence of monomorphisms. These hypotheses make the solution-set condition automatic.
The dual formulation uses a cocomplete category, being a well-copowered category and having a small generating family, and concludes that every small-colimit-preserving functor has a right adjoint.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 22 4 c Solution Created 2026-10-03 Updated 2026-10-06
It is enough to prove the solution-set condition and then apply the general adjoint functor theorem. Fix and a morphism . We will factor it through one object in a set depending only on .
First construct a minimal supported subobject. Among the subobjects for which for some , include and take their intersection . This is a small intersection, by well-poweredness. It exists by completeness as the categorical limit of the diagram consisting of these monomorphisms into . Its map to is a monomorphism: two maps with the same composite to have equal projections to every , since each is monic, and are then equal by the categorical limit property. One can equivalently construct these intersections by pullbacks in a category and small products in a category, using stability of monomorphisms under pullback in a category.
Choose the factorizations . They form a compatible categorical cone into the image diagram under , all with common composite to . Preservation of small categorical limits yields with . If is another subobject through which factors after applying , then is among the original supported subobjects. The intersection property gives with . Since is monic, ; since is monic, also . Thus every supported subobject of is invertible.
For each member of the small cogenerating family, considerThis map is injective. If , preservation of the equalizer of makes factor through the image of that equalizer. Minimality makes its inclusion an isomorphism, so .
Write ; it is a set by local smallness of . Let be the image of . For each , there is exactly one corresponding . These maps define the evaluation embedding into cogenerator productsIt is a monomorphism: if and , cogeneration supplies some distinguishing ; that is one of the projections of , a contradiction. The product is small. It is important to use the subfamilies , since some missing coordinate in need not correspond to a morphism out of .
There are only a set of possible families . For each such family form , choose a set of representatives of its subobjects, and take all pairs . Local smallness of and well-poweredness make their union a set. In the case just constructed, identifies with one chosen representative . Transporting to and composing the inverse identification with gives a factorization of the original through that pair. Therefore these pairs form a weakly initial set in .
The general adjoint functor theorem now supplies a left adjoint to . Conversely, a right adjoint preserves small categorical limits, either by the adjunction hom-set bijections and the hom-set detection of categorical limits, or directly from their universal properties. Hence the stated special theorem is proved in both directions.