Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 119 (Category Theory)

A category is well-powered when the isomorphism classes of monomorphisms into each object form a set. For $P\in [\mathcal{C},\mathbf{Set}]$, every subobject is represented by a subfunctor $S$ with $S(A)\subseteq P(A)$. Since $\mathcal{C}$ is small, all choices lie in the set ${\prod }_A\mathcal{P}(P(A))$, 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 $P(A)$, so they form a subset of ${\prod }_A\mathcal{P}(P(A{)}^2)$; hence it is well-copowered.

A cocone under the identity diagram consists of maps $c_A:A\to L$ satisfying $c_{B}f=c_A$ for every $f:A\to B$. A terminal object supplies the unique such cocone and has the required universal property. Conversely, if $(L,c_A)$ is a colimit of the identity, both $1_L$ and $c_L$ mediate its cocone to itself, so uniqueness gives $c_L=1_L$. For any $f:A\to L$, cocone compatibility gives $f=c_{L}f=c_A$; thus there is exactly one arrow $A\to L$, and $L$ is terminal.

For $A\in \mathcal{C}$, let $\Omega (A)$ be the set of isomorphism classes of quotients of the representable $\mathcal{C}(A,-)$. This is a set by well-copoweredness. A map $u:A\to B$ sends a quotient of $\mathcal{C}(A,-)$ to the image quotient of the composite $\mathcal{C}(B,-)\to \mathcal{C}(A,-)$, making $\Omega$ a functor. For any functor $P$ and $x\in P(A)$, Yoneda gives $\mathcal{C}(A,-)\to P$; factor it as an epimorphism followed by a monomorphism and send $x$ to the resulting quotient class. These maps $P\to \Omega$ agree along every monomorphism. Conversely, a cone with apex $Q$ assigns to a quotient $q:\mathcal{C}(A,-)\twoheadrightarrow R$ the element obtained by applying its leg at $R$ to $q_A(1_A)$. Yoneda and epi-mono factorization show that this is well-defined and is the unique map $\Omega \to Q$. Therefore $\Omega$ is a local state classifier.

An object of $[\mathbb{Z},{\mathbf{Set}}_f]$ is a finite set with a permutation. For each $n>0$, let $A_n=\mathbb{Z}/n$ with trivial action and map a finite $\mathbb{Z}$-set $P$ to $A_n$ by sending every point to the length of its orbit modulo $n$. 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 $L$ existed, its universal map onto every $A_n$ would be surjective because the universal legs are jointly epic. This would force the finite set $L$ to have at least $n$ elements for every $n$, a contradiction.

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An idempotent morphism satisfies $e^2=e$ and splits if $e=ir$ with $ri=1$. Form the relative Karoubi envelope $\mathcal{C}[\check{E}]$: its objects are $(A,e)$ for $e\in E$, andThe functor $A\mapsto (A,1_A)$ is full and faithful. Each $e\in E$ splits through $(A,e)$ using $e$ in both directions. If a functor $F:\mathcal{C}\to \mathcal{D}$ comes with splittings $F(e)=i_e{r}_e$, define its extension by $(A,e)\mapsto$ the splitting object and $u\mapsto r_{f}F(u)i_e$. This gives the required factorization, unique up to the unique compatible natural isomorphism.

A preorder $R$ is reflexive and transitive, so $R\circ R=R$: transitivity gives one inclusion and reflexivity the other. If it is an equivalence relation, the quotient relation $q:X\to X/R$ satisfies $q^{\mathsf{op}}q=R$ and $q{q}^{\mathsf{op}}=1$, so it splits in category of relations $\mathbf{Rel}$. Conversely, writing a split preorder as $R=ir$, $ri=1$ and using reflexivity and transitivity shows $R=R^{\mathsf{op}}$; hence it is an equivalence relation.

Let $B$ be the set of join-inaccessible elements of a completely algebraic lattice $A$, and putThen $ri=1_A$, while $r(S)\le a$ exactly when $S\subseteq i(a)$, so $r⊣i$. Join-inaccessibility gives $i({\bigvee }_j{a}_j)={\bigcup }_{j}i(a_j)$, so $i$ preserves all joins. Conversely, given such $r⊣i$ with $ri=1$, each $r\{b\}$ is join-inaccessible: if $r\{b\}\le \bigvee S$, adjunction and preservation of joins put $b$ in some $i(s)$, whence $r\{b\}\le s$. Finally,so $A$ is completely algebraic.

Under the supplied full embedding $\mathbf{Rel}\to \mathbf{CSLat}$, a set goes to its power-set lattice and a preorder goes to the associated idempotent. Splitting those idempotents gives exactly the retracts of power sets by adjoint join maps characterized above. Therefore $\mathbf{Rel}[\check{E}]$ is equivalent to the full subcategory of complete join-semilattices consisting of completely algebraic lattices.

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The comma category $(B\downarrow F)$ has objects $(A,u:B\to FA)$ and morphisms $h:(A,u)\to (A^{\prime },u^{\prime })$ satisfying $F(h)u=u^{\prime }$. An initial object $(LB,{\eta }_B)$ is precisely a universal arrow from $B$ to $F$. Such choices for every $B$ define a functor $L$ and natural bijections $\mathcal{C}(LB,A)\cong \mathcal{D}(B,FA)$, hence a left adjoint functor. The unit of an existing adjunction supplies the initial objects in the reverse direction.

Suppose $F:\mathcal{C}\to \mathcal{D}$ is final and ${\alpha }_A:D(FA)\to X$ is a cocone. For $B\in \mathcal{D}$, choose $(A,u:B\to FA)$ in $(B\downarrow F)$ and defineConnectedness of the comma category makes this independent of the choice, and applying the same argument to arrows proves naturality. Any extension must have this value, so it is unique. Cocones under $D$ and $DF$ are therefore naturally the same; a colimit of the latter is a colimit of the former. Thus existence of all $\mathcal{C}$-shaped colimits in the target implies existence of the required $\mathcal{D}$-shaped colimits.

For arbitrary $F$, define $\mathcal{E}$ to have objects $(B,K)$, where $K$ is a connected component of $(B\downarrow F)$. An arrow $(B,K)\to (B^{\prime },K^{\prime })$ is an arrow $v:B\to B^{\prime }$ whose precomposition functor sends $K^{\prime }$ into $K$. Let $H(B,K)=B$, and let $G(A)=(FA,K_A)$ where $K_A$ contains $(A,1_{FA})$. Then $F=HG$. Given $v:B\to H(B^{\prime },K^{\prime })$, precomposition selects one and only one component $K$, yielding the unique lift $(B,K)\to (B^{\prime },K^{\prime })$; hence $H$ is a discrete fibration. Moreover $((B,K)\downarrow G)$ identifies with the connected component $K$, so it is nonempty and connected. Thus $G$ is a final functor.

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For an adjunction $F⊣U:\mathcal{D}\to \mathcal{C}$ with monad $T=UF$, the comparison functor sends $D$ to the $T$-algebra $(UD,U{\epsilon }_D)$. The adjunction is monadic when this comparison is an equivalence. The comparison-left-adjoint lemma says that if $\mathcal{D}$ has coequalizers of the reflexive pairs used to present $T$-algebras and $U$ preserves them, then the comparison has a left adjoint. Applying the unit and counit criteria yields the Beck monadicity theorem: $U$ is monadic exactly when it reflects isomorphisms and creates coequalizers of all parallel pairs whose images under $U$ admit split coequalizers.

Let $A_n\subseteq A$ be the fixed-point set of ${\alpha }_n$ and identify $A$ with $A\times \{0\}$. Define $F_{n}A$ to have underlying setKeep the old operations on the first summand, put ${\alpha }_{n+1}(a,0)=(a,1)$ for $a\in A_n$, and put ${\alpha }_1(a,m)=(a,m+1)$ for $m>0$; the remaining higher operations on these new points are undefined. Given $A\to U_{n}B$, the only possible extension sends $(a,m)$ along the iterates of ${\alpha }_1$ beginning at ${\alpha }_{n+1}(a)$. This proves $F_n⊣U_n$.

The forgetful functor $U_n:{\mathcal{C}}_{n+1}\to {\mathcal{C}}_n$ reflects isomorphisms. A $U_n$-split coequalizer carries a unique descended partial operation: splitness prevents any new fixed point of ${\alpha }_n$ from appearing without a representative on which ${\alpha }_{n+1}$ is already prescribed. Hence $U_n$ creates these coequalizers, and Beck's theorem proves the adjunction monadic.

The composite $U_n{U}_{n+1}:{\mathcal{C}}_{n+2}\to {\mathcal{C}}_n$ is not monadic. To see the Beck obstruction explicitly, take $B=\{x,y,z\}$, let all operations through ${\alpha }_n$ be the identity, let ${\alpha }_{n+1}$ swap $x,y$ and fix $z$, and define ${\alpha }_{n+2}(z)=z$. Let $q:B\to Q=\{w,z\}$ identify $x$ and $y$, choose the section $s(w)=x$, $s(z)=z$, and put $h=sq$. Form the kernel pair $A=B{\times }_{Q}B$ with coordinatewise operations and projections $f,g:A\rightrightarrows B$. The map $t:B\to A$, $t(b)=(b,h(b))$, satisfiesso $q$ is a split coequalizer of the underlying pair in ${\mathcal{C}}_n$.

Any lifted structure on $Q$ must have ${\alpha }_{n+1}(w)=w$, so ${\alpha }_{n+2}(w)$ must be defined. If it is $w$, map $x,y$ to a fixed point $d$ and $z$ to a fixed point $e\ne d$ in a target with ${\alpha }_{n+2}(d)=e$ and ${\alpha }_{n+2}(e)=e$; this equalizes $f,g$ but its set-theoretic factor through $Q$ does not preserve ${\alpha }_{n+2}$ at $w$. If instead ${\alpha }_{n+2}(w)=z$, use a target with ${\alpha }_{n+2}(d)=d$ and ${\alpha }_{n+2}(e)=e$ to obtain the same failure. Thus the underlying split coequalizer cannot be created in ${\mathcal{C}}_{n+2}$, and Beck's theorem proves that the composite is not monadic.

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A Lawvere theory is a small category with finite products generated by one object $[1]$, so every object is a finite power $[n]$. A model in a finite-product category $\mathcal{C}$ is a finite-product-preserving functor from the theory to $\mathcal{C}$. For a finitary monad $T$ on sets, take the opposite of the full subcategory of its Eilenberg-Moore category on the finitely generated free algebras; the resulting finite-power category is its Lawvere theory, and its set-valued models are $T$-algebras.

If ${\mathbf{Set}}^T$ is additive, its full subcategory of finite free algebras is additive, and passing to the opposite preserves biproducts and abelian-group enrichment. Thus the associated theory is an additive category. A product-preserving model sends the abelian-group object $[1]$ and its addition, zero, and inverse maps to an internal abelian group in $\mathcal{C}$.

In an additive theory the product $[n]=[1{]}^n$ is also a biproduct. If ${\iota }_i:[1]\to [n]$ and ${\pi }_i:[n]\to [1]$ are its injections and projections, every $n$-ary operation decomposes uniquely aswhich in a model reads $\alpha (x_1,\dots ,x_n)={\sum }_i{\alpha }_i(x_i)$. Put $R=\mathrm{End}([1])$, with addition from the enrichment and multiplication from composition. The unary operations give every model an $R$-module structure, and the displayed decomposition says that all operations are exactly $R$-linear combinations. Conversely every $R$-module interprets them this way. Hence ${\mathbf{Set}}^T\simeq R\text{-}\mathbf{Mod}$, up to the conventional choice of left versus right modules.

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An abelian category is an additive category with all kernels and cokernels in which every monomorphism is a kernel and every epimorphism is a cokernel. If $m:A\rightarrowtail B$ is monic and $c=\mathrm{coker}m$, then $m$ factors through $k=\ker c$. The canonical coimage-to-image morphism is an isomorphism in an abelian category, so this factorization identifies $A$ with $\ker c$; hence $m$ is the kernel of its own cokernel. Dually, every epimorphism is the cokernel of its own kernel. The assignments $m\mapsto \mathrm{coker}m$ and $q\mapsto \ker q$ therefore give inverse bijections between subobjects and quotient objects of a fixed object. Well-poweredness is consequently equivalent to well-copoweredness.

The category ${\mathcal{C}}^{\mathcal{A}}$ of complexes has chain complexes as objects and chain maps as morphisms. Write $Z_n=\ker d_n$ and $Q_n=\mathrm{coker}{d}_{n+1}$. The equation $d_n{d}_{n+1}=0$ gives both an induced map $\mathrm{im}{d}_{n+1}\to Z_n$ and a map $Q_n\to \mathrm{im}{d}_n$. The homology object has the two canonically isomorphic descriptionsPassing to the opposite category exchanges these descriptions, proving self-duality.

Let $S_n(A)=A[n]$. A chain map $A[n]\to C_{\bullet }$ is exactly a map $A\to Z_n(C)$, while a chain map $C_{\bullet }\to A[n]$ is exactly a map $Q_n(C)\to A$. Therefore

The Snake lemma states that a commutative diagram with exact rows yields the exact sequence of the three kernels, followed by its connecting morphism and the three cokernels. Apply it degree by degree to a short exact sequence of complexes $0\to A_{\bullet }\to B_{\bullet }\to C_{\bullet }\to 0$, using the diagrams of cycles, boundaries, and degree objects. The connecting map sends a cycle of $C_n$ to a lift in $B_n$, takes its boundary in $B_{n-1}$, and identifies that boundary with a class in $H_{n-1}(A)$. The Snake lemma gives exactness and producesthe algebraic Mayer-Vietoris theorem.

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