Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 119 (Category Theory)

The Yoneda lemma states that for $A\in \mathcal{C}$ and $P:\mathcal{C}\to \mathbf{Set}$ there is a natural bijectionIf $P\twoheadrightarrow Q$ is an epimorphism in the functor category, it is pointwise surjective, so $P(A)\to Q(A)$ is surjective. Yoneda identifies this map withThus every representable functor is a projective object in a category.

The colimit form of the Special adjoint functor theorem says that a colimit-preserving functor from a locally small, cocomplete, well-copowered category with a small generating family into a locally small category has a right adjoint functor. For small $\mathcal{C}$, the category $[\mathcal{C},\mathbf{Set}]$ is locally small and has colimits pointwise. Quotients of $P$ are represented by compatible equivalence relations on the sets $P(C)$, so they form a set; hence the category is well-copowered. The set of representables $\{\mathcal{C}(C,-):C\in \mathcal{C}\}$ generates it by the Yoneda lemma. The theorem therefore gives a right adjoint to every small-colimit-preserving functorIn particular, product with a fixed functor $P$ is computed pointwise, and $-\times P(C)$ preserves colimits in the Category of sets. Hence $-\times P$ preserves all small colimits and has a right adjoint $(-{)}^P$. Thus $[\mathcal{C},\mathbf{Set}]$ is a cartesian closed category.

Now work in $[{\mathcal{C}}^{\mathrm{op}},\mathbf{Set}]$ and write $yA=\mathcal{C}(-,A)$. If $\mathcal{C}$ has binary products, thenThus exponentiation by $yA$ is precomposition with $-\times A$. Precomposition between functor categories has a right adjoint given by Right Kan extension, so $yA$ is a tiny object.

Conversely, suppose $\mathcal{C}$ has a terminal object $1$ and $F$ is tiny. The representable $y1$ is the terminal presheaf, and the exponential adjunction plus Yoneda givesSince $F$ is tiny, $(-{)}^F$ is a left adjoint and preserves all colimits; evaluation at $1$ also preserves pointwise colimits. Therefore the hom functor $[{\mathcal{C}}^{\mathrm{op}},\mathbf{Set}](F,-)$ preserves coproducts and epimorphisms. Preservation of epimorphisms makes $F$ projective, while preservation of coproducts makes it indecomposable.

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For $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$ with $F⊣G$, the unit and counit of an adjunction areobtained by transposing identity morphisms. They satisfy the triangular identities ${\epsilon }_{F}F\eta =1_F$ and $G\epsilon {\eta }_G=1_G$.

If $G$ is full and faithful, fullness gives a map $u:B\to FGB$ with $G(u)={\eta }_{GB}$; the second triangular identity and faithfulness show that $u$ is inverse to ${\epsilon }_B$. Conversely, if $\epsilon$ is an isomorphism, then for $h:GB\to G{B}^{\prime }$ the unique arrow whose transpose is $h$ iswhich proves that $G$ is full, and the triangular identities prove faithfulness. Hence (i) and (ii) are equivalent. Condition (ii) immediately gives (iii). Conversely, a natural isomorphism $FG\cong 1_{\mathcal{D}}$ combines with the adjunction bijection to show naturally that $\mathcal{D}(B,B^{\prime })\to \mathcal{C}(GB,G{B}^{\prime })$ is bijective; by naturality and the triangular identities this is the map induced by $G$ up to invertible natural conjugation, so $G$ is full and faithful. Thus all three conditions are equivalent.

Suppose $L⊣F⊣R$. If $L$ is full and faithful, the unit $1\to FL$ is an isomorphism. For $A,B$ in the codomain of $F$, the two adjunctions then give natural bijectionsBy the Yoneda lemma, the counit $FRB\to B$ is an isomorphism, so the fully faithful adjoint criterion makes $R$ full and faithful. The converse is dual.

Now assume $F$ is full and faithful. For $L⊣F$, its counit $ϵ:LF\to 1$ is invertible; for $F⊣R$, its unit $\alpha :1\to RF$ is invertible. DefineApplying $F$, then using naturality and the triangular identities, reduces this composite to the same map asSince $F$ is faithful, the two displayed composites are equal.

For a morphism $u:FB\to A$, let $\hat{u}=Ru{\alpha }_B:B\to RA$ be its transpose under $F⊣R$. The identities just proved giveIf every ${\theta }_A$ is monic and $Lu=Lv$, this equation gives $\hat{u}=\hat{v}$, hence $u=v$; thus $L$ is faithful on arrows whose domains are in the image of $F$. Conversely, if ${\theta }_{A}f={\theta }_{A}g$ for $f,g:B\to RA$, naturality of $ϵ$ and the second formula for $\theta$ giveBoth maps have domain $FB$, so the assumed faithfulness gives ${\beta }_{A}Ff={\beta }_{A}Fg$. These are the transposes of $f$ and $g$, hence $f=g$. Therefore $\theta$ is pointwise monic exactly under the stated faithfulness condition.

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A categorical limit of $D:\mathcal{J}\to \mathcal{C}$ is a terminal cone: it consists of an object $L$ and compatible maps $L\to D(j)$ through which every other cone factors uniquely. For a finite diagram, take the product $P={\prod }_{j}D(j)$ and the product $Q={\prod }_{u:i\to j}D(j)$. There are two maps $P\rightrightarrows Q$: on the $u:i\to j$ coordinate one uses respectively $D(u)$ after projection to $D(i)$ and direct projection to $D(j)$. Their equalizer is exactly the compatible-cone object. Hence finite products and equalizers construct every finite limit.

In the category of metric spaces and non-expansive maps, give $X\times Y$ the maximum metricThe projections are non-expansive, and a pair of non-expansive maps into $X$ and $Y$ induces a non-expansive map into this product, proving the universal property.

Let two maps $1\rightrightarrows X$ select $a,b\in X$. On the set quotient identifying $a$ and $b$, define the quotient metric by shortest paths that may jump from $a$ to $b$ at zero cost. Explicitly,This is the largest metric making the quotient map non-expansive. A map $X\to Z$ equalizing $a$ and $b$ factors through the set quotient and remains non-expansive by the path formula, so this is the coequalizer. Its underlying set is the set-theoretic coequalizer.

For $X=\{0,1,5,6\}$ with $0\sim 6$, write $\ast =\{0,6\}$. The quotient hasThus in $Q\times Y$, with $Y=\{0,3\}$,After first taking $X\times Y$, the product of the parallel pair identifies $(0,y)$ with $(6,y)$ separately for $y=0,3$. In its quotient metric, every path from $(1,0)$ to $(5,3)$ has length at least $4$, and length $4$ is attained either directly or via one of those identifications. The canonical bijection from this coequalizer to $Q\times Y$ is therefore not an isometry. Hence $-\times Y$ does not preserve this coequalizer. In a cartesian closed category, $-\times Y$ is a left adjoint and preserves all colimits, so $\mathbf{Met}$ is not cartesian closed.

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For $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$ with $F⊣G$, the monad induced by an adjunction is $T=GF$, with unit the adjunction unit and multiplication $\mu =G\epsilon F$. An algebra for a monad is a map $a:TA\to A$ satisfying $a{\eta }_A=1_A$ and $aT(a)=a{\mu }_A$. The Eilenberg-Moore comparison functor isand sends a morphism $h$ to $Gh$.

Given a $T$-algebra $(A,a)$, form in $\mathcal{D}$ the coequalizerThe pair is reflexive, with common section $F{\eta }_A$, so it exists by hypothesis. A map $J(A,a)\to D$ is equivalently a map $FA\to D$ equalizing the pair. Under adjunction this is exactly a map $A\to GD$ satisfying the $T$-algebra homomorphism equation. Hencenaturally, and $J⊣K$. Iterating these comparison adjunctions gives the monadic tower; the monadic length is the least number of steps required to reach an equivalence.

For a pair of sets $(X_0,X_1)$, defineA pair of maps $X_0\to A_0$, $X_1\to A_1$ extends uniquely to a commutative square from this inclusion to any injection $A_0\hookrightarrow A_1$, proving that $F$ is left adjoint to the forgetful functor $U:\mathbf{Mono}(\mathbf{Set})\to \mathbf{Set}\times \mathbf{Set}$.

The induced monad sends $(X_0,X_1)$ to $(X_0,X_0\bigsqcup X_1)$. Its algebra unit laws show that an algebra is precisely an arbitrary function $X_0\to X_1$, with no injectivity requirement. Thus its Eilenberg-Moore category is the arrow category $[2,\mathbf{Set}]$, and the comparison functor is the full inclusion of injections into all functions. This inclusion is not an equivalence, so the original adjunction is not monadic. It is reflective: a function $f:A_0\to A_1$ maps to its image inclusion $\mathrm{im}f\hookrightarrow A_1$. The supplied result that reflections are monadic says that the next comparison is an equivalence. Therefore the original adjunction has monadic length $2$.

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The statement that limits of shape $\mathcal{I}$ commute with colimits of shape $\mathcal{J}$ means that for every $D:\mathcal{I}\times \mathcal{J}\to \mathcal{C}$, the canonical commutation of limits and colimits mapis an isomorphism whenever the iterated limits and colimits exist.

A filtered category is a nonempty category in which every finite diagram has a cocone. Equivalently, any two objects map to a common object and any parallel pair becomes equal after postcomposition. A weakly filtered category requires cocones only for finite connected diagrams; equivalently, each connected component is filtered.

Write a weakly filtered category as the disjoint union ${\coprod }_s{\mathcal{J}}_s$ of its filtered connected components. Its colimit is the coproduct of the filtered colimits over the ${\mathcal{J}}_s$. In sets, a connected finite limit commutes with coproducts: connectedness forces all coordinates of a compatible tuple to lie in the same coproduct summand. By the assumed theorem, the filtered colimit over each ${\mathcal{J}}_s$ commutes with every finite limit. Applying these two facts successively proves that weakly filtered colimits commute with connected finite limits in $\mathbf{Set}$.

The forgetful functor from abelian group to sets creates finite limits and filtered colimits and reflects isomorphisms. The comparison map for a filtered colimit and a finite limit therefore becomes the corresponding isomorphism of sets, so filtered colimits commute with finite limits in $\mathbf{AbGp}$.

The dual claim fails because inverse limit need not preserve epimorphisms. Take the inverse systemswith identity bonding maps on $A_n$, reduction maps on $B_n$, and levelwise epimorphisms $q_n:A_n\to B_n$. Thenand the induced map $\mathbb{Z}\to {\mathbb{Z}}_2$ is not surjective. Since an epimorphism in abelian groups is a finite-colimit cokernel, cofiltered limits do not commute with finite colimits in $\mathbf{AbGp}$.

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A regular category has finite limits, regular-epimorphism--monomorphism image factorizations, and regular epimorphisms stable under pullback. A relation $A⇝B$ in $\mathbf{Rel}(\mathcal{C})$ is a subobject of $A\times B$; composition forms the pullback over the middle object and then takes its image.

Suppose $R:A⇝B$ has a right adjoint relation $S$, so $1_A\le SR$ and $RS\le 1_B$. In the internal regular logic, the first inequality says that for every $a$ there is a $b$ with $R(a,b)$ and $S(b,a)$. If also $R(a,b^{\prime })$, the second inequality forces $b=b^{\prime }$. Thus $R$ is total and single-valued. Categorically, if $R\hookrightarrow A\times B$ has projections $p:R\to A$ and $q:R\to B$, totality makes $p$ a regular epimorphism and single-valuedness makes it a monomorphism. Hence $p$ is an isomorphism and $R$ is the graph of a morphism as a relation $q{p}^{-1}:A\to B$. Conversely, the graph of any morphism is left adjoint to its converse relation, as the two required inequalities follow directly from equality. This proves the characterization.

Let $L$ be a frame. Composition in the category of matrices valued in a frame isIf $f⊣g$, the diagonal part of $1_A\le gf$ impliesso every row of $f$ joins to $1$. For $b\ne b^{\prime }$, distribute $f(a,b)\wedge f(a,b^{\prime })$ over the displayed join. Every term vanishes by $fg\le 1_B$, first using the factor with column $b$ and then the one with column $b^{\prime }$. HenceConversely, if the rows of $f$ join to $1$ and have pairwise disjoint entries, define $g(b,a)=f(a,b)$. Thenwhile $(fg)(b,b^{\prime })=0$ for $b\ne b^{\prime }$. Thus $1_A\le gf$ and $fg\le 1_B$, proving the stated criterion.

Finally take $L=\Omega (X)$ for a connected topological space $X$. For fixed $a$, the opens $f(a,b)$ are pairwise disjoint and cover $X$. Connectedness forces exactly one of them to be $X$ and all the others to be empty. Hence a left adjoint matrix determines a unique function $\phi :A\to B$ by $f(a,\phi (a))=X$. Conversely every function gives this matrix, and matrix composition agrees with function composition. The left adjoints in $\mathbf{Mat}(\Omega (X))$ therefore form a category isomorphic to $\mathbf{Set}$.

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