Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 20 / Section B

Write $\hat{\mathcal{C}}=[{\mathcal{C}}^{\mathrm{op}},\mathbf{Set}]$ and similarly for $\mathcal{D}$. The geometric morphism induced by a functor hasPrecomposition preserves all pointwise limits and colimits, so in particular it preserves finite limits. The Right Kan extension exists because the categories are small and sets have all small limits; its universal property gives $f^{\ast }⊣f_{\ast }$. Thus these functors define a geometric morphism $\hat{\mathcal{C}}\to \hat{\mathcal{D}}$.

There is also $f_{!}={\mathrm{Lan}}_{F^{\mathrm{op}}}$, a left Kan extension, with $f_{!}⊣f^{\ast }$. The Yoneda lemma identifies $f_{!}(yC)\cong y(FC)$, since for every $P$,This is the representable calculation used in the next parts.

A representable functor $yC$ is an indecomposable projective object. Given an epimorphism ${\coprod }_i{B}_i\twoheadrightarrow yC$, evaluate at $C$. Epimorphisms and coproducts in a presheaf category are pointwise, so $1_C\in yC(C)$ is the image of some element of a particular $B_i(C)$. By the Yoneda lemma, that element defines $s:yC\to B_i$, and its composite into $yC$ corresponds to $1_C$, hence is the identity. The selected component is split epic. More generally, evaluation sends any epimorphism to a surjection, so a map from $yC$ lifts through any epimorphism; this also proves its ordinary projectivity.

Conversely, every presheaf $A$ has the canonical epimorphismwhose component is the natural transformation named by $x$. It is pointwise surjective, since an element at $D$ is reached from its own summand $(D,x)$ at $1_D$. If $A$ is indecomposable projective, one component $r:yC\to A$ has a section $s:A\to yC$. The endomorphism $sr$ of $yC$ is idempotent and therefore corresponds to an idempotent morphism $e:C\to C$.

If idempotents split in $\mathcal{C}$, choose $C\stackrel{p}{\to }D\stackrel{i}{\to }C$ with $ip=e$, $pi=1_D$. Then $A\cong yD$: the mutually inverse maps are $y(p)s:A\to yD$ and $ry(i):yD\to A$. ThusWithout that hypothesis the argument still proves that every such object is a retract of a representable. The initial presheaf is not indecomposable projective, since its identity is the empty-coproduct epimorphism and has no component to select.

The key fact is that the extra left adjoint $f_{!}$ sends each representable $yC$ to an indecomposable projective object. Let ${\coprod }_i{B}_i\twoheadrightarrow f_{!}yC$ be epic. The inverse image $f^{\ast }$ preserves epimorphisms and coproducts, because it is a left adjoint between toposes. Apply it and lift the unit $\eta :yC\to f^{\ast }{f}_{!}yC$ through the resulting epimorphism, using projectivity of $yC$. A map from $yC$ to a coproduct selects one component, by evaluation at $C$ and the Yoneda lemma. Thus for some $i$ we obtain $t:yC\to f^{\ast }{B}_i$ with $f^{\ast }(B_i\to f_{!}yC)t=\eta$.

Transpose $t$ across $f_{!}⊣f^{\ast }$ to $\bar{t}:f_{!}yC\to B_i$. The displayed equality says that its composite back to $f_{!}yC$ is the identity. This proves the required indecomposable-projective property.

Since idempotents split in $\mathcal{D}$, part (ii) supplies objects $FC\in \mathcal{D}$ and isomorphisms $f_{!}yC\cong y(FC)$. Full faithfulness of the Yoneda embedding transports the action of $f_{!}$ on representable arrows to a functor $F:\mathcal{C}\to \mathcal{D}$. For $P\in \hat{\mathcal{D}}$,These identifications are natural in both $C$ and $P$. HenceIts right adjoint is consequently the right Kan extension from part (i), uniquely up to natural isomorphism. Thus the entire geometric morphism is induced by $F$.

The canonical geometric morphism $\hat{\mathcal{C}}\to \mathbf{Set}$ has inverse image the constant-presheaf functor $\Delta$ and direct image the global sections functorIf the presheaf topos is a local topos, $\Gamma$ is also the inverse image of a geometric morphism $g:\mathbf{Set}\to \hat{\mathcal{C}}$. This morphism has an extra left adjoint $\Delta$. Apply part (iii) with source category $1$ and target category $\mathcal{C}$, using its idempotent-splitting hypothesis. Then $g$ is induced by a functor $1\to \mathcal{C}$, choosing an object $C_0$, and $\Gamma$ is naturally evaluation at $C_0$.

Since evaluation at $C_0$ is $\mathrm{Hom}(y{C}_0,-)$, this says $\mathrm{Hom}(1,-)\cong \mathrm{Hom}(y{C}_0,-)$. Uniqueness of representing objects gives $y{C}_0\cong 1$. Thus $\mathcal{C}(C,C_0)$ is a singleton for every $C$: $C_0$ is terminal.

Conversely, if $C_0$ is terminal, $y{C}_0=1$ and $\Gamma (P)=P(C_0)$. Evaluation at that object preserves finite limits and has a right Kan extension as right adjoint, so is an inverse image functor. Therefore, under the permitted idempotent-completeness assumption,

The required condition is the common-refinement condition for nonempty-sieve coverage: for every pair $f:V\to U$, $g:W\to U$, there are arrows $h:T\to V$, $k:T\to W$ withNecessity follows by pulling back the nonempty sieve generated by $f$ along $g$: a member $k$ of the pullback sieve supplies such an $h$. Conversely, this condition makes the pullback of every nonempty sieve on a category nonempty. The maximal sieve is nonempty, and the transitivity axiom holds: if a sieve $S$ is locally covering along every member of a nonempty covering sieve $R$, choose $f\in R$ and then $g\in f^{\ast }S$; their composite is in $S$. Hence the nonempty sieves form a Grothendieck topology, called the atomic topology.

For all functions between nonempty finite sets, the two maps from a singleton to different points of a two-point set have no common refinement. Every potential domain remains nonempty, so the two constant composites cannot agree. The condition fails.

For surjections it holds: $V{\times }_{U}W$ is nonempty and both projections are surjective. Work from now on in a small skeleton $\mathcal{D}$ of nonempty finite sets and surjections. Every morphism $f:P\twoheadrightarrow U$ is a regular epimorphism, with kernel pair $P{\times }_{U}P\rightrightarrows P$, and is the coequalizer of that pair in $\mathcal{D}$.

A matching family in a representable $yV$ on the sieve generated by $f$ is determined by a surjection $t:P\to V$ equalizing that kernel pair. It factors uniquely through a function $U\to V$, which is surjective because $t$ is. This gives the unique amalgamation. A general nonempty covering sieve contains such an $f$; after amalgamating there, common refinements with any other member force agreement on the entire sieve. Therefore every representable is a sheaf, so this atomic site is subcanonical.

For any sheaf $F$, every restriction $F(f)$ is injective: equality after a covering arrow forces equality by the separated part of the sheaf condition. We shall also use descent along any surjection $q:n\twoheadrightarrow k$:is an equalizer of sets. These are the descent identities for the atomic finite-surjection site.

Consider primitive $x\in F(m)$, $y\in F(n)$ with a common restriction along $\alpha :P\twoheadrightarrow m$, $\beta :P\twoheadrightarrow n$. Suppose $a,b\in P$ have $\alpha (a)=\alpha (b)$ but $\beta (a)\ne \beta (b)$. Let $q:n\twoheadrightarrow n-1$ identify just the two points $\beta (a),\beta (b)$. Define the finite nonempty setBoth projections $t_1,t_2:T\twoheadrightarrow P$ are surjective, since $T$ contains every diagonal pair. There is also a surjectionIndeed the target consists of diagonal pairs, which are reached because $\beta$ is surjective, and the two off-diagonal pairs corresponding to $\beta (a),\beta (b)$, reached by $(a,b)$ and $(b,a)$.

Since $\alpha t_1=\alpha t_2$, the common-restriction equality gives $F(\beta t_1)y=F(\beta t_2)y$. If ${\pi }_1,{\pi }_2$ are the target kernel-pair projections, this isInjectivity of $F(h)$ gives the kernel-pair matching condition on $y$. Descent along $q$ then writes $y=F(q)y^{\prime }$, contradicting primitivity. Thus $\ker \alpha \subseteq \ker \beta$; interchange the roles to obtain equality. This is the primitive-element kernel rigidity lemma.

Equal kernels produce a unique bijection $\gamma :m\to n$ with $\beta =\gamma \alpha$. Now $F(\alpha )x=F(\alpha )F(\gamma )y$, and injectivity impliesIn particular equivalent primitive elements have the same cardinality and differ only by transport along a bijection.

Every element $z\in F(p)$ descends to a primitive one: whenever it is not primitive, descend along a surjection reducing the cardinality by one; this process terminates at or before cardinality one. Kernel rigidity shows that all primitive ancestors of $z$ lie in one equivalence class. Let $F_C(P)$ be the elements with primitive-ancestor class $C$. Restriction along a surjection preserves this class, so each $F_C$ is a subfunctor andpointwise. Each $F_C$ is a sheaf. A matching family glues in $F$, and one member along a nonempty covering arrow already determines the primitive class of the glued element; it must be $C$.

Choose a representative primitive $x\in F(m)$ of $C$. The Yoneda map $ym\to F$ named by $x$ has image exactly $F_C$. It reaches all descendants of $x$, and every equivalent primitive ancestor is its transport along a bijection. It is therefore pointwise surjective onto $F_C$ and is epic as a map of sheaves. We obtain the primitive decomposition of an atomic finite-surjection sheafThis includes the empty coproduct for an empty sheaf.

Each nonempty $F_C$ is an atom in a topos. If a sheaf subobject $S\hookrightarrow F_C$ has an element $F(\alpha )x$ at some $P$, membership descends along the covering surjection $\alpha :P\to m$, so $x$ belongs to $S$. All its restrictions then belong to $S$, giving $S=F_C$. Thus every subobject of any $F$ selects entire components of this coproduct, and its complementary selection is again a sheaf subobject. Its characteristic map sends selected components to $\top$ and all others to $\perp$.

The constant two-element presheaf is a sheaf: a matching family on a nonempty sieve has the same value on all its arrows, since any two have a common refinement. The value extends uniquely. It therefore supplies these characteristic maps, with truth the inclusion of the $\top$ value. Equivalently, a J-closed sieve here is either empty or maximal, because every nonempty sieve covers. Hence

Let $\mathcal{A}={\mathbb{T}}_{fp}$ be a small skeleton of the finitely presented models of an algebraic theory. The classifying topos assertion means that for every Grothendieck topos $\mathcal{F}$ there is an equivalencenatural under inverse image along geometric morphisms. On the left, morphisms are transformations between inverse image functors, and on the right they are model homomorphisms. A model in $\mathcal{F}$ interprets the sorts by objects, the operations by arrows, and the equations by equality of the resulting arrows. Finite products suffice for these algebraic operations.

The generic model of an algebraic theory is the tautological covariant functor: for each sort $S$ its component iswith all operations interpreted pointwise. Pulling $U$ back by a geometric morphism gives its classified model. For a single-sorted theory, this is simply the underlying-set functor with its pointwise algebraic structure.

The orientation is important: $[\mathcal{A},\mathbf{Set}]$ is the presheaf topos on ${\mathcal{A}}^{\mathrm{op}}$, and the generic model is covariant on finitely presented algebras. Such an algebra is a finite-generator, finite-relation presentation. In the algebraic syntactic category it corresponds to the formula imposing its relations, with arrows reversed. The finite-presentability/filtered-colimit description of algebraic models gives the above classifying equivalence; a detailed proof is not needed for this part.

Take $\mathcal{A}$ to be the category of finitely presented commutative rings with identity, and put $\mathcal{C}={\mathcal{A}}^{\mathrm{op}}$. In its presheaf topos, the tautological ring $U(A)=A$ is generic. The additional domain axioms are coherent sequents, so they are imposed by a quotient-theory coverage on $\mathcal{C}$.

Concretely, declare the zero ring covered by the empty family. For every finitely presented $A$ and elements $a,b\in A$ with $ab=0$, declare the two opposite quotient arrows associated withto be a covering family at $A$. These quotient rings are finitely presented. Pullback and transitivity generate a Grothendieck topology $J$ from these families. The empty cover forbids $0=1$; the two quotient covers make every zero product locally have a zero factor. Conversely, any internal integral domain satisfies exactly the continuity conditions prescribed by these generating covers. Thusand its generic domain is the associated sheaf $K=a_{J}U$, with the ring operations transported through the left-exact sheaf reflector.

This coverage is not standard, meaning not all representables are sheaves; in modern terminology it is not subcanonical. For an explicit obstruction, use $A=\mathbb{Z}/4\mathbb{Z}$ and $a=b=2$. The two quotient arrows are the same map $A\to \mathbb{Z}/2\mathbb{Z}$, so their generated sieve is a singleton cover. Consider the representable on $\mathcal{C}$ corresponding to $R=\mathbb{Z}[t]$:Its distinct sections $t\mapsto 0$ and $t\mapsto 2$ become equal after restriction to $\mathbb{Z}/2\mathbb{Z}$. Hence this representable is not even separated for $J$. The nilpotent element has to disappear in the generic domain, which explains this failure of standardness.

Work in the domain-classifying site of part (ii), with generic domain $K=a_{J}U$. For finitely presented rings $A$, the object $a_{J}yA$ is an available stage. Tuples of sections of $K$ are locally represented by tuples of actual elements of $A$, because $U\to a_{J}U$ is locally surjective and sheafification preserves finite products. It is therefore enough to prove the requested implication on such representatives.

We first note the empty-cover criterion for the domain-classifying site: $a_{J}yB$ is initial if and only if the finitely presented ring $B$ is the zero ring. One direction is the generating empty cover. Conversely, any nonzero ring has a maximal ideal and hence a homomorphism to a field. That field is a set-based integral domain and defines a point of the classifying topos. The inverse image of $a_{J}yB$ at this point is ${\mathrm{Hom}}_{\mathrm{Ring}}(B,L)$, which is nonempty for the chosen field $L$. It therefore cannot be the inverse image of an initial object. This uses only the ordinary maximal-ideal existence principle externally, not excluded middle in the internal logic.

Suppose $a_1,\dots ,a_n\in A$ represent a tuple lying in the negation of the all-units subobject. Write $t=a_1\cdots a_n$ and consider the finitely presented localization of a ringEvery $a_i$ is invertible in $B$, since $({\prod }_{j\ne i}{a}_j)z$ is an inverse. But the pulled-back tuple still lies in the negation of the all-units subobject. Thus the whole stage $a_{J}yB$ maps into both that subobject and its negation, forcing this stage to be initial. The empty-cover criterion gives $B=0$.

The localization $A[1/t]$ is zero precisely when $t^N=0$ in $A$ for some integer $N\ge 1$, by the equality criterion in localization. Repeated use of the generating zero-product covers now gives a covering familyIndeed, splitting $t^N=t{t}^{N-1}=0$ and inducting first forces $t=0$ locally; splitting the product $a_1\cdots a_n=0$ then forces one $a_i=0$ locally. More formally the two inductions are coherent derivations from the zero-product axiom, so their quotient families belong to the generated topology. The sheaf semantics of disjunction consequently givesThis is the finite-tuple weak-field property of the generic integral domain. The argument is intuitionistically valid inside the topos, although its description of the site uses ordinary external set theory. There is no appeal to preservation of negation by an arbitrary geometric inverse image.

For the converse assertion, let $R$ be any internally nontrivial commutative unital ring satisfying the $n=2$ implication. Assume $xy=0$. If $x$ and $y$ were both units, multiplying by their two inverses would give $0=1$, contradicting nontriviality. ThereforeThe assumed two-variable implication then gives $x=0\vee y=0$. Together with nontriviality this is exactly the internal integral-domain axiom. Hence a nontrivial ring satisfying the two-variable case is an integral domain.

For example, the ordinary integral domain $\mathbb{Z}$ fails the one-variable implication at $2$. This does not contradict the generic result: the displayed implication uses negation and is non-coherent, so it need not survive the inverse image that classifies an arbitrary set-based domain.