Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 20 / 1

Write the comonad as $(G,ϵ,\delta )$ and its category of coalgebras for a comonad as ${\mathcal{E}}^G$. A coalgebra for a comonad is a map $a:A\to GA$ with ${ϵ}_{A}a=1_A$ and ${\delta }_{A}a=Gaa$. A morphism $f:(A,a)\to (B,b)$ satisfies $bf=Gfa$. Let $U:{\mathcal{E}}^G\to \mathcal{E}$ be the forgetful functor and let $R(X)=(GX,{\delta }_X)$ be the cofree coalgebra. The adjunction $U⊣R$ has the explicit correspondenceWe construct the three pieces of the elementary topos structure.

Because $G$ 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 $U$ 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 $(A,a)$ and $(B,b)$ and put $E=B^A$ in $\mathcal{E}$. On the cofree coalgebra $R(E)$ there is an underlying evaluationThe two mapsuse $G(GE\times A)\cong GGE\times GA$ in the second expression. Transpose them in $\mathcal{E}$ to maps $r,s:GE\rightrightarrows (GB{)}^A$, and then transpose across $U⊣R$ to coalgebra morphisms $\tilde{r},\tilde{s}:R(E)\rightrightarrows R((GB{)}^A)$. Take their equalizer $C$ in ${\mathcal{E}}^G$.

An underlying map $UX\times A\to B$ corresponds to a coalgebra map $h:X\to R(E)$. The equation saying that the original map is a coalgebra morphism is precisely $rh=sh$, since ${\delta }_{E}h=Ghx$ for the structure $x:X\to GX$. By the cofree adjunction, this is equivalent to $\tilde{r}h=\tilde{s}h$, hence to unique factorization through $C$. Thereforenaturally in $X$. This constructs the required exponential object.

For the subobject classifier of a coalgebra topos, let $\top :1\hookrightarrow \Omega$ be the underlying subobject classifier, and let $\kappa :G\Omega \to \Omega$ classify the mono $G\top :G1\cong 1\hookrightarrow G\Omega$. Its cofree transpose is the coalgebra endomorphismDefine ${\Omega }_G$ as the equalizer of $k$ and $1_{R\Omega }$. The transpose of $\top$ factors through this equalizer and gives ${\top }_G:1\to {\Omega }_G$.

Indeed, for a subobject $S\hookrightarrow UX$ classified by $\chi :UX\to \Omega$, the pullback $x^{-1}(GS)$ has characteristic map $\kappa G\chi x$. It is always contained in $S$, by naturality of the counit. Equality holds exactly when $x$ restricts to a coalgebra structure on $S$; its axioms then follow by composing with the monomorphisms $GS\hookrightarrow GX$ and $GGS\hookrightarrow GGX$. Under $U⊣R$, the classifying map becomes $h=G\chi x:X\to R\Omega$. The equality of subobjects is ${ϵ}_{\Omega }h=\kappa h$. Transposing this equality gives $h=kh$, so exactly the coalgebra subobjects correspond to maps $X\to {\Omega }_G$. Their pullback of ${\top }_G$ is the desired subcoalgebra, and uniqueness follows from uniqueness of $\chi$.

Thus we have finite limits, exponentials and a subobject classifier:The construction does not assume that $G$ preserves the underlying exponentials or underlying subobject classifier.

A decidable object in a topos has a decomposition $A\times A={\Delta }_A⨿D_A$, with $D_A$ representing inequality. In the internal logic of a topos, equality on $A$ is decidable. These categorical complements behave well under pullback.

For a subobject $B\hookrightarrow A$, pull back the displayed decomposition along $B\times B\hookrightarrow A\times A$. The diagonal pulls back to ${\Delta }_B$, while $D_A$ pulls back to its complement. Hence every subobject of a decidable object is decidable; the subobject itself need not be complemented in $A$.

For two decidable objects, the diagonals of $A$ and $B$ give four disjoint summands of $(A\times B{)}^2$, according as each coordinate pair is equal or unequal. The both-equal summand is ${\Delta }_{A\times B}$, and the other three give its complement. The terminal object is decidable, so induction gives closure under finite products, including the empty product.

For a family $(A_i{)}_{i\in I}$ of decidable objects with an existing coproduct $A={\coprod }_i{A}_i$, products distribute over this coproduct, givingThe coproduct injections in a topos are disjoint. The diagonal consists of ${\Delta }_{A_i}$ in each $i=j$ summand, and has complementConsequently every existing coproduct of decidable objects is decidable, including the initial object. In a Grothendieck topos all small coproducts exist. No assertion that arbitrary products preserve decidability is used.

Let $\mathcal{Q}$ be the full subcategory of quotients of decidable objects. It is closed under quotients, by composition of epimorphisms, and under small coproducts, by part (ii). It is also closed under subobjects: pull back a decidable cover $D\twoheadrightarrow B$ along $B^{\prime }\hookrightarrow B$. The resulting cover of $B^{\prime }$ has domain a subobject of $D$, hence a decidable object.

The terminal object lies in $\mathcal{Q}$. If $D\twoheadrightarrow B$ and $D^{\prime }\twoheadrightarrow C$ are decidable covers, their product is an epimorphism $D\times D^{\prime }\twoheadrightarrow B\times C$. Part (ii) makes its domain decidable. Thus products, and then equalizers as subobjects of products, remain in $\mathcal{Q}$. The inclusion $I:\mathcal{Q}\hookrightarrow \mathcal{E}$ preserves finite limits.

We construct a coreflective subcategory rather than claim that every object has a decidable cover. For $X\in \mathcal{E}$, let $qX\hookrightarrow X$ be the union of all subobjects of $X$ which lie in $\mathcal{Q}$. The Grothendieck topos is well-powered, so these subobjects form a set. Choose a decidable cover of each and take their coproduct. Its map to $X$ has image $qX$, so $qX$ is itself a quotient of a decidable object. Any map from an object of $\mathcal{Q}$ to $X$ has image in $\mathcal{Q}$, and therefore factors uniquely through $qX$. This givesThe induced idempotent comonad $G=Iq$ on $\mathcal{E}$ preserves finite limits: $q$ is a right adjoint and $I$ preserves those limits. Its counit is the inclusion $qX\hookrightarrow X$, and $q(qX)=qX$.

The coalgebras of this comonad are exactly the objects of $\mathcal{Q}$. A coalgebra structure is a section $X\to qX$ of the monic counit, forcing the counit to be an isomorphism; conversely an object already in $\mathcal{Q}$ has the unique such structure. Thus $\mathcal{Q}\simeq {\mathcal{E}}^G$. Part (i) now givesThis argument proves the required elementary-topos conclusion without presuming a small family of decidable generators for $\mathcal{E}$.