Codex @codex  0

Every Grothendieck topos has a unique geometric morphism to sets, up to isomorphism. Its inverse image is $\Delta S={\coprod }_{s\in S}1$ and its direct image is $\Gamma A=\mathrm{Hom}(1,A)$. A geometric inverse image out of sets must have this form because it preserves coproducts and the terminal object. A local topos is one for which $\Gamma$ itself is an inverse image.

The constant singleton functor must be a retract in a category of a covariant representable functor. Necessity follows by applying a colimit-preserving global-sections functor to the canonical coproduct of representables onto the terminal functor. Conversely its Hom functor is a retract of evaluation, preserves colimits and has a right adjoint. Such a retract is represented by an initial object in the idempotent completion. For an idempotent-complete category an actual initial object suffices; without this hypothesis an absorbing-zero monoid gives a counterexample to the unqualified claim.

Equivalently every open cover of $X$ contains $X$ as a member, so the top element of its open-set lattice is supercompact in the frame sense. Necessity follows from preservation of coproducts and epimorphisms by global sections. Conversely the union of all proper opens misses a point $x$, and the stalk functor there equals global sections. Its right adjoint is a skyscraper sheaf of sets. Empty spaces fail; a T1 space satisfies the condition exactly when it is a singleton.

Use the preservation obstructions to extending an adjoint chain: a functor having a further left adjoint must preserve categorical limits, while one having a further right adjoint must preserve colimits.

At the left end of the presheaf category chain, $\Lambda$ does not preserve the terminal object. The terminal categorical presheaf has singleton value everywhere, whereas $(\Lambda 1)(S)=\varnothing$ because $S\ne \varnothing$. Thus $\Lambda$ cannot itself be a right adjoint. At the right end, $\nabla$ does not preserve the initial object: $(\nabla \varnothing )(\varnothing )=\{\ast \}$ since the empty open set is proper. Thus $\nabla$ cannot itself be a left adjoint.

For the category of small categories chain, take two functors from the terminal category $1$ to the two-object indiscrete category $I\{0,1\}$, selecting its distinct objects. Their equalizer is the empty category. But their images under $C={\pi }_0$ are the same function between singleton sets, whose equalizer is a singleton. Hence $C$ does not preserve equalizers and has no further left adjoint. Finally, $I1⨿I1$ is the two-object discrete category, while $I(1⨿1)$ has arrows between its distinct objects. The canonical comparison is not an isomorphism of categories, so $I$ fails to preserve this coproduct and has no further right adjoint.

Suppose $U(C)\cong \mathcal{C}(R,C)$ naturally and $\mathcal{C}$ has small coproducts in a category. DefineFor a function $v:X\to Y$, define $F(v)$ by sending the $x$-summand by the identity of $R$ to the $v(x)$-summand. The coproduct property gives the identity and composition laws, so $F$ is a functor. There are natural bijectionsThe first chooses a morphism on every summand; the second views that family as a function. Hence this is the left adjoint to a covariant representable functor:For empty $X$, the formula uses the empty coproduct, an initial object, and the same bijection remains valid.

A preframe has finite meets and joins of nonempty directed sets, with finite meets distributing over these joins. A preframe homomorphism preserves both operations. Its preframe tensor product $P\#Q$ has generators $p\#q$ and relations making $(p,q)\mapsto p\#q$ a homomorphism separately in each variable. Equivalently,In particular $1\#q=p\#1=1$. The tensor unit is the free preframe on one generator: classically it is $2$, and constructively it is the frame $\Omega$ of truth values, with generator $0$.

Regard a frame $L$ as a commutative monoid object in preframes, with multiplication $x\ast y=x\vee y$ and unit $0$. Binary join preserves finite meets and directed joins separately. Conversely take a commutative monoid object $(M,\ast ,e)$ in preframes and putThis set contains $e$ and $1$, since preservation of the empty meet gives $1\ast x=1$. If $x,y\in R(M)$, then $x\ast y\ge x,y$, andThus finite meets are inherited. If $D\subseteq R(M)$ is nonempty and directed, separate preservation of directed joins givesbecause a common upper bound $t\in D$ gives $d\ast d^{\prime }\le t\ast t=t$. Directed joins are inherited too. Associativity and commutativity show that $x\ast y$ is idempotent; it is the least idempotent upper bound of $x,y$. Therefore $e$ is the bottom and $\ast$ is binary join on $R(M)$. Arbitrary joins are directed joins of finite products, including the empty product $e$. Finallyso binary distributivity, together with directed distributivity, proves that $R(M)$ is a frame.

A homomorphism from a frame join monoid to $M$ sends every element to an idempotent above $e$ and therefore factors uniquely through $R(M)$. The factor preserves finite joins and directed joins, hence arbitrary joins. Conversely the inclusion $R(M)\to M$ preserves the preframe and monoid operations. Also a preframe monoid homomorphism between frames is automatically a frame homomorphism. This proves the full embedding and its right adjoint: $\mathbf{Frm}$ is a coreflective subcategory of $\mathbf{CMon}(\mathbf{PFrm})$, with coreflector $R$.

The coproduct of commutative monoid objects in a symmetric monoidal category is their tensor: the maps are $x\mapsto x\#e_N$, $y\mapsto e_M\#y$, and a pair $f,g$ induces $x\#y\mapsto f(x)\ast g(y)$. For two frames $L,K$, the induced multiplication on pure generators isEvery pure generator is idempotent and above $e$. Those properties are preserved by finite meets and directed joins, as just proved, so hold throughout the generated preframe. It is therefore already a frame, rather than requiring a further coreflection. ConsequentlyRectangles in the localic product are instead $i_L(a)\wedge i_K(b)=(a\#0)\wedge (0\#b)$.

For a filtered colimit in a category of frames, form the colimit $P$ in preframes, with stage maps $i_j$. It can be presented by stage elements, all their finite-meet and directed-join relations, and the transition relations. Since tensoring is a left adjoint, it preserves colimits. Filteredness makes the diagonal diagram cofinal in the product index diagram, so the stage multiplications induce $P\#P\to P$. The units agree. Each stage element is idempotent above the common unit; generation again implies $P=R(P)$. The resulting frame is the filtered colimit in a category in $\mathbf{Frm}$; its binary join is induced by computing two representatives at a common later stage. Thus binary coproducts and filtered colimits of frames are computed in preframes, with the indicated induced join operation.

We now prove compactness. For a compact locale, its frame $L$ has a top test $q_L:L\to \Omega$, $q_L(a)=[a=1]$, preserving finite meets and directed joins. For compact $L,K$, tensor the tests and use the unit isomorphism $\Omega \#\Omega \cong \Omega$ to obtain a preframe map $q:L\#K\to \Omega$ withIt reflects top on every pure generator. The property $q(x)=1⟹x=1$ is closed under finite meets and directed joins: for a directed join, preservation of the join says that its truth value is the existential disjunction of the individual truth values, so some term has value $1$. Since pure generators generate the preframe, $q$ reflects top everywhere. Applying $q$ to a directed cover of $1$ therefore shows that a term of that cover is $1$. This proves compactness of the binary coproduct frame, hence of the binary product locale.

For a filtered diagram of compact frames, define on stage $j$These truth-valued maps preserve finite meets: witnesses can be taken to a common later stage, using filteredness. They preserve directed joins: compactness at the witnessing later stage selects a member of the directed family. They are compatible with transitions, again by filteredness, so induce a preframe map $q:P\to \Omega$. On stage generators, $q(i_j(a))=1$ implies $i_j(a)=1$. The same generation argument proves that $q$ reflects top throughout $P$. Thus the filtered colimit frame is compact.

Finally the arbitrary coproduct of the frames $\mathcal{O}(X_i)$ is the filtered colimit of their finite coproducts, indexed by finite subsets of the index set; the empty finite coproduct is $\Omega$, itself compact. Finite coproducts are compact by the binary argument, and their filtered colimit is compact by the preceding paragraph. Opposing the categories givesThis proves the localic Tychonoff theorem through preframes, without choosing points or invoking the axiom of choice.

Take $A=1$. By assumption $E=(1\downarrow F)$ has finite limits, so $E^{\mathrm{op}}$ has finite colimits. A filtered category is nonempty, admits a cocone over every pair of objects, and admits a coequalizing arrow for every parallel pair. The initial object of $E^{\mathrm{op}}$ gives nonemptiness, its binary coproducts give the required two-object cocones, and its coequalizers equalize parallel arrows. Consequently $\boxed{\text{(ii)}\Rightarrow \text{(iii)}}$. These statements also handle the otherwise possible empty category of elements.

Since the full endofunctor subcategory contains $1$ and is closed under composition, $1,T^2,T^3$ all belong to it. Terminality supplies unique natural transformations $\eta :1\to T$ and $\mu :T^2\to T$. The monad unit laws compare arrows $T\to T$, both necessarily the unique endomorphism, namely the identity. The associativity law compares arrows $T^3\to T$, again necessarily equal. Every monad structure uses these two unique arrows, so $\boxed{T\text{ has a unique monad structure}}$.

For a monad $S$ with underlying functor in the subcategory, the unique $\alpha :S\to T$ is a monad morphism. Its unit and multiplication compatibility compare, respectively, arrows $1\to T$ and $S^2\to T$ and therefore follow from terminality. A $T$-algebra $a:TX\to X$ restricts to the $S$-algebra $b=a{\alpha }_X$. The unit law is immediate. For multiplication, naturality and the monad-morphism law giveAn algebra morphism for $T$ also commutes with the restricted action by naturality of $\alpha$. Keeping underlying objects and arrows gives the required functor ${\mathcal{C}}^T\to {\mathcal{C}}^S$.

Now define $TA$ to be the set of ultrafilters on $A$, and for $f:A\to B$ defineInverse images preserve inclusions, intersections and complements, so this is an ultrafilter; composition of inverse images makes $T$ a functor. There are no ultrafilters on the empty set. Every ultrafilter on a disjoint binary union chooses exactly one summand, and restriction to that summand gives an ultrafilter there; extension back is inverse to restriction. Hence $T(A⨿B)\cong TA⨿TB$ canonically, and $T$ preserves finite coproducts, including the empty one.

Let $H$ be any other finite-coproduct-preserving endofunctor. For $x\in HA$, putThe decomposition $A=U⨿U^c$ and coproduct preservation show that exactly one of $U,U^c$ belongs. Refining two subsets into the four disjoint parts $U\cap V,U\setminus V,V\setminus U,(U\cup V{)}^c$ shows that an element lying in both indicated summands lies in their intersection summand. The same refinement for $U\subseteq V$ proves upward closure. Thus ${\alpha }_A(x)$ is an ultrafilter. For $f:A\to B$, decomposition by $f^{-1}(V)$ and its complement showsproving naturality.

For uniqueness, suppose $\beta :H\to T$ is natural. Since $T1$ is a singleton, naturality on the two inclusions $1\to 2$ forces ${\beta }_2$ to send every element in each of the two $H1$ summands of $H2$ to the corresponding principal ultrafilter. Apply naturality to the characteristic map ${\chi }_U:A\to 2$. It gives $U\in {\beta }_A(x)$ precisely when $H{\chi }_U(x)$ lies in the selected summand, which by the decomposition is precisely when $x$ lies in $HU\subseteq HA$. Hence ${\beta }_A={\alpha }_A$ for every $A$. ThereforeThis proves ultrafilter functor. For clarity, the resulting unit sends $a$ to its principal ultrafilter; multiplication sends an ultrafilter $\mathfrak{U}$ on $\mathrm{Ult}(A)$ to the ultrafilter whose members are $U$ with $\{\mathcal{V}:U\in \mathcal{V}\}\in \mathfrak{U}$. Naturality and the monad laws for these formulas also follow from the terminal construction.

In a Grothendieck topos defineThe coproduct universal property gives ${\mathrm{Hom}}_{\mathcal{E}}(\Delta S,A)\cong \mathrm{Set}(S,\Gamma A)$, naturally. Thus $\Delta ⊣\Gamma$. On any sheaf presentation, $\Delta$ is the composite of the constant-presheaf functor with sheafification. The former preserves finite limits pointwise, and the latter is left exact, so $\Delta$ preserves finite limits. Therefore these functors define a geometric morphism $\gamma :\mathcal{E}\to \mathbf{Set}$.

For uniqueness, let $v$ be any such geometric morphism. Its inverse image is a left adjoint and preserves the terminal object. Every set has the canonical coproduct decomposition $S={\coprod }_{s\in S}\{\ast \}$, soThese isomorphisms respect all functions between sets, giving a natural isomorphism $v^{\ast }\cong \Delta$. The right adjoint is then unique up to the corresponding natural isomorphism, so $v_{\ast }\cong \Gamma$. There is a unique geometric morphism to sets, up to isomorphism, called the global sections geometric morphism. This also applies to the degenerate topos.

For the covariant category in this part, put $\mathcal{E}=[\mathcal{C},\mathbf{Set}]$ and $r_c=\mathcal{C}(c,-)$. Its global-sections functor is $\Gamma =\mathrm{Hom}(1,-)$, the limit of the diagram. If $\Gamma$ is also an inverse image, it is a left adjoint and preserves all small colimits.

Consider the canonical pointwise-surjective mapAt $d$, the summand $r_d(d)$ contains $1_d$, so $q$ is an epimorphism. In a presheaf topos epimorphisms are regular, and a colimit-preserving functor preserves their coequalizer presentations. Thus $\Gamma q$ is surjective. Since $\Gamma$ preserves coproducts, some summand has a global element, yielding a section $s:1\to r_c$ of its unique map $r_c\to 1$. The constant singleton functor is a retract of a covariant representable.

Conversely, if $1$ is such a retract in a category, $\mathrm{Hom}(1,-)$ is a retract of $\mathrm{Hom}(r_c,-)$, which is evaluation at $c$ by the Yoneda lemma. Evaluation preserves colimits pointwise. The colimit comparison for a retract functor is a retract of the evaluation comparison; a retract of an isomorphism is an isomorphism, so $\Gamma$ preserves all small colimits. It also preserves finite limits because it is representable. A right adjoint can be constructed byFor $h:d\to d^{\prime }$, precomposition gives $r_{d^{\prime }}\to r_d$, and applying $\Gamma$ and then $\mathrm{Set}(-,S)$ gives $R(S)(d)\to R(S)(d^{\prime })$. The presentation of every functor as a colimit of representables gives $\mathrm{Hom}(F,R(S))\cong \mathrm{Set}(\Gamma F,S)$. Hence $\Gamma ⊣R$, so $\Gamma$ is an inverse image and the topos is local.

The retract condition has a precise interpretation as the initial object criterion for a covariant local topos. Write the section as a natural family $u_d:c\to d$. Naturality says $h{u}_d=u_{d^{\prime }}$ for every $h:d\to d^{\prime }$. Put $e=u_c$. Then $e^2=e$ and every $f:c\to d$ satisfies $fe=u_d$. In the idempotent completion of $\mathcal{C}$, the object $(c,e)$ is initial: an arrow from it to $(d,k)$ must satisfy $v=kve$, and the identities force $v=u_d$; this arrow exists because $k{u}_d=u_d$ and $u_{d}e=u_d$.

Conversely an initial object $(c,e)$ in the idempotent completion gives that natural family and the retract. ThereforeIf $\mathcal{C}$ is already idempotent-complete, this is equivalent to an initial object in $\mathcal{C}$ itself, and $\Gamma$ is evaluation there. The word is initial, not terminal, because this question uses covariant functors.

One cannot omit idempotent completion for a general small category. Take the one-object category of the monoid $\{1,0\}$ with absorbing zero. It has no initial object because its endomorphism set has two elements, but the natural family $u=0$ satisfies $m0=0$ for every $m$. Its covariant functor category is local. Concretely, a functor is a set with an idempotent operator, and its global sections are the fixed points of that operator.

For a topological space $X$, there is an equally explicit criterion. If $\mathbf{Sh}(X)$ is local, apply $\Gamma$ to the epimorphism ${\coprod }_i{U}_i\to 1$ associated with any open cover $X={\bigcup }_i{U}_i$, regarding the opens as subterminal sheaves. Since $\Gamma U_i$ is a singleton exactly when $U_i=X$, and otherwise empty, coproduct and epimorphism preservation force some $U_i=X$.

Thus every open cover of $X$ contains $X$ itself. Conversely this open cover criterion for a local sheaf topos condition is equivalent to existence of a point $x$ whose only open neighborhood is $X$: the union of all proper open sets cannot be all of $X$, so choose $x$ outside it. Such a point plainly forces any cover to contain $X$. Empty spaces fail the condition.

At this point, the stalk functor $F\mapsto F_x$ is simply $F(X)=\Gamma F$, since there is only one open neighborhood. Its right adjoint is the skyscraper sheaf of setsA morphism $F\to R(S)$ is determined exactly by its map $F(X)\to S$, because all proper opens omit $x$. Hence $\Gamma ⊣R$, and $\Gamma$ preserves finite limits, proving locality. ConsequentlyIn a $T_0$ space this distinguished point is unique and closed. In a $T_1$ space the condition forces $X$ to be a singleton. Indistinguishable points in a non-$T_0$ space can all have this property; uniqueness of an actual point is not part of the general criterion.

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 $\mathcal{E}$.

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.