Reflective Subcategories and the Adjoint Functor Theorems

Suppose item 1 is supplied. By [L3] a supplied left adjoint to $I$ corresponds to an initial object, equivalently a universal arrow, of each comma category $(C\downarrow I)$, and the unit components are exactly those arrows. Hence for each $C$ the unit component ${\eta }_C:C\to IR(C)$ has the universal factorisation property of [L2], and $(R(C),{\eta }_C)$ is the required specified universal arrow.

Conversely, suppose item 2 is supplied. By [L3] the specified initial objects $(R_C,{\eta }_C)$ of $(C\downarrow I)$ determine a functor $R:\mathcal{C}\to \mathcal{A}$ and an adjunction $R⊣I$ whose unit is $({\eta }_C)$; hence $\mathcal{A}$ is reflective by [L1]. No selection beyond the supplied family is made.

Since $\mathcal{A}$ is a full subcategory of $\mathcal{C}$, [L4] gives $\mathcal{A}(A,B)=\mathcal{C}(A,B)=\mathcal{C}(I(A),I(B))$ for all $A,B\in \mathcal{A}$, and the inclusion $I$ acts on those hom-sets as the identity; so every $I_{A,B}$ is bijective and [L1] makes $I$ fully faithful. By its surjectivity there is ${\delta }_A:A\to RI(A)$ with $I({\delta }_A)={\eta }_{I(A)}$, and by its injectivity that ${\delta }_A$ is unique. The first triangle identity in [L2] gives $I({\epsilon }_A\circ {\delta }_A)=I({\epsilon }_A)\circ {\eta }_{I(A)}=1_{I(A)}=I(1_A)$, so faithfulness gives ${\epsilon }_A\circ {\delta }_A=1_A$.

Naturality of the counit at ${\delta }_A$ gives ${\delta }_A\circ {\epsilon }_A={\epsilon }_{RI(A)}\circ RI({\delta }_A)$. Since $I({\delta }_A)={\eta }_{I(A)}$, functoriality gives $RI({\delta }_A)=R({\eta }_{I(A)})$, and the second triangle identity in [L2] makes the right side $1_{RI(A)}$. Thus ${\delta }_A$ is a two-sided inverse of ${\epsilon }_A$, so [L3] proves the claim.

If ${\eta }_C$ is invertible, it itself displays $C$ as isomorphic to the included object $I(R(C))$, so $C$ lies in the essential image.

Conversely, let $u:C\to I(A)$ be an isomorphism. Naturality of $\eta$ gives $IR(u)\circ {\eta }_C={\eta }_{I(A)}\circ u$. The map $IR(u)$ is invertible because a functor sends the inverse of $u$ to its inverse. Applying $R⊣I$ in [L3] at $d=A$ gives $I({\epsilon }_A)\circ {\eta }_{I(A)}=1_{I(A)}$, and $I({\epsilon }_A)$ is invertible because ${\epsilon }_A$ is by [L1] and functors preserve inverses; composing that identity with $I({\epsilon }_A{)}^{-1}$ on the left gives ${\eta }_{I(A)}=I({\epsilon }_A{)}^{-1}$, which is therefore an isomorphism. Hence ${\eta }_C=IR(u{)}^{-1}\circ {\eta }_{I(A)}\circ u$. A composite $g\circ f$ of isomorphisms is an isomorphism, since $f^{-1}\circ g^{-1}$ is a two-sided inverse for it by associativity and the identity laws; applying this twice and using [L2] makes ${\eta }_C$ invertible.

Applying $R$ to the legs and then the counit gives a cone from $R(L)$ to $D$; after inclusion its legs are $I({\epsilon }_{D_j})\circ IR(p_j)$. Naturality of the unit and the triangle identity give $I({\epsilon }_{D_j})\circ IR(p_j)\circ {\eta }_L=p_j$, so ${\eta }_L$ is a morphism from the given cone to this included cone.

By the universal property in [L4], the included cone has a unique map $q:IR(L)\to L$ with $p_j\circ q=I({\epsilon }_{D_j})\circ IR(p_j)$. Cone uniqueness gives $q\circ {\eta }_L=1_L$. Both $1_{IR(L)}$ and ${\eta }_L\circ q$ are maps between reflected objects whose composites with the universal reflection arrow ${\eta }_L$ equal ${\eta }_L$, and by [L5] the unit component ${\eta }_L$ is a universal arrow from $L$ to $I$, so the uniqueness clause of that universal property makes ${\eta }_L\circ q=1_{IR(L)}$. Hence ${\eta }_L$ is invertible and [L1] identifies $L$ with an included object.

Transporting the limiting cone across this isomorphism produces a cone of $D$ in the full subcategory whose image is isomorphic to $(L,p_j)$. Its image is limiting, and since $I$ is a right adjoint, [L2] preserves every source limit; conversely fullness makes any mediating map between included objects a unique map in $\mathcal{A}$, so a source cone with limiting image is limiting. These are exactly the clauses of [L3], including the empty and degenerate indexing categories.

Apply $R$ to the ambient cocone. By [L2], $(R(L),R({\lambda }_j))$ is colimiting for $RID$; precomposing its legs with the inverses ${\epsilon }_{D_j}^{-1}:D_j\to RI(D_j)$ of the isomorphisms supplied by [L1] transports it to a cocone with legs $R({\lambda }_j)\circ {\epsilon }_{D_j}^{-1}:D_j\to R(L)$, also for the empty indexing category.

Transport across isomorphisms preserves the existence and uniqueness clauses in [L3], so the resulting cocone is a colimit of $D$ in $\mathcal{A}$. The construction applies the reflector to $L$ and does not assert that $I(R(L))$ is isomorphic to $L$, so it does not assert preservation by the inclusion.

Let $D$ be any small diagram in $\mathcal{A}$. Completeness of $\mathcal{C}$ gives a limit of $ID$.

By [L1], the inclusion creates from that ambient limit a limit of $D$ in $\mathcal{A}$. This applies to every small $D$, including the empty diagram, so [L2] proves that $\mathcal{A}$ is complete.

Identity factorisations give reflexivity, exchanging the two factor maps gives symmetry, and composing factor maps gives transitivity. The same three operations work dually for epimorphisms.

Suppose $m=n\circ u$ and $n=m\circ v$. Then $m=m\circ v\circ u$, so monicity of $m$ gives $v\circ u=1_A$; similarly monicity of $n$ gives $u\circ v=1_B$. Thus $u$ and $v$ are inverse isomorphisms by [L2], and monicity makes each factor map unique. For epimorphisms the same equations are cancelled on the right, proving the dual claim.

If $m^{\mathrm{\prime }}=m\circ a$ and $m=m^{\mathrm{\prime }}\circ a^{\mathrm{\prime }}$ represent the same subobject, and similarly $n^{\mathrm{\prime }}$ and $n$, then any factorisation $m=n\circ u$ transports by composition to a factorisation of $m^{\mathrm{\prime }}$ through $n^{\mathrm{\prime }}$, and conversely. Thus the relation is independent of representatives by [L1].

Identity factorisations give reflexivity and composites give transitivity. If $[m]\le [n]$ and $[n]\le [m]$, then the representatives mutually factor, so [L1] makes their classes equal. The subobject relation is therefore reflexive, transitive and antisymmetric — the three conditions [L2] names — and by step 1.1 each is a statement about the classes rather than about chosen representatives.

For quotient representatives the same argument is dual, but the order is reversed because $[q]\le [r]$ means that $r$ factors through $q$. Reflexivity, transitivity, and antisymmetry follow by the epic half of [L1].

Nothing in steps 1.1--3.1 quantifies over a collection whose members are subobjects: each is a statement about monomorphisms into $C$ and about the factorisation relation between them, which is what the bracket notation abbreviates. Restricting that relation to a set of monomorphisms into $C$ carrying one representative per class therefore gives a relation on a set satisfying the three conditions of [L2], hence a partial order there.

If $I=\mathrm{\varnothing }$, every subobject of $C$ factors through $1_C:C\to C$, so $1_C$ represents the greatest subobject of $C$; the empty family imposes no lower-bound condition, so $[1_C]$ is its greatest lower bound and hence its intersection by [L4].

Suppose $I$ is nonempty and write $p=m_i\circ p_i$, independent of $i$. If $p\circ x=p\circ y$, then monicity of every $m_i$ gives $p_i\circ x=p_i\circ y$ for all $i$, and joint monicity in [L3] gives $x=y$; hence $p$ is monic. Each equality $p=m_i{p}_i$ makes $[p]\le [m_i]$. If $q:Q\to C$ factors through every $m_i$, its factor maps form a cone and [L1] gives a unique $u:Q\to P$ with $q=p\circ u$, so $[q]\le [p]$. Thus [L4] makes $[p]$ the intersection.

By [L5], representatives of the same subobject mutually factor and their factor maps are unique inverse isomorphisms over $C$. Composing a wide-pullback cone with these isomorphisms gives a cone for the replacement family, and [L1] supplies mutually inverse comparison maps between the two pullback apices. Their induced monomorphisms into $C$ therefore mutually factor, so they represent the same intersection subobject.

If $\mathcal{G}$ is separating and $f,g:X\to Y$ have equal images under every $\mathcal{C}(G,-)$, then $f\circ h=g\circ h$ for every $G\in \mathcal{G}$ and $h:G\to X$; [L4] forces $f=g$. Conversely, joint faithfulness says that distinct $f,g$ have unequal images under some $\mathcal{C}(G,-)$, which means some $h:G\to X$ satisfies $fh\mathrm{\ne }gh$. Thus the separating and joint-faithfulness conditions are equivalent.

Applying the same argument in the opposite category exchanges precomposition with postcomposition and proves that $\mathcal{H}$ is coseparating exactly when the contravariant representables $\mathcal{C}(-,H)$ are jointly faithful.

Regard $\mathcal{S}$ as the full subcategory it spans. Its objects form a set, and by [L2] the union of the hom-sets between them is a set, so this full subcategory is small. By [L1] its inclusion has a limiting cone $(L,p_S:L\to S{)}_{S\in \mathcal{S}}$. This remains valid when $\mathcal{S}$ is empty: then joint weak initiality implies that $\mathcal{C}$ has no objects, so the theorem's hypotheses cannot hold for a category with a target object.

Fix one target $C$. Joint weak initiality supplies some $S_0\in \mathcal{S}$ and one map $h:S_0\to C$, so $h\circ p_{S_0}:L\to C$ exists. The witness is chosen only for this fixed target, not simultaneously for a proper class of targets; hence $L$ is weakly initial.

By [L2] the collection $\mathcal{C}(L,L)$ is a set, so the one-object category whose arrows are the endomorphisms of $L$ is small, and sending its object to $L$ and each arrow to itself is a diagram. By [L1] that diagram has a limit; write its single leg as $j:I\to L$. The cone condition says exactly that $\alpha \circ j=j$ for every $\alpha \in \mathcal{C}(L,L)$, and $j$ is monic, since $j\circ x=j\circ y$ makes $x$ and $y$ both mediate the same cone, so the uniqueness clause of [L4] gives $x=y$.

$I$ is weakly initial: for a target $C$, step 2.1 supplies a map $L\to C$ and composing it with $j$ gives $I\to C$. Again one witness is used for one fixed target.

Let $f,g:I\rightrightarrows C$ and let $m:K\to I$ be their equalizer, which exists by [L1] and is monic by [L6]; it satisfies $fm=gm$ by [L5]. Step 2.1 gives $u:L\to K$, so $j\circ m\circ u:L\to L$ is an endomorphism of $L$ and step 2.2 gives $(j\circ m\circ u)\circ j=j$. Rewriting the left side as $j\circ (m\circ u\circ j)$ and cancelling the monomorphism $j$ by [L7] yields $m\circ u\circ j=1_I$. Hence $m$ is a split epimorphism as well as monic, so $m\circ (u\circ j\circ m)=(m\circ u\circ j)\circ m=m=m\circ 1_K$ and left-cancelling $m$ gives $u\circ j\circ m=1_K$; thus $m$ is an isomorphism by [L7]. From $fm=gm$ and the invertibility of $m$ we get $f=g$. There is therefore exactly one morphism $I\to C$ for every target $C$, so $I$ is an initial object of $\mathcal{C}$.

If $({\eta }_i)$ is a solution set and $(A,f)$ is any comma object, the defining factorisation gives $i$ and $h:A_i\to A$ with $f=U(h){\eta }_i$. By [L1] this is a comma morphism from $(A_i,{\eta }_i)$ to $(A,f)$, so [L2] gives joint weak initiality. If the supplied set is empty, the same assertion says the comma category has no objects.

Conversely, if the corresponding comma objects are jointly weakly initial, apply [L2] to each $(A,f)$. The resulting comma morphism has, by [L1], exactly the equation $f=U(h){\eta }_i$ required by the solution-set condition.

The arrows ${\lambda }_j:C\to U(A_j)$ form a cone over $U\mathrm{\Pi }D$. Since $U(L)$ with legs $U(p_j)$ is limiting, [L2] gives a unique $\lambda :C\to U(L)$ satisfying $U(p_j)\lambda ={\lambda }_j$ for every $j$. By [L1], the same legs $p_j$ are comma morphisms from $(L,\lambda )$.

Given any comma cone with apex $(B,\beta )$, the projected limit supplies a unique $h:B\to L$ with $p_{j}h$ equal to its legs. Both $U(h)\beta$ and $\lambda$ have the same composites with every $U(p_j)$, so uniqueness of the preserved limit gives $U(h)\beta =\lambda$; hence $h$ is the unique comma morphism. The lifted cone is limiting.

The arrow $\lambda$ in step 1.1 is forced by the projected apex and legs, so the lift is unique on the nose. For an empty indexing category, preservation of the terminal object gives the unique map $C\to U(L)$ by the same limit property. Thus all strict-creation clauses in [L3] hold, including the degenerate and empty diagrams.

The comma category is locally small because each of its hom-collections is a subset of a hom-set in $\mathcal{A}$, which is a set by [L2]. Every small projected diagram has a limit by the completeness of $\mathcal{A}$ in the sense of [L1], and $U$ preserves that limit because $U$ is continuous in the sense of [L6], so [L4] creates its limit in the comma category. Hence $(C\downarrow U)$ is complete and locally small, and [L3] supplies a jointly weakly initial set.

Apply [L5] to obtain an initial object of $(C\downarrow U)$. Its construction uses only the supplied solution set for this fixed $C$, so it performs no simultaneous selection over all objects of $\mathcal{C}$.

The supplied initial comma objects satisfy precisely the hypothesis of [L2], so they assemble into a functor $F:\mathcal{C}\to \mathcal{A}$ and an adjunction $F⊣U$.

The assembly uses the supplied family, not merely the separate existential conclusions of [L1]; therefore no unrecorded proper-class choice is hidden in the functor form.

Form the set-indexed product $P={\prod }_{K\in \mathrm{\Phi }}K$. By the Statement's hypothesis in the sense recorded in [L4], the collection of all subobjects of $P$, possibly a proper collection, has an intersection $i:I\to P$. This invokes that order-theoretic greatest lower bound directly and does not form a proper-class diagram.

Fix an object $C$. By [L2], the canonical evaluation map ${\nu }_C:C\to {\prod }_{K\in \mathrm{\Phi }}{\prod }_{f\in \mathcal{C}(C,K)}K$ is a set-indexed product map, and [L3] makes it monic. Repeating each projection of $P$ defines $\mathrm{\Delta }:P\to {\prod }_{K\in \mathrm{\Phi }}{\prod }_{f\in \mathcal{C}(C,K)}K$. Pull back ${\nu }_C$ along $\mathrm{\Delta }$ to obtain $P_C\to P$ with a map $P_C\to C$; that pullback of the monomorphism ${\nu }_C$ is monic by [L7], so $P_C\to P$ is a subobject of $P$. Since $i$ lies below every subobject of $P$, [L4] gives $I\to P_C$, hence a map $I\to C$.

If $f,g:I\rightrightarrows C$, their equalizer $e:E\to I$ is monic by [L5], and the composite $i\circ e:E\to P$ is monic by [L6], hence a subobject of $P$. Minimality of $i$ gives a factorisation $I\to E$ over $P$, so $e$ and $1_I$ — the latter monic by [L6] — represent the same subobject and $e$ is invertible. Therefore $f=g$. Step 2.1 gives existence and this step gives uniqueness for every target, so $I$ is initial.

The comma category is locally small because its hom-collections are subsets of those in $\mathcal{A}$. The set of all comma objects $C\to U(K)$ with $K$ in the supplied coseparating set is again a set by local smallness of $\mathcal{C}$, and it is coseparating by [L5].