Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 21 (Category Theory)

The general adjoint functor theorem says that, for a functor $U:\mathcal{A}\to \mathcal{B}$ between locally small categories, with $\mathcal{A}$ a complete category, $U$ has a left adjoint if and only if it preserves small limits and satisfies the solution-set condition. The solution-set condition means that for each $B\in \mathcal{B}$ there is a set of arrows $b_i:B\to U{A}_i$ such that every $b:B\to UA$ factors as $U(a)b_i$ for some $i$ and some $a:A_i\to A$.

Here is the smallness argument behind the theorem. A complete category that is locally small and has a weakly initial set $(W_i)$ has an initial object. Form the product in a category $W={\prod }_i{W}_i$. It is weakly initial, since a projection $W\to W_i$ followed by a chosen arrow $W_i\to X$ gives an arrow $W\to X$. Take the simultaneous equalizer $e:E\to W$ of all endomorphisms of $W$ with $1_W$; the family is a set by local smallness. The object $E$ is also weakly initial. For parallel arrows $a,b:E\rightrightarrows X$, take their equalizer $j:Y\to E$ and a map $t:W\to Y$. Since $ejt$ is an endomorphism of $W$, the defining property of $e$ gives $ejte=e$. Cancellation of the monomorphism $e$ gives $jte=1_E$. Thus the monomorphism $j$ has a right inverse and is an isomorphism, so $a=b$. Existence of arrows out of $E$ was already established, proving the initial-object lemma for complete categories with a weakly initial set.

If $U$ preserves small categorical limits, the comma category $(B\downarrow U)$ is complete: take the limiting object in $\mathcal{A}$, and use preservation to assemble the arrows from $B$. It is locally small, and the solution-set condition provides a weakly initial set. Its initial object is an arrow ${\eta }_B:B\to ULB$ with a natural bijectionFor a morphism $v:B\to B^{\prime }$, define $Lv$ by $U(Lv){\eta }_B={\eta }_{B^{\prime }}v$. Uniqueness proves functoriality and naturality, giving a left adjoint $L⊣U$. Conversely, a right adjoint preserves categorical limits, because applying the adjunction converts a limiting cone over a diagram into the corresponding limiting cone of hom-sets. Its adjunction unit at $B$ is a singleton solution set. This proves both directions of the general adjoint functor theorem.

Apply it to the inclusion $J:\mathbf{KHaus}\hookrightarrow \mathbf{Top}$, where $\mathbf{KHaus}$ is the full subcategory of compact Hausdorff spaces. A small product in a category of compact Hausdorff spaces is again compact Hausdorff by the Tychonoff theorem. An equalizer of two continuous functions into a Hausdorff space is closed, hence compact Hausdorff. Products and equalizers construct all small categorical limits, and the inclusion preserves these constructions. Both categories are locally small.

Fix a topological space $X$ and a continuous function $f:X\to K$ into a compact Hausdorff space. The closure $K_f=\overline{f(X)}$ is compact Hausdorff and has a dense subset of cardinality at most $|X|$. The permitted cardinal estimate givesFor $X=\varnothing$, use $K_f=\varnothing$. Choose a set of representatives of all compact Hausdorff topologies on underlying sets of cardinality at most this bound, and all continuous functions from $X$ to those representatives. Every $f$ factors through one of them, using the inclusion $K_f\hookrightarrow K$. This is the solution-set condition. Consequently the inclusion of compact Hausdorff spaces has a left adjoint, the compact Hausdorff reflection:The universal arrow $X\to RX$ has dense image: factoring it through its closed image and applying uniqueness supplies an inverse to that image inclusion. It need not be an embedding when $X$ is not sufficiently separated; the assertion is a compact Hausdorff reflection, rather than a claim that every topological space embeds in a compact Hausdorff space.

For completeness, the alternative Special adjoint functor theorem can be proved from the same initial-object argument. Its limit form assumes that $\mathcal{A}$ is complete, locally small and well-powered, with a small cogenerating family $(Q_i{)}_{i\in I}$; then $U:\mathcal{A}\to \mathcal{B}$, with $\mathcal{B}$ locally small, has a left adjoint precisely when it preserves small limits. To prove sufficiency, start with $b:B\to UA$ and intersect all subobjects $M\hookrightarrow A$ through which $b$ factors after applying $U$. Well-poweredness makes this a small intersection of subobjects, and preservation of limits gives a smallest supporting object $(M,b_M)$. For maps $s,t:M\rightrightarrows Q_i$, equality $U(s)b_M=U(t)b_M$ makes their equalizer another supporting subobject. Minimality forces that equalizer to be invertible, so $s=t$. Thereforeis an injective function. The evaluation embedding into cogenerator products embeds $M$ in ${\prod }_i{Q}_i^{S_i}$, where $S_i$ is the realized subset of the fixed set $\mathcal{B}(B,U{Q}_i)$. There are only a set of choices of these subsets, only a set of subobjects of each resulting product, and only a set of arrows from $B$ to each $U$-image. They give a solution set, so the already proved general adjoint functor theorem applies. Necessity is again preservation of limits by a right adjoint. For $\mathbf{KHaus}$, monomorphisms are embeddings onto closed subspaces, and $[0,1]$ is a coseparator because continuous functions to it separate points. Thus the Special adjoint functor theorem gives the same reflection directly.

The Yoneda embedding is an exact representation of a category, but does not by itself reduce the subject to the study of arbitrary functor categories. There are both a size issue and a structure issue.

Let $\mathcal{A}$ be a locally small category and set $yA=\mathcal{A}(-,A)$. For a categorical presheaf $P:{\mathcal{A}}^{\mathrm{op}}\to \mathbf{Set}$, a natural transformation $\theta :yA\to P$ is determined by $x={\theta }_A(1_A)$. Indeed, naturality along $f:B\to A$ forcesConversely this formula defines a natural transformation for each $x\in P(A)$, since functoriality gives naturality along every $g:C\to B$. Taking $P=y{A}^{\prime }$ givesThus the Yoneda embedding is full and faithful, and $\mathcal{A}$ is equivalent to its full subcategory of representable presheaves. The indexing category in the usual version is ${\mathcal{A}}^{\mathrm{op}}$, not necessarily a small category chosen independently of $\mathcal{A}$.

Local smallness makes the values $\mathcal{A}(B,A)$ sets; it does not make the collection of objects a set. For a large indexing category, the collection of natural transformations between arbitrary set-valued functors can itself be a proper class. For example, take the discrete category on the class of ordinals and the constant two-element functor. Its endomorphisms include an independently chosen function $\{0,1\}\to \{0,1\}$ at each ordinal. A small functor category cannot encode this collection without changing universes. The Yoneda lemma remains meaningful when the source is representable because the displayed bijection identifies its natural transformations with a set. It does not make the entire large functor category locally small.

Even for a small category, the representable image is usually a very special part of its presheaf category. If $\mathcal{A}$ is the terminal category, its presheaf category is $\mathbf{Set}$, but its representable image consists only of a singleton. The presheaf category has all small categorical limits and colimits, whereas an arbitrary full subcategory need have almost none. The Yoneda embedding preserves existing limits because representable functors preserve limits. It generally does not preserve colimits. For example, the coproduct of two singleton sets in category of finite sets is a two-element set, but evaluating the proposed presheaf coproduct at a two-element set givesThe pointwise coproduct therefore leaves the representable image. Identifying which presheaves are representable, and which ambient constructions return to the original category, is precisely substantive category theory.

The representation is nonetheless powerful. The Yoneda lemma replaces a morphism by all its probes, converts a universal property into a representation of a functor, and makes the presheaf category a natural place to add formal colimits. A small presheaf category is the free free cocompletion under small colimits: every presheaf is a colimit of representables indexed by its category of elements. But this is a controlled enlargement of the category, not an identification of its original objects with all functors. The stronger conclusion in the proposed assertion therefore does not follow from the full-faithfulness statement.

A balanced category is one in which every morphism that is both a monomorphism and an epimorphism is an isomorphism. A strong monomorphism $m:A\to B$ is a monomorphism right orthogonal to all epimorphisms: whenever $e:X\to Y$ is epic and $a:X\to A$, $b:Y\to B$ satisfy $ma=be$, there is a unique $d:Y\to A$ with $de=a$ and $md=b$. A regular monomorphism is an equalizer of some parallel pair of morphisms. Its equalizer property supplies the indicated lift after epic cancellation, so every regular monomorphism is strong.

If every monomorphism is strong and $m:A\to B$ is also epic, apply strongness to the square whose left and right sides are both $m$, with horizontal arrows $1_A$ and $1_B$. The lift $d:B\to A$ satisfies $dm=1_A$ and $md=1_B$. Thus $m$ is invertible and the category is balanced.

Conversely, suppose the category is balanced and has pullbacks in a category. Given the lifting square above, form $P=A{\times }_{B}Y$ with projections $p:P\to Y$, $q:P\to A$. The pair $(a,e)$ gives $h:X\to P$ with $ph=e$, $qh=a$. The pullback of the monomorphism $m$ is monic, so $p$ is monic. It is also epic: $rp=sp$ implies $re=se$, hence $r=s$ because $e$ is epic. Balancedness makes $p$ invertible. Now $d=q{p}^{-1}$ satisfies both required equations, and monicity of $m$ proves uniqueness. Hence in a balanced category with pullbacks, every monomorphism is strong.

For the finite example, choose a strong monomorphism $m:A\to B$ that is not regular. It is not an isomorphism, since an isomorphism is the equalizer of an identical pair. It is not epic either, since an epic strong monomorphism is invertible by the first argument. Consequently there are distinct arrows $r,s:B\rightrightarrows D$ with $rm=sm$. Since $m$ is not the equalizer of this pair, there is a map $h:X\to B$ with $rh=sh$ that does not factor through $m$. Monicity rules out failure of uniqueness, so failure of existence is the only possible failure of its equalizer property.

Define ${\mathcal{C}}_0$ to have four formally distinct objects $a,b,x,d$ and exactly the following six nonidentity arrows:Their only nontrivial composites areThere are no longer composable strings of nonidentity arrows, so these rules define an associative category. Sending $a,b,x,d$ to $A,B,X,D$, respectively, and the six arrows to $m,h,r,s,rm,rh$, defines a functor. It is a faithful functor: every hom-set has at most one arrow except ${\mathcal{C}}_0(b,d)=\{r_0,s_0\}$, and their images are distinct. Faithfulness does not require the four image objects to be distinct.

The map $m_0$ is monic, by direct inspection of arrows into $a$. Neither $m_0$ nor $h_0$ is epic, since the distinct pair $r_0,s_0$ agrees after either. Every other nonidentity arrow has target $d$ and is epic, because there is only one arrow out of $d$. A lifting square against $m_0$ whose left side is one of these epimorphisms would require an arrow $d\to b$, which does not exist. Squares whose left side is an identity have the evident unique lift. Thus $m_0$ is a strong monomorphism.

It is not a regular monomorphism. The only distinct parallel pair out of $b$ is $r_0,s_0$, and $h_0$ equalizes it but has no factorization through $m_0$, since there is no arrow $x\to a$. An identical pair has $1_b$ as its equalizer and cannot have the noninvertible $m_0$ as its equalizer. Therefore

Write $\eta :1_{\mathcal{C}}\to GF$ and $\epsilon :FG\to 1_{\mathcal{D}}$ for the unit and counit of an adjunction. The induced monad hasThe Eilenberg-Moore comparison functor sends $D\in \mathcal{D}$ to the algebra for a monadand sends $v:D\to D^{\prime }$ to $Gv$. The triangle identities for an adjunction give the algebra unit law, and naturality of the counit gives the algebra multiplication law and the algebra-morphism equation for $Gv$.

If $\mathcal{D}$ has coequalizers, define the Left adjoint to the Eilenberg-Moore comparison functor at an algebra $(A,a:TA\to A)$ byThe parallel pair is reflexive through $F{\eta }_A$. A morphism $L(A,a)\to D$ is equivalently an arrow $b:FA\to D$ with $bFa=b{\epsilon }_{FA}$. Under $F⊣G$, its transpose is $\bar{b}=Gb{\eta }_A:A\to GD$. The transpose of $bFa$ is $\bar{b}a$, while the transpose of $b{\epsilon }_{FA}$ is $G{\epsilon }_{D}T\bar{b}$. Hence the coequalizer condition is exactlythe equation for a morphism of algebras for a monad $(A,a)\to K(D)$. This supplies the natural bijection proving $L⊣K$, and also defines $L$ on algebra morphisms by the coequalizer universal property.

Let $U_n:{\mathcal{C}}_{n+1}\to {\mathcal{C}}_n$ forget the last partial unary operation. For an object $A$ of the nested partial unary operation category ${\mathcal{C}}_n$, setThe nesting rule implies that every earlier operation is defined and fixes such an $x$. To construct a free extension of nested partial unary operations, adjoin a disjoint infinite chain for each $x\in S_n(A)$:Keep the old operations on $A$. On each new chain put ${\alpha }_1(x,k)=(x,k+1)$; every old operation ${\alpha }_i$ with $2\le i\le n$ is undefined there, because the first operation has no fixed points there. Define the new operation only on $S_n(A)$, by ${\alpha }_{n+1}(x)=(x,0)$. These are exactly the fixed points of the old top operation, so the required domain rule is satisfied. The inclusion $A\hookrightarrow U_n{L}_{n}A$ preserves all old defined operations.

For a morphism $f:A\to U_{n}B$, preservation of the old operations makes $f(x)\in S_n(B)$ whenever $x\in S_n(A)$. Any extension to a morphism $\hat{f}:L_{n}A\to B$ must haveThe operation ${\beta }_{n+1}$ is defined at $f(x)$, and ${\beta }_1$ is total, so this formula exists. It preserves the new operation and the chain operation; there are no other defined operations on new points to check. It therefore gives the unique extension. Naturality and functoriality follow from uniqueness. ThusNotice the useful extra fact: the new top operation has no fixed points, since every value $(x,0)$ is new and every input $x$ is old. Hence freely adding any further operations adds no more points; those operations have empty domains.

Put ${\mathcal{C}}_0=\mathbf{Set}$. The initial free construction $L_0:\mathbf{Set}\to {\mathcal{C}}_1$ sends $A$ to $A\times \mathbb{N}$ with ${\alpha }_1(a,k)=(a,k+1)$. A function $f:A\to UB$ extends uniquely by $(a,k)\mapsto {\beta }_1^k(f(a))$. Its induced monad is $T_{0}A=A\times \mathbb{N}$, with unit $a\mapsto (a,0)$ and multiplication $(a,k,l)\mapsto (a,k+l)$. An algebra for a monad is determined by the action of $1\in \mathbb{N}$, so its Eilenberg-Moore category is ${\mathcal{C}}_1$. Thus this first free–forgetful adjunction is a monadic adjunction without needing the given assumption for $n=0$.

The monadic length counts successive Eilenberg-Moore comparison functor steps until the comparison becomes an equivalence. Fix $0\le m<n$. For $m\ge 1$, the free construction in the preceding part adds the $(m+1)$st operation with no fixed points. For $m=0$, the free first operation also has no fixed points. Consequently the left adjoint $F_{n,m}$ to the composite forgetful functor $U_{n,m}:{\mathcal{C}}_n\to {\mathcal{C}}_m$ is obtained by the one-step free construction, followed by empty higher operations. In particular, its underlying object of ${\mathcal{C}}_m$, unit and multiplication are exactly those of the one-step adjunction:as monads, not merely as object functions. The multiplication agrees because both counits evaluate the same first new operation and the same chains; no later operation has a value on a free object.

By the assumed one-step monadic adjunction property, or the explicit argument above when $m=0$, the Eilenberg-Moore category ${\mathcal{C}}_m^{T_m}$ is equivalent to ${\mathcal{C}}_{m+1}$. Under this equivalence the comparison ${\mathcal{C}}_n\to {\mathcal{C}}_m^{T_m}$ is precisely $U_{n,m+1}$. Indeed, its algebra action evaluates each newly adjoined chain using ${\alpha }_1$, and evaluates the first new value using ${\alpha }_{m+1}$; it retains that operation and ignores all higher ones. The comparison has a left adjoint by the same explicit free extension. Thus the whole monadic tower for nested partial unary operations iswith the remaining comparison at stage $m$ equal to $U_{n,m}$, and stage $n$ the identity.

It does not terminate earlier. If $m<n$, take the same underlying two-element set in two objects of ${\mathcal{C}}_n$, making every operation up to $m$ the identity. In the first object make all remaining operations identities; in the second make ${\alpha }_{m+1}$ exchange the two elements, and leave all subsequent operations undefined. Both satisfy the nesting rule. The identity function is a morphism after forgetting to ${\mathcal{C}}_m$, but is not a morphism between the original objects. Thus $U_{n,m}$ is not a full functor, and cannot be an equivalence. Therefore

Let $F:\mathcal{C}\to \mathcal{D}$ be a final functor, and let ${\lambda }_c:GFc\to V$ be a cocone under a diagram $GF$. For $d\in \mathcal{D}$, choose an object $(c,a:d\to Fc)$ of the nonempty comma category $(d\downarrow F)$ and defineIf $t:(c,a)\to (c^{\prime },a^{\prime })$ is a morphism in that comma category, then $Fta=a^{\prime }$, so the cocone equation gives ${\lambda }_{c}G(a)={\lambda }_{c^{\prime }}G(Ft)G(a)={\lambda }_{c^{\prime }}G(a^{\prime })$. Equality also holds along a reversed arrow, and hence along any zigzag. Connectedness makes the definition independent of the chosen object. For $v:d\to d^{\prime }$, use $(c,a:d^{\prime }\to Fc)$ and $(c,av:d\to Fc)$ to obtain ${\bar{\lambda }}_d={\bar{\lambda }}_{d^{\prime }}Gv$. Thus $\bar{\lambda }$ is a cocone under $G$. Taking $(c,1_{Fc})$ shows that it extends $\lambda$.

Every extension must satisfy ${\bar{\lambda }}_d={\lambda }_{c}G(a)$ by its cocone equation, proving uniqueness. The construction commutes with postcomposition by a morphism $V\to V^{\prime }$. Hence restriction gives a natural bijection between the sets of cocones of $G$ and $GF$ with any fixed vertex. A universal cocone for $GF$ therefore extends to a universal cocone for $G$. In particular,whenever the latter exists, and existence of all colimits of shape $\mathcal{C}$ implies existence of all colimits of shape $\mathcal{D}$.

For the factorization of an arbitrary functor $H:\mathcal{C}\to \mathcal{E}$, letthe set of connected components of a category, where components are defined by finite zigzags. Precomposition by $v:e\to e^{\prime }$ gives a functor $(e^{\prime }\downarrow H)\to (e\downarrow H)$ and therefore a map $P(v):P(e^{\prime })\to P(e)$. Identity and composition laws make $P$ a categorical presheaf. This is a set-valued construction when the relevant comma categories are small or have only a set of components, as in the usual small-category convention for this factorization.

Take $\mathcal{D}$ to be the category of elements of this presheaf. Its objects are pairs $(e,S)$ with $S\in P(e)$; a morphism $(e,S)\to (e^{\prime },S^{\prime })$ is an arrow $v:e\to e^{\prime }$ satisfying $P(v)(S^{\prime })=S$. Define $G:\mathcal{D}\to \mathcal{E}$ by forgetting $S$. Given $v:e\to e^{\prime }$ and an object $(e^{\prime },S^{\prime })$, the unique lift into that object has source $(e,P(v)(S^{\prime }))$ and underlying arrow $v$. This proves that $G$ is a discrete fibration.

Define $F(c)=(Hc,[(c,1_{Hc})])$. For $t:c\to c^{\prime }$, take $F(t)=Ht$. The arrow $t$ in $(Hc\downarrow H)$ joins $(c,1_{Hc})$ to $(c^{\prime },Ht)$, so the required component equation holds. It follows that $F$ is a functor and $GF=H$.

Finally, an object of $((e,S)\downarrow F)$ is a pair $(c,v:e\to Hc)$ whose component in $(e\downarrow H)$ is $S$. Its morphisms are exactly the morphisms of that comma category between objects in $S$. Consequently $((e,S)\downarrow F)$ is precisely the full connected component $S$, and is nonempty and connected. Thus $F$ is final, and every such functor factors as a final functor followed by a discrete fibration.

A regular category has finite limits, coequalizers of kernel pairs, and regular epimorphisms stable under pullback. Equivalently, every morphism factors as a regular epimorphism followed by a monomorphism, and these factorizations are stable under pullback. A regular functor preserves finite limits and regular epimorphisms, hence also preserves these image factorizations.

For this question, the capital regular category convention is the one in which every well-supported object is a well-pointed object in a category. Here an object $A$ is well-supported if $A\to 1$ is a regular epimorphism. It is well-pointed if a subobject $m:B\hookrightarrow A$ through which every point $1\to A$ factors must be the whole object. This convention is documented in Definitions 1.1–1.3. It imposes no such point-detection requirement on objects whose support is a proper subobject of $1$. In particular, it should not be replaced here by the stronger assertion that $\mathcal{A}(1,-)$ reflects every isomorphism.

First prove two facts about a capital regular category $\mathcal{A}$. Every well-supported object $A$ has a point. Otherwise $A\times A$ is well-supported and has no points, so its diagonal $A\hookrightarrow A\times A$ contains all points and is invertible. That makes $A$ a subterminal object; the regular epimorphism $A\to 1$ is then also monic and hence invertible, contradicting the absence of a point. If $e:X\to Y$ is a regular epimorphism and $y:1\to Y$ is a point, its pullback in a category is a well-supported object over $1$, so it has a point lifting $y$. Thus the representable functorpreserves regular epimorphisms; it already preserves all existing limits. Hence $\Gamma$ is a regular functor. Second, if $\Gamma X$ is a singleton, its unique point $x:1\to X$ is a monomorphism, $X$ is well-supported, and $x$ contains all points of $X$. Well-pointedness makes $x$ invertible. Thus $\Gamma$ reflects terminal objects, even though it need not reflect arbitrary isomorphisms.

Now let $\mathcal{C}$ be a small category that is a regular category. For each object $B$, its slice category $\mathcal{C}/B$ is small and regular: finite limits, images and the stability of regular epimorphisms are computed using the corresponding constructions in $\mathcal{C}$. By the permitted capitalization result choose a conservative functor that is regular,with ${\mathcal{A}}_B$ a small capital regular category. The base-change functor $B^{\ast }:\mathcal{C}\to \mathcal{C}/B$, given by $X\mapsto (X\times B\to B)$, preserves finite limits and regular epimorphisms. So doesTaking these as the coordinates defines a regular functorThe exponent is a set of indices, so this is a power of the Category of sets, with finite limits and regular epimorphisms computed coordinatewise.

To prove reflection of isomorphisms, suppose $R(f)$ is invertible for $f:A\to B$. In $\mathcal{C}/B$, the object $(A,f)$ is the pullback of $B^{\ast }f:(A\times B\to B)\to (B\times B\to B)$ along the point given by the diagonal $B\to B\times B$. Applying ${\Gamma }_B{J}_B$ shows that ${\Gamma }_B{J}_B(A,f)$ is the fibre of the bijection $R_B(f)$ over this diagonal point, and is a singleton. The preceding lemma gives $J_B(A,f)\cong 1$ through its canonical arrow to $1$. Since $J_B$ reflects isomorphisms, $(A,f)\to (B,1_B)$ is invertible. Its underlying morphism is $f$. Therefore

In the small-category setting of this representation result, the elementary condition for replacing the power by one copy of $\mathbf{Set}$ is:This is almost total support for a regular category. “Strict initial” means initial and every arrow into it is invertible; the condition allows the case in which all objects are well-supported and no initial object exists.

For necessity, let $V:\mathcal{C}\to \mathbf{Set}$ be regular and conservative. If $VA\ne \varnothing$, factor $A\to 1$ through its support of an object in a regular category $S\hookrightarrow 1$. Preservation of images gives $VS=1$. Conservativity makes $S\to 1$ invertible, so $A$ is well-supported. If $VA=\varnothing$, then for every $X$ the projection $A\times X\to A$ is sent to the bijection $\varnothing \to \varnothing$, hence is invertible. Its inverse followed by the other projection supplies an arrow $A\to X$. For any two such arrows, their equalizer is sent to a bijection onto $VA$, hence is invertible, proving uniqueness. Thus $A$ is initial. If $X\to A$ exists, then $VX\to \varnothing$ exists, so $VX=\varnothing$ and that arrow is sent to a bijection. Conservativity makes it invertible. Hence $A$ is strict initial.

For sufficiency, choose the permitted conservative regular functor $J:\mathcal{C}\to \mathcal{A}$ into a small capital regular category, and put $V=\Gamma J$. This is regular. A well-supported object of $\mathcal{C}$ is sent to a well-supported object of $\mathcal{A}$, which has a point. A non-well-supported object is strict initial by the condition, and is consequently a proper subterminal object. The functor $J$ preserves subterminality and reflects the noninvertibility of its arrow to $1$; its image is a proper subterminal object and has no points. Thus $VA$ is empty exactly on the non-well-supported objects.

Suppose $Vf$ is a bijection for $f:A\to B$. If both sets are empty, $A$ and $B$ are strict initial objects and $f$ is invertible. Otherwise $A,B$ are well-supported. Factor $Jf$ as a regular epimorphism followed by a monomorphism $m:I\hookrightarrow JB$. Surjectivity of $Vf$ makes every point of $JB$ factor through $m$; capitality of $\mathcal{A}$ makes $m$ invertible. Hence $Jf$ is a regular epimorphism. Its kernel pair $P=JA{\times }_{JB}JA$ is well-supported, because $P\to JA$ and $JA\to 1$ are regular epimorphisms. Injectivity of $Vf$ implies that every point of $P$ has equal coordinates, and therefore factors through the diagonal $JA\hookrightarrow P$. Since $P$ is well-pointed, this diagonal is invertible. Thus $Jf$ is also monic, hence an isomorphism, and conservativity of $J$ makes $f$ invertible. This proves sufficiency with actual point and image arguments, rather than asserting that global sections of every capital regular category is conservative.