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Identify the reflective subcategory with its full replete image, writing ${\eta }_X:X\to LX$ for the reflection unit. An object $Z$ is fixed by $L$ exactly when precomposition with every ${\eta }_X$ gives a bijection $\mathcal{E}(LX,Z)\to \mathcal{E}(X,Z)$. One implication is the reflection adjunction; for the other, take $X=Z$ to obtain a retraction of ${\eta }_Z$, and uniqueness of maps into the fixed object $LZ$ makes it a two-sided inverse.

An ambient finite limit of fixed objects has this same property: its hom-sets are limits of the hom-sets of the diagram, and a limit of these bijections is a bijection. Thus the ambient terminal object, products and equalizers are fixed and retain their universal properties in the full subcategory. The reflective subcategory is Cartesian, with finite limits computed in the ambient topos.

Let $J:\mathcal{C}\hookrightarrow \mathcal{B}$ exhibit a reflective subcategory, with reflector $K$, and let $E:I\to \mathcal{C}$ be a small diagram in a category. Completeness of $\mathcal{B}$ gives a categorical limit $L$ of $JE$. For every $B^{\prime }\in \mathcal{B}$, the universal property of this categorical limit and the reflection adjunction giveThe composite is precomposition with ${\rho }_{B^{\prime }}$, so $L$ satisfies the hom-set condition from the preceding part. Its proof of invertibility of ${\rho }_L$ did not require repleteness. Thus $L\cong JKL$ whether or not the chosen full reflective subcategory is replete.

Transport the ambient limit cone along ${\rho }_L^{-1}:JKL\to L$. Its legs lie in the full subcategory, and their ambient universal property, restricted to objects of $\mathcal{C}$, is exactly the internal categorical limit property. Hence every small diagram in the reflective subcategory has a limit:For a replete subcategory, the ambient limit object $L$ itself belongs to $\mathcal{C}$. This argument includes the empty diagram and requires no limit-preservation hypothesis on the reflector.

A monomorphism $m:A\to B$ is left cancellable: $mu=mv$ implies $u=v$ for all parallel arrows into $A$. A strong monomorphism is a monomorphism with the following lifting property: if $e:X\to Y$ is an epimorphism and $u:X\to A$, $v:Y\to B$ satisfy $mu=ve$, there is a diagonal $d:Y\to A$ with $de=u$ and $md=v$. It is unique by monicity. An extremal monomorphism is a monomorphism $m$ for which every factorization $m=he$ with $e$ an epimorphism forces $e$ to be an isomorphism. We use the strong lifting formulation for the paper's terminology; below we also justify its equivalence with the extremal formulation under the complete, well-powered hypotheses. A regular monomorphism is an equalizer of two parallel arrows.

First, a strong monomorphism is extremal. In a factorization $m=he$, apply its lifting property with upper arrow $1_A$ and lower arrow $h$. This gives $d$ with $de=1_A$. Since $e$ is an epimorphism, $ede=e$ implies $ed=1$, so $e$ is an isomorphism.

If $m:A\to B$ is an equalizer of $r,s:B\rightrightarrows C$, consider a lifting square as above. The equation $rve=smu=sve$ and the epimorphism property give $rv=sv$. The equalizer property therefore gives a unique $d$ with $md=v$. Monicity then gives $de=u$. Hence every regular monomorphism is a strong monomorphism.

For the intersection of strong subobjects, let $m_i:A_i\to B$ be strong and let $p_i:P\to A_i$ exhibit their intersection, with $m=m_1{p}_1=m_2{p}_2$. Given $mu=ve$ with $e$ an epimorphism, the two strong monomorphisms supply diagonals $d_i:Y\to A_i$ satisfying $d_{i}e=p_{i}u$ and $m_i{d}_i=v$. The pullback in a category property supplies $d:Y\to P$ with $p_{i}d=d_i$. Its two projections show $de=u$, and $md=v$. Thus the intersection is strong. The same argument works for every existing small intersection of strong monomorphisms. Also, composites of strong monomorphisms are strong: first lift through the outer one, then through the inner one.

Write $\mathcal{S}$ for the full subcategory formed by every saturated object with respect to anodyne morphisms, using the paper's anodyne morphism in a category convention. Let $m:C\to B$ be a strong monomorphism and $B$ saturated. For an anodyne morphism in a category $e:X\to Y$ and $u:X\to C$, saturation extends $mu$ to $v:Y\to B$. Since $e$ is an epimorphism, the strong lifting property supplies $d:Y\to C$ extending $u$. Therefore every strong subobject of a saturated object is saturated.

Now assume $\mathcal{C}$ is a complete category and a well-powered category, with the given saturated embeddings. Choose a monomorphism $a:A\to B$ into a saturated object. Intersect all strong subobjects of $B$ through which $a$ factors. They form a set by the well-powered category hypothesis, and the identity of $B$ is one of them. Completeness constructs their intersection, so we obtainBy the intersection of strong subobjects, $m$ is strong; hence $LA$ is saturated. Since $a$ is a monomorphism, so is ${\eta }_A$.

To prove that ${\eta }_A$ is an epimorphism, take $u,v:LA\rightrightarrows D$ with $u{\eta }_A=v{\eta }_A$. Their equalizer $j:E\to LA$ is a regular monomorphism, and ${\eta }_A$ factors through it. The composite $mj:E\to B$ is strong, so it belongs to the family defining $LA$. Minimality gives $t:LA\to E$ with $mjt=m$. Cancelling $m$ gives $jt=1_{LA}$; since $j$ is a monomorphism, also $tj=1_E$. Thus $j$ is an isomorphism and $u=v$. We have provedThis is the saturated reflection from a strong-subobject intersection.

The construction of a least strong subobject and the equalizer argument work for any morphism into any object, without requiring the source map to be monic or the target to be saturated. They give an epimorphism followed by a strong monomorphism. Applying this factorization to an extremal monomorphism makes the first factor invertible; thus the extremal and strong formulations agree under the present complete, well-powered hypotheses.

For every saturated $S$, saturation extends every $f:A\to S$ across the anodyne morphism in a category ${\eta }_A$ to $\bar{f}:LA\to S$. Its extension is unique because ${\eta }_A$ is an epimorphism. Consequentlynaturally in $S$ and $A$. In particular, for $h:A\to A^{\prime }$ define $Lh$ by $Lh{\eta }_A={\eta }_{A^{\prime }}h$; uniqueness proves the identity and composition laws. This constructs the reflector and proves that $\mathcal{S}$ is a reflective subcategory.

Finally we prove that $\mathcal{S}$ is a balanced category. If $e:U\to V$ is epic in $\mathcal{S}$ and $u,v:V\to X$ satisfy $ue=ve$, embed $X$ by a monomorphism $j:X\to Y$ into a saturated object. Fullness makes $ju,jv$ arrows of $\mathcal{S}$, so $ju=jv$ and hence $u=v$. Thus an epimorphism in $\mathcal{S}$ is also epic in $\mathcal{C}$.

A monomorphism $f:U\to V$ in $\mathcal{S}$ is also monic in $\mathcal{C}$: for $a,b:X\to U$, extend them uniquely along ${\eta }_X$ to $\bar{a},\bar{b}:LX\to U$. If $fa=fb$, cancellation of the ambient epimorphism ${\eta }_X$ gives $f\bar{a}=f\bar{b}$, so monicity in $\mathcal{S}$ gives $\bar{a}=\bar{b}$ and $a=b$. Now if $f$ is both monic and epic in $\mathcal{S}$, it is an anodyne morphism in a category in $\mathcal{C}$. Saturation of $U$ extends $1_U$ across $f$ to $r:V\to U$ with $rf=1_U$. Ambient epicity gives $fr=1_V$. Fullness puts this inverse in $\mathcal{S}$, proving

A monomorphism is a left-cancellable morphism. A strong monomorphism $m:S\to A$ has the right lifting property against every epimorphism: from a commutative squarewith $e$ epic, one obtains $d:Y\to S$ satisfying $de=u$ and $md=v$. A regular monomorphism is an equalizer of a parallel pair.

Suppose $m:S\to A$ equalizes $r,s:A\rightrightarrows C$. In the square above,Since $e$ is epic, $rv=sv$, so the universal property of the equalizer gives the required $d$. Thus every regular monomorphism is strong.

Let $m_i:S_i\to A$ be strong for $i=1,2$, and let $P=S_1{\times }_A{S}_2$. Given a lifting square against $P\to A$, compose its top map with each projection. Strength of $m_i$ produces maps $d_i:Y\to S_i$ with $m_i{d}_i=v$. The pair $(d_1,d_2)$ has equal composites to $A$, hence induces $d:Y\to P$. Therefore intersections of strong subobjects are strong.

Call a morphism anodyne when it is both monic and epic, and call an object saturated when it is injective with respect to every such morphism. Let $m:S\to B$ be a strong subobject of a saturated object. Given an anodyne $e:X\to Y$ and $u:X\to S$, saturation of $B$ extends $mu$ to some $v:Y\to B$. Strength of $m$ applied to $mu=ve$ lifts $v$ to $d:Y\to S$ with $de=u$. Hence $S$ is saturated.

Now embed $A$ as a subobject of a saturated object $B$. Since the category is well-powered, the strong subobjects of $B$ through which $A\to B$ factors form a set; completeness supplies their intersection $j:R\to B$. The same coordinatewise lifting argument used for two factors shows that $j$ is strong, so $R$ is saturated.

The induced map $\eta :A\to R$ is monic. To prove it epic, let $f,g:R\rightrightarrows C$ satisfy $f\eta =g\eta$. Their equalizer $q:E\to R$ is regular and hence strong. Composites of strong monomorphisms are strong, so $jq:E\to B$ is a strong subobject containing $A$. Minimality of the intersection forces $R\to B$ to factor through $E$, which implies $f=g$. Thus $\eta$ is epic and therefore anodyne.

For every saturated $S$, each map $a:A\to S$ extends across $\eta$ to a map $\bar{a}:R\to S$. This extension is unique because $\eta$ is epic. Consequently $A\mapsto R$ is left adjoint to the inclusion of saturated objects: the full subcategory $\mathcal{S}$ is reflective. This is the saturated reflection from a strong-subobject intersection.

It remains to prove that $\mathcal{S}$ is balanced. First let $e:X\to Y$ be epic in $\mathcal{S}$. If $f,g:Y\rightrightarrows Z$ are morphisms in the ambient category, embed $Z$ into a saturated object $T$. Equality $fe=ge$ then implies equality after composing with $Z\to T$; epicity in the full subcategory gives equality there, and monicity of $Z\to T$ gives $f=g$. Thus $e$ is epic in the ambient category.

If $e:X\to Y$ is also monic in $\mathcal{S}$, it is monic in the ambient category as well. Indeed, for $u,v:W\rightrightarrows X$ with $eu=ev$, reflect $W$ by an anodyne map $W\to RW$. Saturation extends $u$ and $v$ to $\bar{u},\bar{v}:RW\rightrightarrows X$; ambient epicity of $W\to RW$ and monicity of $e$ inside $\mathcal{S}$ give $\bar{u}=\bar{v}$, hence $u=v$. Therefore $e$ is anodyne in the ambient category. Saturation of $X$ extends $1_X$ across $e$ to a retraction $r:Y\to X$. Since $e$ is epic, $ere=e$ implies $er=1_Y$, so $e$ is an isomorphism. Hence $\mathcal{S}$ is balanced.

Let $B$ be a subterminal object in a Cartesian closed category. For every $X$ and $A$, the exponential object adjunction givesThe set on the right has at most one element because $B$ is subterminal. Hence $B^A$ is subterminal, proving that $\mathrm{Sub}(1)$ is an exponential ideal.

Now let $\mathcal{D}\subseteq \mathcal{C}$ be reflective, with reflector $L$ and unit ${\eta }_A:A\to LA$. Recall the useful form of its universal property: an object $D$ lies in $\mathcal{D}$ precisely when every map $f:A\to D$ factors uniquely through ${\eta }_A$.

Suppose first that $\mathcal{D}$ is an exponential ideal. Reflective subcategories are closed under ambient limits, so $LA\times LB$ lies in $\mathcal{D}$. Given $f:A\times B\to D$ with $D\in \mathcal{D}$, curry in the first variable to obtain $A\to D^B$. Since $D^B\in \mathcal{D}$, this factors uniquely through ${\eta }_A$ and uncurries to $LA\times B\to D$. Curry once more, now in $B$; since $D^{LA}\in \mathcal{D}$, the result factors uniquely through ${\eta }_B$. Thus every $f$ factors uniquely throughThis makes $LA\times LB$ a reflection of $A\times B$, so uniqueness of reflections gives

Conversely, suppose $L$ preserves binary products, and take $D\in \mathcal{D}$. To prove $D^A\in \mathcal{D}$, start with $f:X\to D^A$ and let $\tilde{f}:X\times A\to D$ be its transpose. Since $D$ is reflective, $\tilde{f}$ factors uniquely throughComposing the resulting map $LX\times LA\to D$ with $1_{LX}\times {\eta }_A$ and currying produces an extension $LX\to D^A$ of $f$. Product preservation identifies the unit on $LX\times A$ with $1_{LX}\times {\eta }_A$, so the same universal property proves uniqueness. Therefore $D^A$ is reflective, and $\mathcal{D}$ is an exponential ideal. This proves the reflector product criterion for an exponential ideal.

Finally consider the arrow category $[2,\mathbf{Set}]$. For arrows $a:A_0\to A_1$ and $b:B_0\to B_1$, letThen the exponential $b^a$ is the arrowIndeed, a commutative square from $c:C_0\to C_1$ to this arrow is, after currying, exactly a commutative square $a\times c\to b$. This establishes the required exponential adjunction and proves that $[2,\mathbf{Set}]$ is cartesian closed.

If $b$ is injective, two elements $(u_0,u_1)$ and $(u_0^{\prime },u_1)$ of $F$ haveso injectivity gives $u_0=u_0^{\prime }$. Hence $F\to B_1^{A_1}$ is injective. The category of injective functions is therefore an exponential ideal in the arrow category. Its terminal object and binary products are inherited pointwise, so these same exponential objects make $\mathrm{Mono}(\mathbf{Set})$ cartesian closed.

The reflector of a reflective subcategory is the left adjoint $L$ to its inclusion. Its unit ${\eta }_A:A\to LA$ is universal among morphisms from $A$ to objects of the subcategory.

For a reflective subcategory $\mathcal{D}$ of a Cartesian closed category $\mathcal{C}$, with reflector $L$, the subcategory $\mathcal{D}$ is an exponential ideal if and only if the canonical comparisonis an isomorphism for every $A,B$. The proof repeatedly curries a map into an object of $\mathcal{D}$ and factors it through the unit of the reflection.