= Solution
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>, $ed e=e$ 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 \b[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 $m=m_1p_1=m_2p_2$. Given $mu=ve$ with $e$ an <epimorphism>, the two <strong monomorphisms> supply diagonals $d_i:Y\to A_i$ satisfying $d_ie=p_iu$ and $m_id_i=v$. The <pullback in a category> property supplies $d:Y\to P$ with $p_id=d_i$. Its two projections show $de=u$, and $md=v$. Thus \b[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 \b[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 obtain
$$
A\xrightarrow{\eta_A}LA\xrightarrow{m}B,\qquad m\eta_A=a.
$$
By 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 proved
$$
\boxed{\eta_A:A\to LA\text{ is both monic and epic, and }LA\text{ is saturated}.}
$$
This 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>. Consequently
$$
\boxed{\mathcal S(LA,S)\cong\mathcal C(A,S),}
$$
naturally in $S$ and $A$. In particular, for $h:A\to A'$ define $Lh$ by $Lh\,\eta_A=\eta_{A'}h$; 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 \b[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
$$
\boxed{\mathcal S\text{ is balanced}.}
$$
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