A monomorphism is left cancellable: implies for all parallel arrows into . A strong monomorphism is a monomorphism with the following lifting property: if is an epimorphism and , satisfy , there is a diagonal with and . It is unique by monicity. An extremal monomorphism is a monomorphism for which every factorization with an epimorphism forces 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 , apply its lifting property with upper arrow and lower arrow . This gives with . Since is an epimorphism, implies , so is an isomorphism.
If is an equalizer of , consider a lifting square as above. The equation and the epimorphism property give . The equalizer property therefore gives a unique with . Monicity then gives . Hence every regular monomorphism is a strong monomorphism.
For the intersection of strong subobjects, let be strong and let exhibit their intersection, with . Given with an epimorphism, the two strong monomorphisms supply diagonals satisfying and . The pullback in a category property supplies with . Its two projections show , and . 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 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 be a strong monomorphism and saturated. For an anodyne morphism in a category and , saturation extends to . Since is an epimorphism, the strong lifting property supplies extending . Therefore every strong subobject of a saturated object is saturated.
Now assume is a complete category and a well-powered category, with the given saturated embeddings. Choose a monomorphism into a saturated object. Intersect all strong subobjects of through which factors. They form a set by the well-powered category hypothesis, and the identity of is one of them. Completeness constructs their intersection, so we obtain
By the intersection of strong subobjects, is strong; hence is saturated. Since is a monomorphism, so is .
To prove that is an epimorphism, take with . Their equalizer is a regular monomorphism, and factors through it. The composite is strong, so it belongs to the family defining . Minimality gives with . Cancelling gives ; since is a monomorphism, also . Thus is an isomorphism and . We have proved
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 , saturation extends every across the anodyne morphism in a category to . Its extension is unique because is an epimorphism. Consequently
naturally in and . In particular, for define by ; uniqueness proves the identity and composition laws. This constructs the reflector and proves that is a reflective subcategory.
Finally we prove that is a balanced category. If is epic in and satisfy , embed by a monomorphism into a saturated object. Fullness makes arrows of , so and hence . Thus an epimorphism in is also epic in .
A monomorphism in is also monic in : for , extend them uniquely along to . If , cancellation of the ambient epimorphism gives , so monicity in gives and . Now if is both monic and epic in , it is an anodyne morphism in a category in . Saturation of extends across to with . Ambient epicity gives . Fullness puts this inverse in , proving