Epimorphism

An epimorphism $e:A\to B$ is right-cancellative: $fe=ge$ implies $f=g$. In the Category of sets and in abelian groups, epimorphisms are precisely the surjective homomorphisms.

An epimorphism $e$ is extremal when a factorization $e=mh$ with $m$ a monomorphism forces $m$ to be an isomorphism. Every regular epimorphism is extremal: its coequalizer property factors $h$ back through $e$, giving an inverse to $m$. In a regular category, the converse follows by applying extremality to the regular-epi/mono image factorization.

Map two disjoint walking arrows to the two mutually inverse arrows of the walking isomorphism. Their images generate the entire target, so the functor is an extremal epimorphism. Its kernel pair only identifies the matching endpoint objects, not the inverse-composition relations. Mapping both arrows to a nonidentity idempotent in a one-object category equalizes that kernel pair but cannot factor through the target groupoid. Thus the functor is not a regular epimorphism, although the category of small categories has finite limits.

A split epimorphism $r:B\to A$ has a right inverse $s:A\to B$ with $rs=1_A$. Every functor preserves split epimorphisms.

A regular epimorphism is a morphism that is the coequalizer of some parallel pair. Every regular epimorphism is a strong epimorphism.

The coequalizer universal property supplies the diagonal filler against any monomorphism, so every regular epimorphism is strong. In a regular category, every strong epimorphism is regular: factor it as a regular epimorphism followed by a monomorphism and use the lifting property to make the monomorphism an isomorphism.

A strong epimorphism has the left lifting property with respect to every monomorphism. Thus every commutative square with the strong epimorphism on the left and a monomorphism on the right has a diagonal filler.

A map $e:X\to Y$ has this property if every commutative square with $e$ as its left side and a monomorphism $m:A\to B$ as its right side has a diagonal filler $t:Y\to A$. The filler is unique because $m$ is monic. This definition does not itself stipulate that $e$ is an epimorphism; binary products imply that conclusion by testing against the categorical diagonal.

In a category with binary products in a category, a map with the left lifting property against monomorphisms is an epimorphism. If $ue=ve$, lift the square with right side $\Delta$ and bottom side $⟨u,v⟩$. The two projections of its filler give $u=v$. This argument does not require equalizers.

If $gf$ has the left lifting property against monomorphisms, so does $g$. Precompose a lifting square for $g$ with $f$ and lift the composite. To check the remaining triangle, cancel the monic side of the square, rather than cancelling $f$. In particular, a monic right factor of a lifting-only strong morphism is an isomorphism.

If $e:X\to Y$ is a monomorphism and has the left lifting property against monomorphisms, apply its property to the square with both vertical arrows $e$ and both horizontal arrows identities. The filler is a two-sided inverse, so $e$ is an isomorphism.

Suppose a category has pullbacks and coequalizers and every pullback of a regular epimorphism is epic. Coequalizing the kernel pair of any morphism produces a factorization as a regular epimorphism followed by a monomorphism. Consequently every strong epimorphism is regular.

A functor between small categories is a strong epimorphism precisely when its image generates its codomain as a subcategory. It is a regular epimorphism precisely when it is surjective on objects and full. Hence a functor can be strong without being regular when its image arrows generate additional composites that have no single preimage.

A strong quotient of an object $A$ is the codomain of a strong epimorphism out of $A$. A class of objects is closed under strong quotients when every such codomain remains in the class up to isomorphism.