Let $\mathcal{C}$ be a category. Recall that the equalizer of a pair of morphisms is monomorphic. Call a monomorphism $f:A\to B$ a regular monomorphism if it is the equalizer of a pair of morphisms.

The dual notion of this is that of a regular epimorphism: a morphism that is the coequalizer of a pair of morphisms. As above, a regular epimorphism is an epimorphism.

For example, in Set, the category of sets, every monomorphism (epimorphism) is regular.

Proof. If the monomorphism $f:A\to B$ is split, then there is a morphism $g:B\to A$ such that

Then $f$ is the equalizer of $f\circ g, 1_B:B\to B$ . First, $f$ equalizes $f\circ g$ and $1_B$ :

Furthermore, if $h:C\to B$ also equalizes $f\circ g$ and $1_B$ :

then by defining $x: C\to A$ by $x:=g\circ h$ , we see that $h=(f\circ g)\circ h = f\circ x$ , or $h$ factors through $f$ . Furthermore, $x$ is uniquely determined by $g$ and $h$ , showing that $f$ is indeed the equalizer of $f\circ g$ and $1_B$ .

Proof. Suppose $f:A\to B$ is the equalizer of $s,t:B\to E$ , and we have the following commutative diagram with $g$ epimorphic:

Now we do some diagram chasing. Since $s\circ f = t\circ f$ , we have $s\circ f\circ x=t\circ f\circ x$ . But $f\circ x = y\circ g$ , we get $s\circ y\circ g = t\circ y\circ g$ . Since $g$ is epimorphic, $s\circ y=t\circ y$ . Since $f$ is the equalizer of $s$ and $t$ , there is a unique morphism $u:D\to A$ such that the following triangle is commutative:

As a result, $f\circ x = y\circ g= f\circ u \circ g$ . Since $f$ is monomorphic, $x=u\circ g$ , yielding the following commutative diagram:

which is the precise statement that $f$ is strong.

Bibliography

1

F. Borceux Basic Category Theory, Handbook of Categorical Algebra I, Cambridge University Press, Cambridge (1994)