Regular epimorphisms are coequalizers of their kernel pairs

A regular epimorphism is the coequalizer of some pair of morphisms. It turns out that, in many cases, a canonical pair is available.

Lemma.

Let $f : X \to Y$ be a regular epimorphism in a category $\C$ that has a kernel pair $p_1,p_2 : X \times_Y X \rightrightarrows X$ Then $f$ is the coequalizer of $p_1,p_2$ i.e. $f$ is an effective epimorphism.

Proof. Say $f$ is the coequalizer of $a,b : A \rightrightarrows X$ Since $f \circ a = f \circ b$ there is a morphism $c : A \to X \times_Y X$ with $p_1 \circ c = a$ and $p_2 \circ c = b$ Now let $g : X \to T$ be a morphism with $g \circ p_1 = g \circ p_2$ Then $g \circ a = g \circ p_1 \circ c = g \circ p_2 \circ c = g \circ b,$ so $g$ factors uniquely through $f$ Since also $f \circ p_1 = f \circ p_2$ this shows that $f$ is the coequalizer of $p_1,p_2$ $\square$

Remark. The dual statement is that any regular monomorphism that has a cokernel pair is the equalizer of this cokernel pair, and hence an effective monomorphism.

Context

This page is referenced by the following categories.

This page is referenced by the following functor implications.

preserves coreflexive equalizers implies preserves regular monomorphisms