Quotients of effective congruences

Recall that a congruence $E \rightrightarrows X$ is called effective if it is the kernel pair of some morphism with domain $X$ The next result shows that, in many situations, this morphism can be chosen canonically, namely as the quotient of the congruence, provided that it exists.

Lemma.

An effective congruence whose coequalizer exists is the kernel pair of its coequalizer. More precisely, let $f,g : E \rightrightarrows X$ be an effective congruence. If $f,g$ have a coequalizer $p : X \to X/E$ then $f,g$ is the kernel pair of $p$ That is, the square $\begin{CD} E @> f >> X \\ @V g VV @VV p V \\ X @>> p > X/E \end{CD}$ is a pullback.

Proof. Let $h : X \to Z$ be a morphism such that $f,g$ is the kernel pair of $h$ By the universal property of the coequalizer, there is a unique morphism $\bar h : X/E \to Z$ such that $h = \bar h \circ p$ Now suppose we have morphisms $a, b : T \rightrightarrows X$ such that $p \circ a = p \circ b$ Then $h \circ a = \bar h \circ p \circ a = \bar h \circ p \circ b = h \circ b.$ Since $f,g$ is the kernel pair of $h$ this means that the pair $a, b$ factors uniquely through the pair $f,g$ $\square$

This result is similar to the lemma that a regular monomorphism is the equalizer of its cokernel pair, provided that the cokernel pair exists.

Context

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This page is referenced by the following category implications.

generalized variety implies effective congruences