nLab
reflexive coequalizer

Definitions

A reflexive pair is a parallel pair $f,g:A\rightrightarrows B$ having a common section, i.e. a map $s:B\to A$ such that $f\circ s=g\circ s=1_B$. A reflexive coequalizer is a coequalizer of a reflexive pair. A category has reflexive coequalizers if it has coequalizers of all reflexive pairs.

Dually, a reflexive coequalizer in the opposite category $C^{\mathrm{op}}$ is called a coreflexive equalizer in $C$.

Remarks

• Reflexive coequalizers should not be confused with split coequalizers, a distinct concept.

• Any congruence is a reflexive pair, so in particular any quotient of a congruence is a reflexive coequalizer.

Properties

Theorem

If $T$ is a monad on a cocomplete category $C$, then the category $C^T$ of Eilenberg Moore algebras is cocomplete if and only if it has reflexive coequalizers. This is the case particularly if $T$ preserves reflexive coequalizers.

This is due to (Linton).

Proposition

If $F:C\times D\to E$ is a functor of two variables which preserves reflexive coequalizers in each variable separately (that is, $F(c,-)$ and $F(-,d)$ preserve reflexive coequalizers for all $c\in C$ and $d\in D$), then $F$ preserves reflexive coequalizers in both variables together.

Remark

This is emphatically not the case for arbitrary coequalizers.

This result is particularly interesting when $F$ is the tensor product of a cocomplete closed monoidal category $C$. In this case it implies that the free monoid monad on such a category preserves reflexive coequalizers, and thus (by Linton’s theorem) the category of monoid objects in $C$ is cocomplete.

Corollary

Reflexive coequalizers in Set commute with finite products:

the $n$-fold product functors ${\mathrm{Set}}^n\stackrel{\prod }{\to }\mathrm{Set}$ preserve reflexive coequalizers.

Of course, the diagonal functor $\Delta :\mathrm{Set}\to {\mathrm{Set}}^n$, being left adjoint to the product functor, preserves reflexive coequalizers; therefore the composite

$$\mathrm{Set}\stackrel{\prod \Delta }{\to }\mathrm{Set}:x\mapsto \mathrm{hom}(n,x)$$Set \stackrel{\prod \Delta}{\to} Set: x \mapsto \hom(n, x)

also preserves reflexive coequalizers.

This has a further consequence which is technically very convenient:

Theorem

If $T$ is a finitary monad on $\mathrm{Set}$, then $T$ preserves reflexive coequalizers.

Proof

We have a coend formula for $T$:

$$T(-)\cong {\int }^{n\in \mathrm{Fin}}T(n)\times \mathrm{hom}(n,-)$$T(-) \cong \int^{n \in Fin} T(n) \times \hom(n, -)

and since this is a colimit of functors $\mathrm{hom}(n,-)$ which preserve reflexive coequalizers, $T$ must also preserve reflexive coequalizers.

Applications

• Reflexive coequalizers figure in the crude monadicity theorem.

References

• Fred Linton?, Coequalizers in categories of algebras