reflexive coequalizer

Reflexive coequalisers



A reflexive pair is a parallel pair f,g:ABf,g\colon A\rightrightarrows B having a common section, i.e. a map s:BAs\colon B\to A such that fs=gs=1 Bf \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 opC^{op} is called a coreflexive equalizer in CC.







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

This is due to (Linton).


Suppose C TC^T has reflexive coequalizers. Then C TC^T certainly has coproducts, because if A iA_i is a collection of TT-algebras, then we can form the coequalizer in C TC^T of the reflexive pair




since the left adjoint FF preserves coproducts, assumed to exist in CC. That this reflexive coequalizer is the coproduct iA i\sum_i A_i in C TC^T is routine.



is a parallel pair, then the coequalizer of the reflexive pair


(note both maps are retracts of the inclusion BA+BB \to A + B) also exists, and gives the coequalizer of the first pair. This completes the proof.


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


This is emphatically not the case for arbitrary coequalizers.


of proposition 1 Suppose given two reflexive coequalizers


and let c ijc_{i j} denote F(c i,d j)F(c_i, d_j) so that we have a diagram



Proposition 1 is particularly interesting when FF is the tensor product of a cocomplete closed monoidal category CC. 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 CC is cocomplete.


Reflexive coequalizers in Set commute with finite products:

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


This follows from prop. 1 as well as from the fact that the diagram category {0d 1s 0d 01}\{ 0 \stackrel{\overset{d_0}{\to}}{\stackrel{\overset{s_0}{\leftarrow}}{\underset{d_1}{\to}}} 1\} with d 0s 0=d 1s 0=idd_0 \circ s_0 = d_1 \circ s_0 = id is a sifted category.

Of course, the diagonal functor Δ:SetSet n\Delta: Set \to Set^n, being left adjoint to the product functor, preserves reflexive coequalizers; therefore the composite


also preserves reflexive coequalizers.

This has a further consequence which is technically very convenient:


If TT is a finitary monad on SetSet, then TT preserves reflexive coequalizers.


We have a coend formula for TT:


and since this is a colimit of functors hom(n,)\hom(n, -) which preserve reflexive coequalizers, TT must also preserve reflexive coequalizers.

Since finitary monads TT preserve reflexive coequalizers, it follows that the monadic functor U:Set TSetU \colon Set^T \to Set reflects reflexive coequalizers, and so since SetSet has reflexive coequalizers, Set TSet^T must as well. Therefore, by proposition 1, Set TSet^T is cocomplete. This is actually true for infinitary monads TT on SetSet as well, at least if we assume the axiom of choice (see here for a proof), but the argument just given is a choice-free proof for the case of finitary monads.



