# nLab reflexive coequalizer

Reflexive coequalisers

### Context

#### Limits and colimits

limits and colimits

# Reflexive coequalisers

## Definitions

###### Definition

A reflexive pair is a parallel pair $f,g\colon A\rightrightarrows B$ having a common section, i.e. a map $s\colon B\to A$ 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 $C^{op}$ is called a coreflexive equalizer in $C$.

###### Remark

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

###### Example

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).

###### Proof

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

$\sum_i F U F U A_i \underoverset {\sum_i F U \varepsilon A_i} {\sum_i \varepsilon F U A_i} {\rightrightarrows} \sum_i F U A_i$

using the fact that the displayed coproducts exist because, for example,

$\sum_i F U A_i \cong F(\sum_i U A_i)$

since the left adjoint $F$ preserves coproducts, assumed to exist in $C$. That this reflexive coequalizer is the coproduct $\sum_i A_i$ in $C^T$ is routine.

Finally, a category with coproducts and reflexive coequalizers is cocomplete. It suffices that general coequalizers exist, but it is easily seen that if

$f, g \colon A \stackrel{\to}{\to} B$

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

$A + B \stackrel{\overset{(f, 1_B)}{\to}}{\underset{(g, 1_B)}{\to}} B$

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

###### Proposition

If $F\colon C\times D\to E$ 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.

###### Proof

of proposition Suppose given two reflexive coequalizers

$c_0 \stackrel{\to}{\to} c_1 \to c_2$
$\,$
$d_0 \stackrel{\to}{\to} d_1 \to d_2$

and let $c_{i j}$ denote $F(c_i, d_j)$ so that we have a diagram

$\array{ c_{0 0} & \stackrel{\to}{\to} & c_{0 1} & \to & c_{0 2} \\ \downarrow \downarrow & & \downarrow \downarrow & & \downarrow \downarrow \\ c_{1 0} & \stackrel{\to}{\to} & c_{1 1} & \to & c_{1 2} \\ \downarrow & & \downarrow & & \downarrow \\ c_{2 0} & \stackrel{\to}{\to} & c_{2 1} & \to & c_{2 2} }$

in which all rows and columns are reflexive coequalizers (using preservation of reflexive coequalizers in separate variables), and all squares are serially commutative. According to Toposes, Triples, Theories, lemma 4.2 page 248, the diagonal is also a (reflexive) coequalizer, as claimed. (See also the lemma on page 1 of Johnstone’s Topos Theory.)

Proposition 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.

###### Proposition

Reflexive coequalizers in Set commute with finite products:

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

###### Proof

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

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

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

also preserves reflexive coequalizers.

This has a further consequence which is technically very convenient:

###### Theorem

If $T$ is a finitary monad on $Set$, then $T$ preserves reflexive coequalizers.

###### Proof

We have a coend formula for $T$:

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

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

Since finitary monads $T$ preserve reflexive coequalizers, it follows that the monadic functor $U \colon Set^T \to Set$ reflects reflexive coequalizers, and so since $Set$ has reflexive coequalizers, $Set^T$ must as well. Therefore, by proposition , $Set^T$ is cocomplete. This is actually true for infinitary monads $T$ on $Set$ 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.

## Applications

• Fred Linton, Coequalizers in categories of algebras, Seminar on Triples and Categorical Homology Theory, Lecture Notes in Mathematics Vol. 80 (1969), 75-90.
• Peter Johnstone, Topos Theory, London Mathematical Society Monographs no. 10, Academic Press, 1977.