# nLab colimits in categories of algebras

### Context

#### Limits and colimits

limits and colimits

## (∞,1)-Categorical

### Model-categorical

#### Higher algebra

higher algebra

universal algebra

# Contents

## Introduction

Let $T$ be a monad on a category $C$, and let $C^T$ denote the Eilenberg-Moore category of $T$. Let

$U: C^T \to C$

be the usual underlying or forgetful functor, with left adjoint $F: C \to C^T$, unit $\eta: 1_C \to U F$, and counit $\varepsilon: F U \to 1_{C^T}$. It is well-known that $U$ reflects limits, so that if $C$ is complete, then $C^T$ is also complete and $U$ is continuous.

The situation with regard to colimits is more complicated. It is not generally true that if $C$ is cocomplete, then $C^T$ is also. In this article we collect some partial results which address these issues.

## Reflexive coequalizers and cocompleteness

A simple but basic fact is the following. Suppose $J$ is a small category, and suppose that the monad $T$ preserves colimits over $J$, that is, suppose that for every $F: J \to C$ the canonical map

$colim_J T \circ F \to T(colim_J F)$

is an isomorphism.

###### Proposition

Under these hypotheses, $U: C^T \to C$ reflects colimits over $J$.

Here are some sample applications of this proposition which arise frequently in practice. Let $J$ be the generic reflexive fork, having exactly two objects $0, 1$, generated by three non-identity arrows

$0 \to 1 \stackrel{\to}{\to} 0,$

and subject to the condition that the two composites from $0$ to $0$ are the identity. A colimit over $J$ is called a reflexive coequalizer. It frequently happens that a monad $T: C \to C$ preserves reflexive coequalizers; in this case, if $C$ has reflexive coequalizers, then so does $C^T$.

###### Theorem

If $C$ is cocomplete and $C^T$ has reflexive coequalizers, then $C^T$ is cocomplete.

###### Proof

First observe that if $(c, \xi: T c \to c)$ is a $T$-algebra, then $\xi$ is the coequalizer of the reflexive fork

$F U c \stackrel{F \eta U c}{\to} F U F U c \stackrel{\overset{\varepsilon F U c}{\to}}{\underset{F U \varepsilon c}{\to}} F U c$

To show $C^T$ has coproducts, let $(c_i, \xi_i)$ be a collection of algebras. Then $F(\sum_i U c_i)$ is the coproduct $\sum_i F U c_i$ in $C^T$ (since $F$ preserves coproducts and $C$ has them). We have a reflexive fork

$\sum_i F U c_i \stackrel{\sum_i F \eta U c_i}{\to} \sum_i F U F U c_i \stackrel{\overset{\sum_i \varepsilon F U c_i}{\to}}{\underset{\sum_i F U \varepsilon c}{\to}} \sum_i F U c_i$

and it is not difficult to show that the coequalizer in $C^T$ of this diagram is the coproduct $\sum_i c_i$.

Finally, general coequalizers in $C^T$ are constructed from coproducts and reflexive coequalizers: given a parallel pair $f, g: c \stackrel{\to}{\to} d$ in $C^T$, the coequalizer of $f$ and $g$ is the colimit of the reflexive fork

$d \to c + d \stackrel{\overset{(f, 1_d)}{\to}}{\underset{(g, 1_d)}{\to}} d$

where the first arrow is the coproduct coprojection.

###### Corollary

If $T$ is a monad on $Set$, then $Set^T$ is cocomplete.

###### Proof

It is enough to show that $Set^T$ has coequalizers. Suppose given a pair of algebra maps $f, g: A \stackrel{\to}{\to} B$ whose coequalizer we wish to construct. Let $R$ be the $T$-algebra relation

$R = \langle f, g \rangle: A \to B \times B$

and then let $E$ be the smallest $T$-congruence (equivalence relation that is a $T$-subalgebra map $E \hookrightarrow B \times B$) through which $R$ factors. (This is the intersection of all $T$-congruences through which $R$ factors, and may be calculated in $Set$, where it is reflected in $T$-$Alg$.) The coequalizer as calculated in $Set$,

$E \stackrel{\overset{\pi_1}{\to}}{\underset{\pi_2}{\to}} B \stackrel{p}{\to} Q$

is a split coequalizer, because every quotient of an equivalence relation in $Set$ is a split coequalizer. (This requires the axiom of choice. A splitting is given by any splitting $i: Q \to B$ of $p$, which picks a representative in each equivalence class, together with $\langle i p, 1 \rangle: B \to E$.) It is therefore an absolute colimit, which the monad $T$ preserves. Hence the top row in

$\array{ T E & \stackrel{\overset{T\pi_1}{\to}}{\underset{T\pi_2}{\to}} & T B & \stackrel{T p}{\to} & T Q \\ \downarrow & & \downarrow & & \downarrow \\ E & \stackrel{\overset{\pi_1}{\to}}{\underset{\pi_2}{\to}} & B & \to & Q}$

(the first two vertical arrows being algebra structure maps) is a coequalizer in $Set^T$. The last vertical arrow making the diagram commute gives $Q$ a $T$-algebra structure, and the split coequalizer in the bottom row is reflected in $Set^T$.

###### Corollary

If $T$ is a monad on a complete and cocomplete category $C$ that preserves reflexive coequalizers, then $C^T$ is complete and cocomplete.

The hypotheses of the preceding corollary hold when $T$ is a monad on a complete, cocomplete, cartesian closed category that is induced from a finitary algebraic theory. (The key observation being that the finitary power functors $x \mapsto x^n$ preserve reflexive coequalizers.)

Here is a more difficult result given in Toposes, Theories, and Triples (theorem 3.9, p. 267):

###### Proposition

If $C$ has coequalizers and equalizers of arbitrary sets of parallel morphisms, and if a monad $T: C \to C$ preserves colimits of countable chains $\omega \to C$, then $C^T$ has coequalizers.

###### Corollary

If $C$ is complete and cocomplete and $T: C \to C$ preserves filtered colimits, then $C^T$ is complete and cocomplete.

## Relatively free functors

###### Theorem

Suppose that $\theta: S \to T$ is a morphism of monads on $C$, and suppose that $C^T$ has coequalizers. Then the forgetful functor

$C^\theta: C^T \to C^S$

###### Proof

Since the following diagram is commutative:

$\begin{array}{cccc}C^T & \overset{C^{\theta}}{\to} & C^S \\ ^{U^T}\downarrow & & \downarrow^{U^S} \\ C & = & C \end{array}$

(using an obvious notation), it follows immediately from a corollary to the adjoint lifting theorem that if $C^T$ has coequalizers of reflexive pairs, then $C^{\theta}$ has a left adjoint and is, in fact, monadic.

This completes the proof, but it is interesting to observe that there is a concrete description of the left adjoint in this case: The left adjoint sends an $S$-algebra $(c, \xi: S c \to c)$ to the (reflexive) coequalizer of the pair

$T S c \stackrel{T \xi}{\to} T c \qquad T S c \stackrel{T \theta c}{\to} T T c \stackrel{\mu c}{\to} T c$

where $\mu: T T \to T$ is the monad multiplication. (If $u: 1_C \to S$ is the unit of $S$, then $T u c: T c \to T S c$ is a common right inverse of both arrows of the pair.) This coequalizer is analogous to the construction of the left adjoint $B \otimes_A -$ to the functor $Ab^f: Ab^B \to Ab^A$ between module categories given by restricting scalar multiplication; the coequalizer will be denoted $T \circ_S c$ to underline the analogy.

To see that $T \circ_S -$ is the left adjoint, let $(d, \alpha: T d \to d)$ be a $T$-algebra. Any map $f: c \to d$ in $C$ induces a unique $T$-algebra map $\phi: T c \to d$:

$\phi = (T c \stackrel{T f}{\to} T d \stackrel{\alpha}{\to} d)$

and the claim is that $f: c \to d$ is an $S$-algebra map $c \to C^\theta(d)$ if and only if $\phi$ coequalizes the pair of above, i.e., if $\phi$ factors (uniquely) through a $T$-algebra map $T \circ_S c \to d$.

Indeed, the condition that $f$ is an $S$-algebra map is satisfaction of the equation

$(S c \stackrel{\xi}{\to} c \stackrel{f}{\to} d) = (S c \stackrel{S f}{\to} S d \stackrel{\theta d}{\to} T d \stackrel{\alpha}{\to} d)$

and now we have a long train of equations

$\array{ \alpha \circ T f \circ \mu c \circ T \theta c & = & \alpha \circ \mu d \circ T T f \circ T \theta c \\ & = & \alpha \circ \mu d \circ T \theta d \circ T S f \\ & = & \alpha \circ T \alpha \circ T \theta d \circ T S f \\ & = & \alpha \circ T(\alpha \circ \theta d \circ S f)\\ & = & \alpha \circ T(f \circ \xi) \\ & = & \alpha \circ T f \circ T \xi }$

which gives $\phi \circ \mu c \circ T \theta c = \phi \circ T \xi$. This completes one direction. In the other direction, suppose $\phi \circ \mu c \circ T \theta c = \phi \circ T \xi$. Then

$\array{ \alpha \circ \theta d \circ S f & = & \alpha \circ \T f \circ \theta c \\ & = & \phi \circ \theta c \\ & = & \phi \circ \mu c \circ \eta T c \circ \theta c \\ & = & \phi \circ \mu c \circ T \theta c \circ \eta S c \\ & = & \phi \circ T \xi \circ \eta S c \\ & = & \phi \circ \eta c \circ \xi \\ & = & \alpha \circ T f \circ \eta c \circ \xi \\ & = & \alpha \circ \eta d \circ f \circ \xi \\ & = & f \circ \xi }$

as desired. This completes the proof.

## References

Revised on April 25, 2014 04:49:53 by Todd Trimble (67.81.95.215)