# 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

$U:{C}^{T}\to C$U: C^T \to C

be the usual underlying or forgetful functor, with left adjoint $F:C\to {C}^{T}$, unit $\eta :{1}_{C}\to UF$, and counit $\epsilon :FU\to {1}_{{C}^{T}}$. 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

${\mathrm{colim}}_{J}T\circ F\to T\left({\mathrm{colim}}_{J}F\right)$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,$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 $\left(c,\xi :Tc\to c\right)$ is a $T$-algebra, then $\xi$ is the coequalizer of the reflexive fork

$FUc\stackrel{F\eta Uc}{\to }FUFUc\stackrel{\stackrel{\epsilon FUc}{\to }}{\underset{FU\epsilon c}{\to }}FUc$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 $\left({c}_{i},{\xi }_{i}\right)$ be a collection of algebras. Then $F\left({\sum }_{i}U{c}_{i}\right)$ is the coproduct ${\sum }_{i}FU{c}_{i}$ in ${C}^{T}$ (since $F$ preserves coproducts and $C$ has them). We have a reflexive fork

$\sum _{i}FU{c}_{i}\stackrel{\sum _{i}F\eta U{c}_{i}}{\to }\sum _{i}FUFU{c}_{i}\stackrel{\stackrel{\sum _{i}\epsilon FU{c}_{i}}{\to }}{\underset{\sum _{i}FU\epsilon c}{\to }}\sum _{i}FU{c}_{i}$\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

$d\to c+d\stackrel{\stackrel{\left(f,{1}_{d}\right)}{\to }}{\underset{\left(g,{1}_{d}\right)}{\to }}d$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 $\mathrm{Set}$, then ${\mathrm{Set}}^{T}$ is cocomplete.

###### Proof

It is enough to show that ${\mathrm{Set}}^{T}$ 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

$R=⟨f,g⟩:A\to B×B$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 $E↪B×B$) through which $R$ factors. (This is the intersection of all $T$-congruences through which $R$ factors, and may be calculated in $\mathrm{Set}$, where it is reflected in $T$-$\mathrm{Alg}$.) The coequalizer as calculated in $\mathrm{Set}$,

$E\stackrel{\stackrel{{\pi }_{1}}{\to }}{\underset{{\pi }_{2}}{\to }}B\stackrel{p}{\to }Q$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 $\mathrm{Set}$ 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 $⟨ip,1⟩:B\to E$.) It is therefore an absolute colimit, which the monad $T$ preserves. Hence the top row in

$\begin{array}{ccccc}TE& \stackrel{\stackrel{T{\pi }_{1}}{\to }}{\underset{T{\pi }_{2}}{\to }}& TB& \stackrel{Tp}{\to }& TQ\\ ↓& & ↓& & ↓\\ E& \stackrel{\stackrel{{\pi }_{1}}{\to }}{\underset{{\pi }_{2}}{\to }}& B& \to & Q\end{array}$\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 ${\mathrm{Set}}^{T}$. 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 ${\mathrm{Set}}^{T}$.

###### 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 $x↦{x}^{n}$ 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

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

###### Proof

Since the following diagram is commutative:

$\begin{array}{ccc}{C}^{T}& \stackrel{{C}^{\theta }}{\to }& {C}^{S}\\ {}^{{U}^{T}}↓& & {↓}^{{U}^{S}}\\ C& =& C\end{array}$\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 $\left(c,\xi :Sc\to c\right)$ to the (reflexive) coequalizer of the pair

$TSc\stackrel{T\xi }{\to }Tc\phantom{\rule{2em}{0ex}}TSc\stackrel{T\theta c}{\to }TTc\stackrel{\mu c}{\to }Tc$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 $\mu :TT\to T$ is the monad multiplication. (If $u:{1}_{C}\to S$ is the unit of $S$, then $Tuc:Tc\to TSc$ 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 ${\mathrm{Ab}}^{f}:{\mathrm{Ab}}^{B}\to {\mathrm{Ab}}^{A}$ 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 $\left(d,\alpha :Td\to d\right)$ be a $T$-algebra. Any map $f:c\to d$ in $C$ induces a unique $T$-algebra map $\varphi :Tc\to d$:

$\varphi =\left(Tc\stackrel{Tf}{\to }Td\stackrel{\alpha }{\to }d\right)$\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 $c\to {C}^{\theta }\left(d\right)$ if and only if $\varphi$ coequalizes the pair of above, i.e., if $\varphi$ 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

$\left(Sc\stackrel{\xi }{\to }c\stackrel{f}{\to }d\right)=\left(Sc\stackrel{Sf}{\to }Sd\stackrel{\theta d}{\to }Td\stackrel{\alpha }{\to }d\right)$(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

$\begin{array}{ccc}\alpha \circ Tf\circ \mu c\circ T\theta c& =& \alpha \circ \mu d\circ TTf\circ T\theta c\\ & =& \alpha \circ \mu d\circ T\theta d\circ TSf\\ & =& \alpha \circ T\alpha \circ T\theta d\circ TSf\\ & =& \alpha \circ T\left(\alpha \circ \theta d\circ Sf\right)\\ & =& \alpha \circ T\left(f\circ \xi \right)\\ & =& \alpha \circ Tf\circ T\xi \end{array}$\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 $\varphi \circ \mu c\circ T\theta c=\varphi \circ T\xi$. This completes one direction. In the other direction, suppose $\varphi \circ \mu c\circ T\theta c=\varphi \circ T\xi$. Then

$\begin{array}{ccc}\alpha \circ \theta d\circ Sf& =& \alpha \circ Tf\circ \theta c\\ & =& \varphi \circ \theta c\\ & =& \varphi \circ \mu c\circ \eta Tc\circ \theta c\\ & =& \varphi \circ \mu c\circ T\theta c\circ \eta Sc\\ & =& \varphi \circ T\xi \circ \eta Sc\\ & =& \varphi \circ \eta c\circ \xi \\ & =& \alpha \circ Tf\circ \eta c\circ \xi \\ & =& \alpha \circ \eta d\circ f\circ \xi \\ & =& f\circ \xi \end{array}$\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 8, 2013 19:57:49 by Todd Trimble (67.81.93.26)