colimits in categories of algebras


Limits and colimits

Higher algebra



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

U:C TCU: C^T \to C

be the usual underlying or forgetful functor, with left adjoint F:CC TF: C \to C^T, unit η:1 CUF\eta: 1_C \to U F, and counit ε:FU1 C T\varepsilon: F U \to 1_{C^T}. It is well-known that UU reflects limits, so that if CC is complete, then C TC^T is also complete and UU is continuous.

The situation with regard to colimits is more complicated. It is not generally true that if CC is cocomplete, then C TC^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 JJ is a small category, and suppose that the monad TT preserves colimits over JJ, that is, suppose that for every F:JCF: J \to C the canonical map

colim JTFT(colim JF)colim_J T \circ F \to T(colim_J F)

is an isomorphism.


Under these hypotheses, U:C TCU: C^T \to C reflects colimits over JJ.

Here are some sample applications of this proposition which arise frequently in practice. Let JJ be the generic reflexive fork, having exactly two objects 0,10, 1, generated by three non-identity arrows

010,0 \to 1 \stackrel{\to}{\to} 0,

and subject to the condition that the two composites from 00 to 00 are the identity. A colimit over JJ is called a reflexive coequalizer. It frequently happens that a monad T:CCT: C \to C preserves reflexive coequalizers; in this case, if CC has reflexive coequalizers, then so does C TC^T.


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


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

FUc FηUc FUFUc FUεcεFUc FUc\array{ 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 TC^T has coproducts, let (c i,ξ i)(c_i, \xi_i) be a collection of algebras. Then F( iUc i)F(\sum_i U c_i) is the coproduct iFUc i\sum_i F U c_i in C TC^T (since FF preserves coproducts and CC has them). We have a reflexive fork

iFUc i iFηUc i iFUFUc i iFUεc iεFUc i iFUc i\array{ \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 TC^T of this diagram is the coproduct ic i\sum_i c_i.

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

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

where the first arrow is the coproduct coprojection.


If TT is a monad on SetSet, then Set TSet^T is cocomplete.


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

R=f,g:AB×BR = \langle f, g \rangle: A \to B \times B

and then let EE be the smallest TT-congruence (equivalence relation that is a TT-subalgebra map EB×BE \hookrightarrow B \times B) through which RR factors. (This is the intersection of all TT-congruences through which RR factors, and may be calculated in SetSet, where it is reflected in TT-AlgAlg.) The coequalizer as calculated in SetSet,

Eπ 2π 1BpQE \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 SetSet is a split coequalizer. (This requires the axiom of choice. A splitting is given by any splitting i:QBi: Q \to B of pp, which picks a representative in each equivalence class, together with ip,1:BE\langle i p, 1 \rangle: B \to E.) It is therefore an absolute colimit, which the monad TT preserves. Hence the top row in

TE Tπ 2Tπ 1 TB Tp TQ E π 2π 1 B p Q\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 & \stackrel{p}{\to} & Q }

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


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

The hypotheses of the preceding corollary hold when TT 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 xx nx \mapsto x^n preserve reflexive coequalizers.)


If C\mathbf{C} is a regular category or exact category in which regular epimorphisms split, and TT is any monad on C\mathbf{C}, then C T\mathbf{C}^T is a regular category (or exact category, respectively).


For regularity, we first construct coequalizers of kernel pairs in C TC^T. So suppose π 1,π 2:EB\pi_1, \pi_2: E \rightrightarrows B is the kernel pair of some f:BCf: B \to C in C T\mathbf{C}^T. The kernel pair Uπ 1,Uπ 2:UEUBU\pi_1, U\pi_2: U E \to U B of UfU f in C\mathbf{C} has a coequalizer q:UBQq: U B \to Q in C\mathbf{C}, and of course Uπ 1,Uπ 2U\pi_1, U\pi_2 is the kernel pair of qq as well. It follows that the fork

UE Uπ 2Uπ 1 UB q Q\array{ U E & \stackrel{\overset{U\pi_1}{\to}}{\underset{U\pi_2}{\to}} & U B & \stackrel{q}{\to} & Q}

splits in C\mathbf{C}, hence this coequalizer diagram is preserved by TT and hence lifts to a coequalizer diagram in C T\mathbf{C}^T; cf. Beck’s monadicity theorem. Thus kernel pairs in C T\mathbf{C}^T have coequalizers.

That regular epis in C T\mathbf{C}^T are stable under pullback follows a similar line of reasoning: let p:BPp: B \to P be a regular epi in C T\mathbf{C}^T. It is the coequalizer of its kernel pair π 1,π 2:EB\pi_1, \pi_2: E \rightrightarrows B. We just calculated that the coequalizer q:UBQq: U B \to Q of Uπ 1,Uπ 2U\pi_1, U\pi_2 in C\mathbf{C} lifts to C T\mathbf{C}^T, so that QQ is identified with UPU P and qq with UpU p. Thus UpU p is a regular epi in C\mathbf{C}. Now if f:APf: A \to P is a map in C T\mathbf{C}^T, and b=f *pb = f^\ast p is the pullback of pp along ff (with kernel pair ker(b)\ker(b)), then UbU b is the pullback of UpU p along UfU f since UU preserves pullbacks, and so UbU b is a regular epi since C\mathbf{C} is regular. This UbU b is the coequalizer of its kernel pair, and splits, so that again as we argued before, its lift bb is the coequalizer of ker(b)\ker(b). Thus regular epis in C T\mathbf{C}^T are stable under pullback.

For Barr-exactness, suppose π 1,π 2:EB\pi_1, \pi_2: E \rightrightarrows B is an equivalence relation (or congruence) in C T\mathbf{C}^T. Then Uπ 1,Uπ 2:UEUBU\pi_1, U\pi_2: U E \rightrightarrows U B is an equivalence relation in C\mathbf{C}, and hence a kernel pair since C\mathbf{C} is exact. It is the kernel pair of its coequalizer qq in C\mathbf{C}. By the argument we have used several times, the split coequalizer

\array{ U E & \stackrel{\overset{U\pi_1}{\to}}{\underset{U\pi_2}{\to}} & U B & \stackrel{q}{\to} & Q

lifts to a coequalizer diagram in C T\mathbf{C}^T, and since kernel pairs are preserved and reflected by U:C TCU: \mathbf{C}^T \to \mathbf{C}, we conclude that π 1,π 2\pi_1, \pi_2 is the kernel pair of the lifted regular epi over qq.


If TT is a monad on a slice category Set/XSet/X, then the category of TT-algebras is Barr-exact.

Returning now to existence of general coequalizers, here is a more difficult and arcane result given in Toposes, Theories, and Triples (theorem 3.9, p. 267):


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


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

Relatively free functors


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

C θ:C TC SC^\theta: C^T \to C^S

has a left adjoint.


Since the following diagram is commutative:

C T C θ C S U T U S C = C \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 TC^T has coequalizers of reflexive pairs, then C θC^{\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 SS-algebra (c,ξ:Scc)(c, \xi: S c \to c) to the (reflexive) coequalizer of the pair

TScTξTcTScTθcTTcμcTcT 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 μ:TTT\mu: T T \to T is the monad multiplication. (If u:1 CSu: 1_C \to S is the unit of SS, then Tuc:TcTScT 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 AB \otimes_A - to the functor Ab f:Ab BAb AAb^f: Ab^B \to Ab^A between module categories given by restricting scalar multiplication; the coequalizer will be denoted T ScT \circ_S c to underline the analogy.

To see that T ST \circ_S - is the left adjoint, let (d,α:Tdd)(d, \alpha: T d \to d) be a TT-algebra. Any map f:cdf: c \to d in CC induces a unique TT-algebra map ϕ:Tcd\phi: T c \to d:

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

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

Indeed, the condition that ff is an SS-algebra map is satisfaction of the equation

(Scξcfd)=(ScSfSdθdTdαd)(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

αTfμcTθc = αμdTTfTθc = αμdTθdTSf = αTαTθdTSf = αT(αθdSf) = αT(fξ) = αTfTξ\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 ϕμcTθc=ϕTξ\phi \circ \mu c \circ T \theta c = \phi \circ T \xi. This completes one direction. In the other direction, suppose ϕμcTθc=ϕTξ\phi \circ \mu c \circ T \theta c = \phi \circ T \xi. Then

αθdSf = αTfθc = ϕθc = ϕμcηTcθc = ϕμcTθcηSc = ϕTξηSc = ϕηcξ = αTfηcξ = αηdfξ = fξ\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.


Revised on October 3, 2015 19:14:20 by Todd Trimble (