nLab colimits in categories of algebras

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Limits and colimits

Higher algebra

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Introduction

Let TT be a monad on a category CC, and let C TC^T denote the Eilenberg-Moore category of TT, i.e., the category of TT-algebras. 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. (See Adámek & Koubek, III.10, for an example when C=C = Pos.) In this article we collect various results that guarantee existence of colimits of algebras.

Reflexive coequalizers and cocompleteness

A simple but basic fact is the following. Suppose JJ is a small category, and suppose that the functor 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.

Proposition

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

For a proof, see Proposition 4.3.2 of Handbook of Categorical Algebra (Volume 2). In fact, what is shown there is a tighter result: if D:JC TD:J\to C^T is any diagram such that UDU \circ D has a colimit preserved by both TT and TTT \circ T, then DD has a colimit preserved by UU.

Here preservation by just TT is not sufficient. A counterexample is given by the algebraic theory generated by a 1-ary operation ff and a 3-ary operation gg with the following laws:

  • f(f(x))=xf(f(x)) = x.
  • g(x,x,y)=g(x,y,x)=g(y,x,x)=xg(x,x,y) = g(x,y,x) = g(y,x,x) = x.
  • g(x,f(x),y)=g(x,y,f(x))=g(y,x,f(x))=yg(x,f(x),y) = g(x,y,f(x)) = g(y,x,f(x)) = y.

The corresponding monad on Set\mathrm{Set} preserves the coproduct 1+11 + 1. However, in the category of algebras we have 1+1=11 + 1 = 1.

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{\longrightarrow}{\longrightarrow} 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.

The following very useful observation was first made by Linton.

Theorem

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

Proof

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 FξεFUc FUc\array{ F U c & \stackrel{F \eta U c}{\longrightarrow} & F U F U c & \stackrel{\overset{\varepsilon F U c}{\longrightarrow}}{\underset{F \xi}{\longrightarrow}} & 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 iFξ i 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}{\longrightarrow}}{\underset{\sum_i F \xi_i}{\longrightarrow}} & \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{\longrightarrow}{\longrightarrow} 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.

Corollary

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 CC is a complete, cocomplete, cartesian closed category and TT is the monad corresponding to a finitary algebraic theory. (The key observation being that the finitary power functors xx nx \mapsto x^n preserve reflexive coequalizers if CC is cartesian closed.)

Corollary

Assuming the axiom of choice, if TT is a monad on Set, then Set TSet^T is cocomplete. Similarly upon replacing SetSet by a slice Set/XSet XSet/X \simeq Set^X, or by Vect.

Proof

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 since U:Set TSetU: Set^T \to Set reflects arbitrary intersections.) The coequalizer as calculated in SetSet,

UE Uπ 2Uπ 1 UB p Q\array{ U E & \stackrel{\overset{U \pi_1}{\longrightarrow}}{\underset{U \pi_2}{\longrightarrow}} & U B & \stackrel{p}{\longrightarrow} & 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:QUBi: Q \to U B of pp, which picks a representative in each equivalence class, together with ip,1:UBUE\langle i p, 1 \rangle: U B \to U E.) The proof is completed by the following lemma.

Lemma

In a category C\mathbf{C}, given a pair π 1,π 2:EB\pi_1, \pi_2: E \rightrightarrows B in C T\mathbf{C}^T such that Uπ 1,Uπ 2:UEUBU\pi_1, U\pi_2: U E \rightrightarrows U B has a split coequalizer UBQU B \to Q in C\mathbf{C}, the pair π 1,π 2\pi_1, \pi_2 has a coequalizer in C T\mathbf{C}^T (reflected by the split coequalizer).

Proof

(Cf. monadicity theorem.) A split coequalizer is an absolute colimit, which the functor TT preserves. Hence the top row in

TUE TUπ 2TUπ 1 TUB Tp TQ UE Uπ 2Uπ 1 UB p Q\array{ T U E & \stackrel{\overset{T U\pi_1}{\to}}{\underset{T U\pi_2}{\to}} & T U B & \stackrel{T p}{\longrightarrow} & T Q \\ \downarrow & & \downarrow & & \downarrow \\ U E & \stackrel{\overset{U\pi_1}{\longrightarrow}}{\underset{U\pi_2}{\longrightarrow}} & U B & \stackrel{p}{\longrightarrow} & Q }

(the first two vertical arrows being algebra structure maps) is a coequalizer in C T\mathbf{C}^T. The last vertical arrow making the diagram commute gives QQ a TT-algebra structure, and the split coequalizer in the bottom row is thereby reflected in C T\mathbf{C}^T (i.e., lifts to a coequalizer in C T\mathbf{C}^T, albeit not necessarily to one that is itself split).

Categories of algebras are Barr exact

Theorem

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).

Proof

For regularity, we first construct coequalizers of kernel pairs in C T\mathbf{C}^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 from the splitting hypothesis that the fork

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

splits in C\mathbf{C}, hence by Lemma this diagram lifts to a coequalizer for π 1,π 2:EB\pi_1, \pi_2: E \rightrightarrows B in C T\mathbf{C}^T. 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 by Lemma , 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 Lemma , the split coequalizer

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

Corollary

If TT is a monad on a slice category Set/XSet/X, then the category of TT-algebras is (Barr-)exact. If TT is a monad on VectVect, then Vect TVect^T is exact.

For functors preserving filtered colimits

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

Proposition

Let CC be a category with finite colimits, colimits of chains κC\kappa \to C (for κ\kappa an infinite cardinal), and equalizers of arbitrary sets of endomorphisms. If a monad T:CCT : C \to C preserves colimits of κ\kappa-chains, then C TC^T admits coequalizers.

Corollary

If CC is complete and cocomplete and T:CCT: C \to C preserves filtered colimits, or even just colimits of ω\omega-chains, then C TC^T is complete and cocomplete.

For locally presentable categories

If CC is a locally presentable category and TT is an accessible monad (aka a bounded monad, aka a monad with rank) on CC, then C TC^T is also locally presentable and in particular cocomplete. Details may be found in Locally presentable and accessible categories.

Algebraic functors have left adjoints

Theorem

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 induced algebraic functor

C θ:C TC SC^\theta: C^T \to C^S

(pulling back a TT-algebra (c,ξ:Tcc)(c, \xi: T c \to c) to the SS-algebra (c,ScθcTcξc)(c, S c \stackrel{\theta c}{\longrightarrow} T c \stackrel{\xi}{\longrightarrow} c), thus remembering only underlying SS-algebra structure) has a left adjoint.

Proof

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 \\ \mathllap{U^T} \downarrow & & \downarrow \mathrlap{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 actually completes the proof, but here is a concrete description of the left adjoint to C θC^\theta: it sends an SS-algebra (c,ξ:Scc)(c, \xi: S c \to c) to the (reflexive) coequalizer of the pair

(1) TTc Tθc μc TSc Tξ Tc\array{ & & T T c & & \\ & \mathllap{T \theta c} \nearrow & & \searrow \mathrlap{\mu c} \\ T S c & & \stackrel{\; \; \; \; \; \; \; \; T \xi \; \; \; \; \; \; \; \; }{\longrightarrow} & & 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 “restriction” functor Ab f:Ab BAb AAb^f: Ab^B \to Ab^A between module categories (restricting scalar multiplication on a BB-module along a ring map f:ABf: A \to B); given an AA-module (M,α M:AMM)(M, \alpha_M: A \otimes M \to M), the BB-module B AMB \otimes_A M is the coequalizer in Ab BAb^B of

BBM BfM mult BM BAM Bα M BM,\array{ & & B \otimes B \otimes M & & \\ & \mathllap{B \otimes f \otimes M} \nearrow & & \searrow \mathrlap{mult_B \otimes M} \\ B \otimes A \otimes M & & \stackrel{\; \; \; \; \; \; \; \; B \otimes \alpha_M \; \; \; \; \; \; \; \; }{\longrightarrow} & & B \otimes M, }

and so the coequalizer of (1) 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 in (1), i.e., if ϕ\phi factors (uniquely) through a TT-algebra map T ScdT \circ_S c \to d.

Indeed, assume ff is an SS-algebra map, so we have a commutative diagram

(2)Sc ξ c Sf f Sd αθd d.\array{ S c & \stackrel{\xi}{\to} & c \\ \mathllap{S f} \downarrow & & \downarrow \mathrlap{f} \\ S d & \underset{\alpha \circ \theta d}{\to} & d. }

That ϕ=αTf\phi = \alpha \circ T f coequalizes the pair of (1) follows by expanding the diagram

(3)TSc Tθc TTc Tξ μc Tc Tf Td α d \array{ T S c & \stackrel{T \theta c}{\to} & T T c & & & & \\ & \mathllap{T \xi} \searrow & \downarrow \mathrlap{\mu c} & & & & \\ & & T c & \underset{T f}{\to} & T d & \underset{\alpha}{\to} & d }

to

(4) TSd TSf (nat) Tθd TSc Tθc TTc TTf TTd Tα Td Tξ μc (nat) μd (alg) α Tc Tf Td α d\array{ & & T S d & & & & \\ & \mathllap{T S f} \nearrow & \; \; (nat) & \searrow \mathrlap{T \theta d} & & & \\ T S c & \stackrel{T \theta c}{\to} & T T c & \stackrel{T T f}{\to} & T T d & \stackrel{T \alpha}{\to} & T d \\ & \mathllap{T \xi} \searrow & \downarrow \mathrlap{\mu c} & \; \; (nat) & \downarrow \mathrlap{\mu d} & \; \; (alg) & \downarrow \mathrlap{\alpha} \\ & & T c & \underset{T f}{\to} & T d & \underset{\alpha}{\to} & d }

where using (2), the path along the top may be replaced by TfTξ:TScTdT f \circ T \xi: T S c \to T d, reducing the desired coequalizing of (3) to the tautology αTfTξ=αTfTξ\alpha \circ T f \circ T \xi = \alpha \circ T f \circ T \xi.

Conversely, assuming the coequalizing of (3), the perimeter of (4) commutes, and on top of that we stack naturality diagrams for the monad unit η\eta of TT:

(5) Sd Sf ηSd θd Sc (nat) TSd (nat) Td α d ηSc TSf (nat) Tθd ηTd (nat) ηd TSc Tθc TTc TTf TTd Tα Td Tξ μc (nat) μd (alg) α Tc Tf Td α d\array{ & & S d & & & & \\ & _\mathllap{S f} \nearrow & _\mathllap{\eta S d} \downarrow & \searrow _\mathrlap{\theta d} & & & \\ S c & (nat) & T S d & (nat) & T d & \stackrel{\alpha}{\to} & d \\ _\mathllap{\eta S c} \downarrow & _\mathllap{T S f} \nearrow & \; \; (nat) & \searrow _\mathrlap{T \theta d} & \downarrow_\mathrlap{\eta T d} & (nat) & \downarrow _\mathrlap{\eta d}\\ T S c & \stackrel{T \theta c}{\to} & T T c & \stackrel{T T f}{\to} & T T d & \stackrel{T \alpha}{\to} & T d \\ & \mathllap{T \xi} \searrow & \downarrow \mathrlap{\mu c} & \; \; (nat) & \downarrow \mathrlap{\mu d} & \; \; (alg) & \downarrow \mathrlap{\alpha} \\ & & T c & \underset{T f}{\to} & T d & \underset{\alpha}{\to} & d }

The vertical composite on the right is 1 d1_d by a unit equation for a TT-algebra, and thus we may simplify the perimeter. Retaining the (simplified) perimeter of (5), and inserting some naturality squares and a unit diagram inside, we arrive at the commutative diagram

Sc Sf Sd θd Td ηSc ξ α TSc (nat) c f d 1 d d Tξ ηc (nat) ηd (unit) 1 d Tc Tf Td α d\array{ S c & \stackrel{S f}{\to} & S d & \stackrel{\theta d}{\to} & T d & & \\ _\mathllap{\eta S c} \downarrow & \searrow\; _\mathllap{\xi} & & & & \searrow \mathrlap{\alpha} & \\ T S c & (nat) & c & \stackrel{f}{\to} & d & \stackrel{1_d}{\to} & d \\ & \mathllap{T \xi} \searrow & \downarrow _\mathrlap{\eta c} & (nat) & \downarrow _\mathrlap{\eta d} & (unit) & \downarrow \mathrlap{1_d} \\ & & T c & \underset{T f}{\to} & T d & \underset{\alpha}{\to} & d }

where the commutativity of the unlabeled polygonal region is just the commutativity of (2). This completes the proof of the claim.

References

  • Fred Linton, Coequalizers in categories of algebras, Seminar on Triples and Categorical Homology Theory: ETH 1966/67. Berlin, Heidelberg: Springer Berlin Heidelberg, 2006.

  • Michael Barr and Charles Wells, Toposes, Triples, and Theories, Reprints in Theory and Applications of Categories (2005), 1-289. (online pdf)

  • Jiří Adámek and Václav Koubek, Are colimits of algebras simple to construct?, Journal of Algebra, Volume 66, Issue 1 (September 1980), 226-250. (link)

Last revised on August 17, 2025 at 04:42:26. See the history of this page for a list of all contributions to it.