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colimits in categories of algebras

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Limits and colimits

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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. 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.

Proposition

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.

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

FUcFηUcFUFUcFUεcεFUcFUcF 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\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

dc+d(g,1 d)(f,1 d)dd \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 SetSet, then Set TSet^T is cocomplete.

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.) 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 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 & \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.

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 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.)

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

Proposition

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.

Corollary

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

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 forgetful functor

C θ:C TC SC^\theta: C^T \to C^S

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 \\ ^{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.

References

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