Limits and colimits
limits and colimits
limit and colimit
limits and colimits by example
commutativity of limits and colimits
connected limit, wide pullback
preserved limit, reflected limit, created limit
product, fiber product, base change, coproduct, pullback, pushout, cobase change, equalizer, coequalizer, join, meet, terminal object, initial object, direct product, direct sum
end and coend
Algebras and modules
Model category presentations
Geometry on formal duals of algebras
Let be a monad on a category , and let denote the Eilenberg-Moore category of , i.e., the category of -algebras. Let
be the usual underlying or forgetful functor, with left adjoint , unit , and counit . It is well-known that reflects limits, so that if is complete, then is also complete and is continuous.
The situation with regard to colimits is more complicated. It is not generally true that if is cocomplete, then is also. (See Adámek & Koubek, III.10, for an example when 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 is a small category, and suppose that the monad preserves colimits over , that is, suppose that for every the canonical map
is an isomorphism.
Under these hypotheses, reflects colimits over .
Here are some sample applications of this proposition which arise frequently in practice. Let be the generic reflexive fork, having exactly two objects , generated by three non-identity arrows
and subject to the condition that the two composites from to are the identity. A colimit over is called a reflexive coequalizer. It frequently happens that a monad preserves reflexive coequalizers; in this case, if has reflexive coequalizers, then so does .
The following very useful observation was first made by Linton.
If is cocomplete and has reflexive coequalizers, then is cocomplete.
First observe that if is a -algebra, then is the coequalizer of the reflexive fork
To show has coproducts, let be a collection of algebras. Then is the coproduct in (since preserves coproducts and has them). We have a reflexive fork
and it is not difficult to show that the coequalizer in of this diagram is the coproduct .
Finally, general coequalizers in are constructed from coproducts and reflexive coequalizers: given a parallel pair in , the coequalizer of and is the colimit of the reflexive fork
where the first arrow is the coproduct coprojection.
If is a monad on a complete and cocomplete category that preserves reflexive coequalizers, then is complete and cocomplete.
The hypotheses of the preceding corollary hold when is a complete, cocomplete, cartesian closed category and is the monad corresponding to a finitary algebraic theory. (The key observation being that the finitary power functors preserve reflexive coequalizers if is cartesian closed.)
If is a monad on Set, then is cocomplete (under the axiom of choice). Similarly upon replacing by a slice , or by Vect.
It is enough to show that has coequalizers. Suppose given a pair of algebra maps whose coequalizer we wish to construct. Let be the -algebra relation
and then let be the smallest -congruence (equivalence relation that is a -subalgebra map ) through which factors. (This is the intersection of all -congruences through which factors, and may be calculated in , where it is reflected in - since reflects arbitrary intersections.) The coequalizer as calculated in ,
is a split coequalizer, because every quotient of an equivalence relation in is a split coequalizer. (This requires the axiom of choice. A splitting is given by any splitting of , which picks a representative in each equivalence class, together with .) The proof is completed by the following lemma.
In a category , given a pair in such that has a split coequalizer in , the pair has a coequalizer in (reflected by the split coequalizer).
(Cf. monadicity theorem.) A split coequalizer is an absolute colimit, which the functor preserves. Hence the top row in
(the first two vertical arrows being algebra structure maps) is a coequalizer in . The last vertical arrow making the diagram commute gives a -algebra structure, and the split coequalizer in the bottom row is thereby reflected in (i.e., lifts to a coequalizer in , albeit not necessarily to one that is itself split).
Categories of algebras are Barr exact
If is a regular category or exact category in which regular epimorphisms split, and is any monad on , then is a regular category (or exact category, respectively).
For regularity, we first construct coequalizers of kernel pairs in . So suppose is the kernel pair of some in . The kernel pair of in has a coequalizer in , and of course is the kernel pair of as well. It follows from the splitting hypothesis that the fork
splits in , hence by Lemma 1 this diagram lifts to a coequalizer for in . Thus kernel pairs in have coequalizers.
That regular epis in are stable under pullback follows a similar line of reasoning: let be a regular epi in . It is the coequalizer of its kernel pair . We just calculated that the coequalizer of in lifts to , so that is identified with and with . Thus is a regular epi in . Now if is a map in , and is the pullback of along (with kernel pair ), then is the pullback of along since preserves pullbacks, and so is a regular epi since is regular. This is the coequalizer of its kernel pair, and splits, so by Lemma 1, its lift is the coequalizer of . Thus regular epis in are stable under pullback.
For Barr-exactness, suppose is an equivalence relation (or congruence) in . Then is an equivalence relation in , and hence a kernel pair since is exact. It is the kernel pair of its coequalizer in . By Lemma 1, the split coequalizer
lifts to a coequalizer diagram in , and since kernel pairs are preserved and reflected by , we conclude that is the kernel pair of the lifted regular epi over .
If is a monad on a slice category , then the category of -algebras is (Barr-)exact. If is a monad on , then 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, Theories, and Triples (theorem 3.9, p. 267):
If has coequalizers and equalizers of arbitrary sets of parallel morphisms, and if a monad preserves colimits of countable chains , then has coequalizers.
If is complete and cocomplete and preserves filtered colimits, or even just colimits of -chains, then is complete and cocomplete.
For locally presentable categories
If is a locally presentable category and is an accessible monad (aka a bounded monad, aka a monad with rank) on , then is also locally presentable and in particular cocomplete. Details may be found in Locally presentable and accessible categories.
Relatively free functors
Suppose that is a morphism of monads on , and suppose that has coequalizers. Then the relative “forgetful” functor
(pulling back a -algebra to the -algebra , thus remembering only underlying -algebra structure) has a left adjoint.
Since the following diagram is commutative:
(using an obvious notation), it follows immediately from a corollary to the adjoint lifting theorem that if has coequalizers of reflexive pairs, then 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 : it sends an -algebra to the (reflexive) coequalizer of the pair
where is the monad multiplication. (If is the unit of , then is a common right inverse of both arrows of the pair.) This coequalizer is analogous to the construction of the left adjoint to the “restriction” functor between module categories (restricting scalar multiplication on a -module along a ring map ); given an -module , the -module is the coequalizer in of
and so the coequalizer of (1) will be denoted to underline the analogy.
To see that is the left adjoint, let be a -algebra. Any map in induces a unique -algebra map :
and the claim is that is an -algebra map if and only if coequalizes the pair in (1), i.e., if factors (uniquely) through a -algebra map .
Indeed, assume is an -algebra map, so we have a commutative diagram
That coequalizes the pair of (1) follows by expanding the diagram
where using (2), the path along the top may be replaced by , reducing the desired coequalizing of (3) to the tautology .
Conversely, assuming the coequalizing of (3), the perimeter of (4) commutes, and on top of that we stack naturality diagrams for the monad unit of :
The vertical composite on the right is by a unit equation for a -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
where the commutativity of the unlabeled polygonal region is just the commutativity of (2). This completes the proof of the claim.
- Jiří Adámek and Václav Koubek, Are colimits of algebras simple to construct?, Journal of Algebra, Volume 66, Issue 1 (September 1980), 226-250. (link)