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 . 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. 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 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 .
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 , then is cocomplete.
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 -.) 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 .) It is therefore an absolute colimit, which the monad 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 reflected in .
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 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 preserve reflexive coequalizers.)
Here is a more difficult 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, then is complete and cocomplete.
Relatively free functors
Suppose that is a morphism of monads on , and suppose that has coequalizers. Then the forgetful functor
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 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 -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 functor between module categories given by restricting scalar multiplication; the coequalizer 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 of above, i.e., if factors (uniquely) through a -algebra map .
Indeed, the condition that is an -algebra map is satisfaction of the equation
and now we have a long train of equations
which gives . This completes one direction. In the other direction, suppose . Then
as desired. This completes the proof.