Completeness and cocompleteness of the Kleisli category

Question

If (T,η,μ) is a monad over a category C, which is complete and cocomplete, then what about the Kleisli category? And also, if C is cartesian closed, what about the Kleisli one?

Answer 1

In general, the Kleisli category will be neither complete nor cocomplete nor cartesian closed, even if $C$ is.

You can think of the Kleisli category as the full subcategory of algebras whose objects are the free algebras. So take for example $T$ to be the monad coming from the free-forgetful adjunction between abelian groups and sets. In order for $\mathbb{Z} \times -: FreeAb \to FreeAb$ to have a right adjoint (as required by cartesian closure), it would have to preserve all colimits, for example coproducts. But $\mathbb{Z} \times (\mathbb{Z} \oplus \mathbb{Z})$ is not isomorphic to $(\mathbb{Z} \times \mathbb{Z}) \oplus (\mathbb{Z} \times \mathbb{Z})$.

Now let's tackle completeness. We can use the same example; the basic idea is that free abelian groups are not closed under (infinite) products. (In fact, a famous but nontrivial result, due to Kurosh I believe, is that the countably infinite power of $\mathbb{Z}$ is not free abelian.) But to apply this idea, one should first check that any limit in $Kl(T)$ really is constructed just as it would be in $Alg(T)$ or in $C$. That's not hard: first note that the underlying functor $U: Kl(T) \to C$ is representable (in fact, $U \cong \hom(F(1), -)$), so $U$ preserves any limits that exist in $Kl(T)$. Also, $U$ reflects isomorphisms (i.e., if $U(g)$ is an isomorphism in $C$, then $g$ is an isomorphism in $Kl(T)$). It follows that $U$ both preserves and reflects limits, and so any limit in $Kl(T)$ is constructed as it would be in $Alg(T)$ or $C$ (here, $Set$).

Finally, cocompleteness. This time I don't think the example above works quite as easily, but one example that does work is to consider the Kleisli category for the free-forgetful adjunction between $\mathbb{Z}_6$-modules and sets. I claim that the coequalizer of the pair $id, \mu_3: \mathbb{Z}_6 \to \mathbb{Z}_6$ ($\mu_3$ is multiplication by 3) does not exist in the category of free $\mathbb{Z}_6$-modules. I chose this particular coequalizer because it is an example of "splitting an idempotent"; here the idempotent is $\mu_3$. The virtue of splittings of idempotents is that they are preserved by any functor whatsoever, and reflected by any functor that reflects isomorphisms. Given this fact, it follows that splittings of idempotents in $Kl(T)$, if they exist, are constructed just as they would be in $Alg(T)$ or in $C$. But the idempotent splitting of this pair of maps in $\mathbb{Z}_6$-Mod is $\mathbb{Z}_2$, which is not free. It follows that the coequalizer doesn't exist in the Kleisli category.

Answer 2

The article on limits and colimits in the Kleisli category shows that (with $C_T$ denoting the Kleisli category, and $C^T$ denoting the Eilenberg-Moore category):

1) If $C$ is complete, and the canonical functor $C_T\to C^T$ has a right adjoint, then $C_T$ is complete. 2) If $C$ has coproducts, then $C_T$ has coproducts (already mentioned in the comments to another answer to the question). 3) Cocompleteness is more subtle, and under some condition on partial algebras cocompleteness of $C_T$ is equivalent to the existence of a left adjoint of the canonical functor $C_T\to C^T$.