Let $j: A \to B$ be a fully faithful functor.

When $j$ has a left adjoint $L$, the codensity monad $\text{Ran}_jj$ will coincide with the monad $jL$ and thus will be idempotent, because $A$ is reflective in $B$.

Rem 1. Let $j$ be a fully faithful functor with a left adjoint. Then the codensity monad $\text{Ran}_jj$ is idempotent.

I was wondering if this is still true removing the assumption of having a left adjoint.

In a relatively trivial way, one can reformulate what I said in the following way.

Rem 2. Let $j$ be a fully faithful functor. TFAE:

  • $\text{Ran}_jj$ preserve itself.
  • $\text{Ran}_jj$ is idempotent.

So, the question is finally the following,

Q1. Let $j$ be a fully faithful functor, is it true that one of the two equivalent conditions in Rem 2 is verified? You can still assume that $j$ preserve all limits.

Q2. If not, are there some natural assumptions that will make it true?


The following addresses Q1 but not Q2.

Let $j: A \to B$ be the inclusion $Set_{\leq 3} \to Set$ of sets of cardinality $\leq 3$ into $Set$. Then the codensity monad for $j$ is the ultrafilter monad (as I learned from Tom Leinster-- I think this observation goes back to Isbell). This monad is not idempotent.

  • $\begingroup$ Hello Tim, I was partially aware of this counterexample, and of course, you are right. I will upvote the answer. I hope you will understand that I am much more into Q2, and that's why I can't accept this answer. $\endgroup$ – Ivan Di Liberti Oct 19 '18 at 17:36

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