For a category $C$, let $C-Set$ denote the category of set-valued functors $\delta\colon C\to Set$. Given categories $C$ and $D$, and a functor $F\colon C\to D$, composition with $F$ yields a functor that I'll denote by $$\Delta_F\colon D-Set\longrightarrow C-Set.$$ The functor $\Delta_F$ has both a left adjoint, which I'll denote by $\Sigma_F\colon C-Set\longrightarrow D-Set$, and a right adjoint $\Pi_F\colon C-Set\longrightarrow D-Set$. One then has a monad $M = \Delta_F\Sigma_F$ on $C-Set$ and a monad $N=\Pi_F\Delta_F$ on $D-Set$. Let $M-alg=(C-Set)^M$ denote the category of $M$-algebras on $C-Set$, and similarly, let $N-alg=(D-Set)^N$ denote the category of $N$-algebras on $D-Set$.

I'll say that $\Delta_F$ is monadic if the obvious functor $D-Set\longrightarrow M-alg$ is an equivalence of categories, and I'll say that $\Pi_F$ is monadic if the obvious functor $C-Set\longrightarrow N-alg$ is an equivalence of categories.

Questions:

1. Under what conditions on $F$ is $\Delta_F$ monadic?

2. Under what conditions on $F$ is $\Pi_F$ monadic?

Thanks!

The functor $\Delta_F$ verifies automatically nearly all the conditions of the monadicity theorem:

• it is a right adjoint;
• it is a left adjoint with cocomplete domain, and thus coequalizers exist in the source and are preserved by $\Delta_F$.

It remains to see when the functor $\Delta_F$ is conservative, i.e. reflects isomorphisms. This is true, for example, in each of the following two cases:

• $F$ is essentially surjective;
• $\Delta_F$ is full and faithful. This condition holds if $F$ induces an equivalence of Cauchy completions, i.e. $F$ is full and faithful and every object in the codomain is a retract of an object in the image. $\Delta_F$ is also full and faithful when $F$ is a reflection onto a reflective subcategory, or more generally when $F$ formally inverts some arrows (such as the map from a model category to its homotopy category).

I've been wondering about the same thing recently and here's my best guess. Let $\widetilde C$ denote the Cauchy completion of $C$. if the associated functor $\widetilde F : \widetilde C \to \widetilde D$ induces a bijection of the sets isomorphism classes of objects in $\widetilde C$ and $\widetilde D$, then $\Delta_F$ is monadic.

Here's a rough idea why this is true. The monad $M$ is given by a coend formula $M X = \int^c X c \times D(F c, F-)$. For simplicity let's assume that $C$ and $D$ are Cauchy complete (we can do this since $C$ and $\widetilde C$ are Morita equivalent) and $F$ is actually bijective on objects (replace $C$ and $D$ by their skeleta). Now a map $M X \to X$ is a transformation $X c \times D(F c, F c') \to X c'$ dinatural in $c$ and natural in $c'$ i.e. a transformation $D(F c, F c') \to \mathrm{Set}(X c, X c')$ natural in both in $c$ and $c'$. Since $F$ is bijective on objects we can rewrite it as $D(d, d') \to \mathrm{Set}(X d, X d')$ where $X (F c) = X c$. At this point this is just a family of maps and I do not claim any compatibility with composition in $D$. However, if $M X \to X$ is an $M$-algebra structure on $X$, then this family actually obtains a structure of a functor $D \to \mathrm{Set}$. It shouldn't be to difficult to verify (I admit I haven't done it in full detail myself) that this is an inverse to the canonical functor $\mathrm{Set}^D \to M$-$\mathrm{Alg}$.

I also have reasons to believe that the essential surjectivity is necessary. Here's a very rough idea why I suspect this. If $F$ is not essentially surjective, then we seem not to be able to control the behavior of functors $D \to \mathrm{Set}$ by means of $M$-algebra structure. If I wanted to try to actually verify this, I would try to show that in that case $\Delta_F$ fails to create limits. (This is what fails when $C$ is the terminal category and $D$ is a two-point discrete category.)

On the other hand I don't know whether the injectivity on the isomorphism classes is necessary.

I also don't know any answer to your second question and in fact I don't see any non-trivial example of $F$ such that $\Pi_F$ is monadic (if you have such an example, please share it).