(When) does a morphism of monad induce adjoint functors between categories of algebras?

Question

For monads $S$ and $T$ on a fixed Abelian category $C$, a morphism of monads $\sigma: S\rightarrow T$ induces a functor between Eilenberg-Moore categories $\sigma^*:C^T\rightarrow C^S$. This functor sends a $T$-algebra $(A,\rho:TA\rightarrow A)$ to an $S$-algebra $(A,\rho\circ \sigma_A:SA\rightarrow A)$. This functor commutes with forgetful functors $U^T:C^T\rightarrow C$ and $U^S:C^S\rightarrow C$.

Are there any known results that would also provide a left adjoint $\sigma_!$ to $\sigma^*$? In addition, I would like this left adjoint to commute with the free algebra functors for $T$ and $S$, i.e. $F^T=\sigma_!\circ F^S$.

Does it even make sense to expect such a functor to exist?

The reason I'm looking for a left adjoint is that I want to work with a category of Abelian categories whose morphisms are pairs of adjoint functors. Monads $S$ and $T$ are induced by an adjoint pair, there is a canonical $free \dashv forgetful$ adjunction for EM categories, so I would like to have an adjoint for $\sigma^*$ too.

Answer 1

First note that the "in addition" part is automatic because left adjoints compose and by design $U^S\sigma^* = U^T$.

Second, it is not unreasonable to expect $\sigma_!$ to exist : $\sigma^*$ preserves all limits that exist in $C$, so under common set-theoretic assumtpions, $\sigma_!$ will exist for free. But in this specific situation, one can be more explicit and relax the assumptions: as I said (and as you required), the value of $\sigma_!$ on free algebras is determined, and free algebras "generate" all algebras.

In particular, if $C^T$ admits reflexive coequalizers, then $\sigma_!$ exists.

These will exist if, e.g. $C$ has them (it does if $C$ is abelian as in your question) and $T$ preserves them (all monads preserve a certain amount of reflexive coequalizers, but not necessarily all). In that case, $U^T : C^T\to C$ in fact creates reflexive coequalizers, but this need not happen in general, even if $C^T$ has them.

In a different direction, let me spell out the set theory: if $C$ is presentable and $T$ is accessible, then $C^T$ is presentable and thus admits reflexive coequalizers (although in this case, the adjoint functor theorem also kicks in).

Answer 2

To complement the answer of Maxime with a useful reference, my go-to article for this circle of questions is "Coequalizers in categories of algebras" by Fred Linton (LNM 80, Seminar on Triples and Categorical Homology Theory, ETH 1966/67).