Zoran Skoda compatible idempotent monads

Given two idempotent monads in a category $A$, we say that they are mutually (strongly) compatible if there is an invertible distributive law between them. Such a distributive law is automatically unique and can be given by a formula.

Motivated by the localization theory, denote $Q_\mu = Q_{\mu *}Q_\mu^*$ and $Q_\lambda = Q_{\lambda*}Q_\lambda^*$ the underlying endofunctors of the monads, where $Q_\mu^*$ and $Q_\lambda^*$ are the free functors and $Q_{\mu*},Q_{\lambda*}$ the forgetful functors between the Eilenberg-Moore categories $A_\lambda = A^{Q_\lambda}$, $A_\mu = A^{Q_\mu}$ and $A$.

Regardless the compatibility, in this situation define the category $A_{\mu\lambda}$ as the equalizer

Suppose there is an invertible distributive law $Q_\lambda Q_\mu \cong Q_\mu Q_\lambda$, then one has the lift $\bar{Q}_\lambda : A^{Q_\mu}\to A^{Q_\mu}$ and the composed monad $Q_\mu Q_\lambda$ with $A^{Q_\mu Q_\lambda}\cong (A^{Q_\mu})^{Q_\lambda}$ and the decomposition of $\bar{Q}_\lambda$ as

$$A^{Q_\mu}\stackrel{\bar{Q}_\lambda^*}\longrightarrow (A^{Q_\mu})^{Q_\lambda} \stackrel{\bar{Q}_{\lambda*}}\longrightarrow A^{Q_\mu}$$

The free functor $\bar{Q}_\mu^*$ above is a localization, hence in particular essentially surjective on objects, and $\bar{Q}_{\lambda*}$ is fully faithful, thus $A^{Q_\mu Q_\lambda}$ is the essential image of $\bar{Q}_\lambda$.

Claim (ZŠ, GB) Under the invertible compatibility, the equalizer above, the essential image of $\bar{Q}_\lambda$, and the consecutive EM category are equivalent:

$$A_{\mu\lambda} \cong (A^{Q_\mu})^{Q_\lambda} \cong EssIm \bar{Q}_\lambda.$$

Furthermore, under these conditions, $A_{\lambda\mu}\cong A_{\mu\lambda}$ and the latter equivalence commutes with the forgetful functor to $A$. See also compatible localization.