This is mainly about distributive laws between a monad and an endofunctor, where the monad is idempotent. It is partly overlaping to the context of entry compatible localization in *n*lab.

*Warning: under construction!! I noticed some errors in reasoning below, and I am redoing it – Zoran*

If $Q^*=Q^*_\Sigma:A\to B$ is a localization in general sense, that is a universal functor inverting some family $\Sigma$ of arrows in $A$, and $G:Q^*\to Q^*$ an endofunctor, then Lunts and Rosenberg say that $G$ is compatible with the localization if $G(\Sigma)\subset \Sigma$; it follows that $Q^* G$ also inverts all arrows in $\Sigma$. By the universal property of the localization, there exist a unique functor $G_\Sigma:B\to B$ such that $Q^* G = G_\Sigma\circ Q^*$. Given any endofunctor $G: A\to A$, such that there is an endofunctor $G':B\to B$ and a natural *isomorphism* $\xi: Q^* G \Rightarrow G' Q^*$, one can show that $G$ is compatible with the localization: indeed, $G'Q^*$ inverts all morphisms in $\Sigma$, as already $Q^*$ does; and the natural isomorphisms of functors preserve the inverted class (indeed, for $f:a\to a'$, consider the equality $\alpha_{a'}\circ Q^* G (f)= G'Q^*(f)\circ\alpha_a$ with invertible $\alpha_a,\alpha_{a'}$, clearly $Q^*G(f)$ is invertible iff $G'Q^*(f)$ is).

Let now $Q^*$ be a localization having a right adjoint $Q_*$. Notice that $G$ is compatible iff the natural transformation $Q^* G\eta: Q^* G\Rightarrow Q^* G Q_* Q^*$ is an isomorphism. Indeed, in one direction just set $G'=Q^* G Q_*$ in the statement above. On the other hand, in the other direction $Q^*G Q_* Q^* = G_\Sigma Q^* Q_* Q^*$, hence by the triangle identities for adjunction, $G_\sigma \epsilon Q^*$ is a left inverse of $Q^* G\eta$, and by the full invertibility of $\epsilon$, also the right inverse.

**Proposition.** A distributive law $l:Q_* Q^* G\to GQ_* Q^*$ exists iff $G$ is compatible with the localization, and then it is invertible and unique.

*Proof.* The monad corresponding to a localization having a right adjoint is idempotent; i.e. the counit is an isomorphism. A characterization of idempotent monads is that every module over such a monad has invertible action, say $\nu: Q_* Q^* M\to M$. A distributive law $l$ provides a lift $G^l$, $(M,\nu)\mapsto (G M,G(\nu)\circ l)$; $G(\nu)\circ l$ is here an action itself so it is invertible as well. Now $G(\nu)$ is iso and $G(\nu)\circ l$ is iso, hence $l$ is iso.

Now uniqueness in compatible case. By invertibility of $\epsilon$ we can invert the vertical arrows in the distributivity pentagon, with a morphism involving eta; then we can use the unit triangles to splice the pentagon into 3 triangles. All arrows are in the diagram are now invertible, hence the commutativity of the upper triangle reads a formula for $l$, determining it uniquely.

**Remark.** I do not know if the distributive laws in opposite direction $G Q_* Q^*\Rightarrow Q_* Q^* G$ need to be invertible, hence inverses of the distributive laws above, hence unique. Namely in that case we can still invert the vertical arrows but the splicing of the pentagon does not help as we maybe do not have invertibility of $Q_* Q^* G\eta$. Is there an actual counterexample there, or there is an argument ruling it out ?

Last revised on September 9, 2019 at 15:25:49. See the history of this page for a list of all contributions to it.