Take a Cartesian (or monoidal) closed category; define Reader monad for a given object $E$ as
$X \mapsto X^E$; and take a strong monad $M$ (strong means preserves product or tensor product).


Now the composition $M\circ E$ ($M$ followed by $E$) is a monad again. 


At least I believe it is true; and I wonder if it is a known fact, or plain wrong.



Seems like Reader monad composed with a strong monad produces a monad, am I right?
Although I believe the question needs cleaning up, what I believe to be the desired statement is true. I cannot find a direct reference for it at the moment, but the statement should be equivalent to something familiar from the literature.


First, the "Reader monad" for a fixed object $E$ crucially uses cartesian (not general monoidal) products; the canonical comonoid structure on $E$, with counit the unique map $!: E \to 1$ to the terminal object and comultiplication the diagonal map $\delta: E \to E \times E$, induces the monad structure on the Reader monad, with unit and multiplication given respectively by


$$X \cong X^1 \stackrel{X^!}{\to} X^E,$$


$$(X^E)^E \cong X^{E \times E} \stackrel{X^\delta}{\to} X^E.$$


So, let $\mathcal{C}$ be a cartesian closed category; for an object $E$ of $\mathcal{C}$, we'll denote the associated reader monad by $\mathbf{E}$.


Next, there are standard notions of strong endofunctors on $\mathcal{C}$ and strong monads $M$ on $\mathcal{C}$; it is not that $M$ preserves products, but rather that there is a tensorial strength, i.e., a natural transformation


$$\theta\_{X, Y}: X \times MY \to M(X \times Y)$$


satisfying a number of coherence conditions. These conditions involve only finite products, but if we are dealing (as we are here) with a cartesian closed category $\mathcal{C}$ so that $\mathcal{C}$ is enriched in itself in the usual way, then strong endofunctors $M: \mathcal{C} \to \mathcal{C}$ are tantamount to $\mathcal{C}$-enriched endofunctors where the enrichment


$$A^B \to MA^{MB}$$


is derived from the strength by currying the composite


$$A^B \times MB \stackrel{\theta\_{A^B, B}}{\to} M(A^B \times B) \stackrel{M(eval\_{A, B})}{\to} MA;$$


there is a similar inverse procedure for deriving the strength from the enrichment. Similarly, a *strong monad* means we are dealing with an enriched monad, meaning that we have an enriched endofunctor $M: \mathcal{C} \to \mathcal{C}$ and enriched natural transformations $m: MM \to M$, $u: 1\_{\mathcal{C}} \to M$.


Next, the question seems to be about a canonical monad structure on the $\mathbf{E} \circ M$ (which is $M$ followed by $\mathbf{E}$), not $M \circ \mathbf{E}$ as confusingly written. (The notation $\circ$ should be used only for the traditional right-to-left order of composition; the left-to-right "followed by" is usually denoted by a semicolon $;$ instead of $\circ$, to avoid confusion.)


It is well-known that such a monad structure on the composite is guaranteed by a [distributive law](http://ncatlab.org/nlab/show/distributive+law) between monads. Here the distributive law is a natural transformation $\sigma: M \circ \mathbf{E} \to \mathbf{E} \circ M$, in other words a transformation


$$\sigma\_X: M(X^E) \to (MX)^E$$


natural in $X$, satisfying a number of compatibility conditions between the two monad structures. Given such a distributive law, the monad multiplication on $\mathbf{E} \circ M$ is given by


$$\mathbf{E} \circ M \circ \mathbf{E} \circ M \stackrel{\mathbf{E} \circ \sigma \circ M}{\to} \mathbf{E} \circ \mathbf{E} \circ M \circ M \stackrel{m\_{\mathbf{E}} \circ m\_M}{\to} \mathbf{E} \circ M.$$


Now the punchline is that the desired distributive law is tantamount to the strength on the monad, i.e., $\sigma\_X$ is obtained by currying


$$M(X^E) \times E \to M(X^E \times E) \stackrel{eval\_{X, E}}{\to} MX$$


where the first arrow is related to the strength $\theta$ (as written above) by applying some symmetry isomorphisms (in the first and third arrows) as follows:


$$M(X^E) \times E \cong E \times M(X^E) \stackrel{\theta}{\to} M(E \times X^E) \cong M(X^E \times E)$$


I cannot find a suitable reference where the (routine) diagram chase is carried out, but compare the remarks after definition 6.1.1 (page 14) of


* Brookes, S., Van Stone, K.: Monads and Comonads in Intensional Semantics. Tech. Rep. CMUCS-93-140, Pittsburgh, PA, USA (1993)


(see [here](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.45.5695)), where instead of working with distributivity over the reader monad $\mathbf{E}$, the authors relate the tensorial strength to distributivity of a strong monad $M$ over the associated *comonad* $E \times -$ that is adjoint to $(-)^E$ (see pages 12-13 for the notion of distributivity between a monad and comonad).


