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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.

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I suggest that you edit the question to include all the corrections to the question that Todd Trimble starts his answer off with. –  Omar Antolín-Camarena Dec 11 '12 at 17:07

1 Answer 1

up vote 5 down vote accepted

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 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 this paper, 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).

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