I'm learning about monads in Riehl's Category Theory in Context, and after reading Example 5.1.4.ii, about the free monoid monad (also known as the list monad to computer scientists) on the monoidal category $(\mathsf{Set}, \times, *)$, I've reached Exercise 5.1.i, meant to demonstrate that this monad can in fact be defined on any monoidal category with
"coproducts that distribute over the monoidal product." To summarize the example, the list monad is the functor with action $T(A) = \coprod_{n \geq 0} A^n$ induced by the free $\dashv$ forgetful adjunction between sets and monoids, where the unit components are the coproduct inclusions and the multiplication components are concatenations. (e.g. $((a,b),(c)) \mapsto (a, b, c)$.)
In the exercise, the reader is given a generic monoidal category $(V, \otimes, *)$ with finite coproducts which $\otimes$ "preserves in each variable." (A footnote clarifies this with the equality $(v \sqcup v') \otimes (w \sqcup w') = v \otimes w \sqcup v' \otimes w \sqcup v \otimes w' \sqcup v' \otimes w'$. Isn't this "the monoidal product distributing over the coproduct," the reverse of the author's claim?) Then the reader is asked to define the unit and multiplication ($\eta : 1_V \Rightarrow T$ and $\mu : T^2 \Rightarrow T$) so that $T(X) = \coprod_{n \geq 0} X^{\otimes n}$ is a monad. (Tangent: shouldn't $V$ be required to have countable rather than just finite coproducts for this to make sense?) In analogy with the example, it's clear to me that the the unit components should again be the coproduct inclusions, but I'm having a really hard time defining the multiplication.
If I understand correctly, in the example the singleton $*$ is both the unit object and the terminal object. That would allow me to take the morphism $T(A)^{k-1} \to *$ and apply $T(A) \times (-)$ to get $T(A)^k \to T(A) \times * \cong T(A)$ for any $k$, defining a cone from the diagram $\emptyset, T(A), T(A)^2, \dots$ to $T(A)$ and thus inducing a unique morphism $T^2(A) \to T(A)$ by the universal property which I can take to be $\mu_A$. But not only am I unable to verify that this coincides with the example's concatenation, this also doesn't seem to be a useful analogy for the exercise since, as far as I can tell, there's no guarantee that $V$ has a terminal object at all!
EDIT: To clarify my response to Kevin Arlin's comment, the coproduct inclusions $A^k \xrightarrow{\iota} T(A)$ are split monomorphisms in $\mathsf{Set}$, (any injective function restricts to a bijection onto its image) giving us morphisms $T(A) \xrightarrow{r} A^k$ for any $k$ such that $r \circ \iota = \mathrm{id}_{A^k}$. This allows us to form diagrams as in the following example:
Here, $r \circ (a) = a$ and $r \circ (b) = b$, so it should be that $\tilde{r} \circ ((a), (b)) = (a,b)$, which is exactly the desired concatenation for pairs of singlet lists. Similar constructions define concatenations of triples, quadruples, etc. of lists into $T(A)$, and hopefully induce a unique map $T^2(A) \to T(A)$. But now my concerns become:
- This construction is specific to pairs of singlet lists. Each different combination of list lengths requires a unique such morphism. Can this collection of morphisms in $\mathsf{Set}(T(A)^2, T(A))$ define a single morphism?
- Are coproduct inclusions in $V$ still split monic?
Best Answer
Here's an alternative take on the problem: one may use Theorem 2.1 from nLab's article on the free monoid (note: $\operatorname{Mon}(\mathsf{C})$ means the category of monoids in a monoidal category $\mathsf{C}$) and then apply Lemma 5.1.3 of Riehl's book (saying that any functor adjunction induces a monad on the domain of the left adjoint). A proof of the mentioned Theorem 2.1 of the nLab's article may be read in Mac Lane's Categories for the Working Mathematician, VII.3, Theorem 2. Anyway, here's a proof outline:
We define a left adjoint $F:\mathsf{C}\to\mathrm{Mon}(\mathsf{C})$ to the forgetful functor $\mathrm{Mon}(\mathsf{C})\to\mathsf{C}$. It sends $c$ to $Fc=\sum_{n\geq 0}c^{\otimes n}$. The unit $\eta=i_0:*\to Fc$ is just the inclusion in degree $0$. Distributivity of the tensor product with respect to countable coproducts means that the tensored inclusions $c^{\otimes n}\otimes c^{\otimes k}\to Fc\otimes Fc$ give rise to a canonical isomorphism $$ \label{iso}\tag{1} \sum_{n,k\geq 0}c^{\otimes n}\otimes c^{\otimes k}\cong Fc\otimes Fc. $$ We will now define the multiplication $\mu:Fc\otimes Fc\to Fc$. By \eqref{iso}, this amounts to define a map $\mu_{nk}:c^{\otimes n}\otimes c^{\otimes k}\to Fc$, for each $n,k\geq 0$. If $n,k\geq 1$, then $c^{\otimes n}\otimes c^{\otimes k}=c^{\otimes n+k}$ and $\mu_{nk}=i_{n+k}:c^{\otimes n+k}\to Fc$ is the inclusion in degree $n+k$. If one of $n$ or $k$ is zero (WLOG suppose $n=0$), then we define $\mu_{0k}$ to be $*\otimes c^{\otimes k}\xrightarrow{\lambda_k}c^{\otimes k}\xrightarrow{i_k} Fc$, where $\lambda_k$ is the component of the left unitor of (the monoidal structure on) $\mathsf{C}$ at $c^{\otimes k}$.
With this unit and multiplication, $Fc$ is a monoid in $\mathsf{C}$: it is associative [ref]. Unitality is expressed by the commutativity of the outer square in the following diagram:
(Here, $i_{nk}:c^{\otimes n}\otimes c^{\otimes k}\to\sum_{n,k\geq 0}c^{\otimes n}\otimes c^{\otimes k}$ denotes the inclusion in degree $n,k$.) Note that, in the diagram, the arrows $\nwarrow$ and $\uparrow$ are isos by distributivity of the tensor product with respect to countable coproducts. Hence, it suffices to see that the four internal triangles commute. The top, bottom and right triangles commute basically by definition. The left triangle commutes by naturality of $\lambda$. Hence, $Fc$ is left-unital. Showing that $Fc$ is right-unital is analogous.
This finishes the definition of the action of $F$ on objects. We now define the action on morphisms. A morphism $c\to d$ in $\mathsf{C}$ induces a canonical map $Fc\to Fd$, not difficult to verify to be a morphism in $\operatorname{Mon}(\mathsf{C})$.
Let $m$ be a monoid in $\mathsf{C}$. It is left to see that the natural map \begin{align*} \tag{2}\label{nat} \operatorname{Mon}(\mathsf{C})(Fc,m)&\rightarrow\mathsf{C}(c,m)\\ (Fc\to m)&\mapsto(c\to Fc\to m) \end{align*} is a bijection. We first see it is onto: a morphism $c\to m$ induces a morphism $c^{\otimes n}\to m^{\otimes n}$. Post-composing with the $n$-ary multiplication $m^{\otimes n}\to m$ (defined up to unique natural isomorphism, see Mac Lane's book, VII.3, Proposition 1) and summing everything up gives a map $Fc\to m$. One may verify that this constitutes a morphism of monoids.
To see injectivity, given a morphism of monoids $Fc\to m$, it suffices to see that for $n\geq 1$, the map $c^{\otimes n}\to Fc\to m$, is completely determined by $c\to Fc\to m$. The diagram $$ \require{AMScd} \begin{CD} (Fc)^{\otimes n}@>>> Fc\\ @VVV@VVV\\ m^{\otimes n}@>>> m \end{CD} $$ commutes by hypothesis. Precomposing the diagram with the canonical map $j:c^{\otimes n}\to (Fc)^{\otimes n}$ (equal to the composite $c^{\otimes n}\to\sum_{k_1,\dots,k_n\geq 0}c^{\otimes k_1+\cdots +k_n}\cong(Fc)^{\otimes n}$, where the latter is canonical and the former is the inclusion in degree $(k_1,\dots,k_n)=(1,\dots,1)$) gives the desired result.