# Recovering $\mathsf{Ab}$ from $\mathbb{Z}$-mod in a closed symmetric monoidal category

Given the usual definition of a module over a ring it is trivial to show that a $\mathbb{Z}$-module is an abelian group (it is just by definition). But my question concerns recovering this idea in a more abstract setting.

Let $$(\mathcal{C},\otimes,1)=(\mathsf{Set},\times,\{*\})$$ be the closed symmetric monoidal category of sets with the cartesian product. Then we have the notion of a commutative monoid in $\mathcal{C}$: an object $A\in\mathcal{C}$ along with morphisms $\mu\colon A\times A\to A$ and $\eta\colon 1\to A$ satisfying certain commutative diagrams.

If we have such a monoid $A$ then we can define a module over $A$: an object $M\in\mathcal{C}$ along with an action $\sigma\colon A\times M\to M$ satisfying certain commutative diagrams.

Now, by closedness of $\mathcal{C}$, we know that specifying an action $\sigma\colon A\times M\to M$ is equivalent to specifying a morphism $\varphi\colon A\to\mathrm{Hom}_{\mathsf{Set}}(M,M)$ given by currying. That is, $$\varphi(a)(-)=\sigma(a,-).$$

It is just a matter of diagram chasing to then show that the axioms that make $\sigma$ an action give us certain constraints on $\varphi$. In particular, they ensure that $\varphi$ is a morphism of monoids (where the monoid structure on $\mathrm{Hom}_{\mathsf{Set}}(M,M)$ is given by composition, and $\eta\colon 1\to\mathrm{id}_M$).

Taking $A=\mathbb{Z}$ (since $\mathsf{Comm}(\mathsf{Set})=\mathsf{CMon}$, the category of commutative monoids in the usual sense), we see that $M\in\mathsf{Set}$ can always be seen as a $\mathbb{Z}$-module by taking the trivial action $\sigma\colon(n,x)\mapsto x$. This corresponds to having $\varphi\colon n\mapsto\mathrm{id}_M$. More generally though, an action $\sigma$ determines a monoid morphism $\varphi\colon\mathbb{Z}\to\mathrm{Hom}_\mathsf{Set}(M,M)$ satisfying

1. $\varphi(m+n)=\varphi(m)\circ\varphi(n)$;
2. $\varphi(0)=\mathrm{id}_M$.

My (simple) question: how does this allow us to consider some $\mathbb{Z}$-module $M$ as an abelian group - what is our binary operation $M\times M\to M$?

Edit: thinking about it more, it seems like it might be more logical to think of $\mathbb{Z}\in\mathsf{CMon}$ as a multiplicative monoid, since the natural $\mathbb{Z}$ action on a group comes from $$nx:=\underbrace{x+\ldots+x}_{n\text{ times}}$$ in which case we would have $\varphi(mn)=\varphi(m)\circ\varphi(n)$ and $\varphi(1)=\mathrm{id}_M$, but I still can't quite piece it all together.

Edit (follow-up question): if we set $(\mathcal{C},\otimes,1)=(\mathsf{Ab},\otimes,e)$ the be the category of abelian groups with the usual tensor product then, by definition, we see that, since $\mathbb{Z}$ is a ring, and $\mathsf{Comm}(\mathcal{C})$ is the category of commutative rings, $\mathbb{Z}$-mod is exactly the category of abelian groups again. So do we get different $\mathbb{Z}$-mods depending on how much structure of $\mathbb{Z}$ we use?

• Just because you obfuscate an argument does not mean you're not using the argument... Even if you manage to write down enough abstract nonsense to prove the result you're after, once you unwind all the definitions you will get the same proof. It's just like that "topological" proof on the infinite number of primes. – Najib Idrissi Feb 3 '16 at 14:23

Your first construction is a category of sets a set-monoid acts on. Its objects are called monoid actions, and are exactly what you wrote down - either sets equipped with a nice operation $A × M → M$, or a way to interpret elements of $A$ as transformations of $M$ (ie. a nice function $A → \mathrm{End}(M))$. These things certainly don't come automatically equipped with a natural binary operation for any monoid $A$.
What is usually meant by $ℤ$-modules are on the other hand the category of abelian groups the $(\mathrm{Ab}, ⊗)$-module $ℤ$ acts on, ie. exactly your second construction.
To make your last comment on the amount of structure precise, note that the forgetful functor $U : \mathrm{Ab} → \mathrm{Set}$ is monoidal, and every monoidal functor sends monoids to monoids, and furthermore lifts to a functor between the categories of modules over those monoids. In particular, $(\mathrm{Ab}, ⊗)$-module/ring $ℤ$ gets sent to $(\mathrm{Set}, ×)$-monoid $(ℤ, \cdot)$, and every abelian group gets sent to its underlaying $ℤ$-action.
• That's cleared up quite a few things, thanks! Two more questions though: (1) does this forgetful functor have an adjoint? if so, can we use this to remedy the problem (i.e. use the first construction and then 'lift' to obtain a $\mathbb{Z}$-module in the usual ($\mathsf{Ab}$) sense)? (2) when we talk of $\mathbb{N}$-modules we usually mean commutative monoids, so what category do we have to think of $\mathbb{N}$ as sitting inside to recover this? is it exactly the category of commutative monoids? – Tim Feb 2 '16 at 22:34
• Which functor? The forgetful functor ℤ-Mod → ℤ-Set from Abelian groups to ℤ-actions has an adjoint by abstract nonsense. Given a ℤ-action $\cdot : ℤ × X → X$, I'm guessing it's something like the free Abelian group $FX$ on $X$ quotiented by the subgroup generated by $nx - n \cdot x$. – user54748 Feb 2 '16 at 23:55
• As for your second question, I've never heard anyone talk of ℕ-modules and mean commutative monoids by it, so I'm not even sure I'm understanding you correctly, but yes, I guess this is true. ℕ-modules in Set are just ℕ-actions, but in (CMod, ⊗) it should be that a) monoids are rigs b) ℕ-modules are just commutative monoids again. In fact the latter is true in every monoidal category $(C, ⊗, I)$. Every object of $C$ is an $I$-module in a unique, obvious way. – user54748 Feb 3 '16 at 0:03