Recover the monoidal structure on $\mathbb{Ab}$ from the monad over Set

Question

The category of abelian groups $\mathbb{Ab}$ has a monoidal (closed) structure $(\otimes, \mathbb{Z})$. Moreover, it is monadic over the category of sets via the free abelian group monad $$\mathbb{Z}[\_]: \text{Set} \to \text{Set}.$$

Q: I wonder (and I believe) if it is possible to recover the monoidal structure from the monad and the underlying monoidal structure on the category of sets.

The question might seem vague, an accepted answer would look like "of course it is possible, $A \otimes B$ is some algebra structure on the coequalizer of some diagram of the form $\mathbb{Z}[\mathsf{U}A \times \mathsf{U}B] \rightrightarrows \mathbb{Z}[\mathsf{U}A \times \mathsf{U}B]$", or something similar where the cartesian product of Set and the monad appear.

Answer 1

Under some mild assumptions (which I don't remember at the moment) it is possible. So please consider my answer as the one that provides hints. You need to check the details yourself, which is rather tedious.

Note that your monad is monoidal (either lax or oplax I forgot which one and that is certainly important assumption) i.e. for sets $X,Y$ you have natural transformation

$$T(X) \times T(Y) \rightarrow T(X\times Y)$$

that makes certain diagrams involving the rest of monad data commutative.

Now for free algebras $(T(X),\mu_X)$ and $(T(Y),\mu_Y)$ define

$$(T(X),\mu_X)\otimes (T(Y),\mu_Y) = \left(T(X\times Y),\mu_{T(X\times Y)}\right)$$

This defines monoidal structure on Keisli category.

Note next that if $(A,h)$ is an algebra, then $(A,h)$ is a cokernel in the category of EM algebras of Kleisli algebras (this is called the canonical free presentation). Indeed, consider a pair of arrows $Th,\mu_A:(T^2(A),\mu_{TA})\rightarrow (T(A),\mu_A)$ in Eilenberg-Moore category, then $(A,h)$ is its cokernel. Define $(TX,\mu_X)\otimes (A,h)$ as a cokernel of a pair $(Th\otimes 1_{TX},\mu_A \otimes 1_{TX})$.

Finally for two algebras $(A,h)$ and $(B,g)$ you can tensor $(A,h)$ by the canonical free presentation of $(B,g)$ and pick a cokernel of the corresponding pair.

Eventually, this makes Eilenberg-Moore adjoint functors $F^T\perp G^T$ monoidal (lax or oplax, as I said I don't remember) and such that $G^TF^T = T$ as monoidal functors.

Answer 2

I don't know if that answers your question but it's too long for a comment :

Recall the usual construction of the tensor product : you take the free module on $A\times B$ and mod out by the submodule generated by the things that say that $(a,b)\mapsto a\otimes b$ is bilinear.

Here you can be more economical since you're tensoring over $\mathbb Z$, so it's even simpler : you only need $(a+a',b)- (a,b)-(a',b)$ and $(a,b+b')-(a,b)-(a,b')$.

So you get a map $\mathbb Z[UA\times UA\times UB \cup UA\times UB\times UB] \to \mathbb Z[UA\times UB]$ whose cokernel is $A\otimes B$ (in abelian groups), so if you take this map of sets, and the $0$ map of sets, then the coequalizer is $U(A\otimes B)$