# Comonoids in the category of monoids

Let us give the category of monoids $$\mathbf{Mon}$$ a monoidal structure with $$\otimes = \sqcup$$ (coproduct). How can we classify $$\mathbf{CoMon}(\mathbf{Mon})$$, the category of comonoids of monoids?

This category has (at least) two different descriptions, namely it is also the category of cocontinuous functors $$\mathbf{Mon} \to \mathbf{Mon}$$ (given by tensoring with a comonoid), but also the opposite of the category of continuous functors $$\mathbf{Mon} \to \mathbf{Mon}$$ (the functors represented by comonoids). From the description of cocontinous functors $$\mathbf{Mon} \to \mathbf{Mon}$$ we also get a structure of a cocomplete monoidal category (with $$\otimes=\circ$$ being cocontinous in each variable) with a zero object.

So far I have found three basic comonoids: $$\langle x \rangle$$ with $$\nabla(x) = x_1 x_2$$, $$\langle x^{\pm 1} \rangle$$ with $$\nabla(x)=x_1 x_2$$, and $$\langle x \rangle$$ with $$\nabla(x)=x_2 x_1$$. The continuous functors $$\mathbf{Mon} \to \mathbf{Mon}$$ represented by them are the identity functor, the group of units functor (which factors over $$\mathbf{Grp}$$), and the opposite monoid functor. Their left adjoints are the identity $$\mathbf{1}$$, the group completion $$K$$ and the opposite monoid functor $$D$$. Hence, every coproduct of them is also an example. Apart from the unit $$\eta : \mathbf{1} \to K$$, I think that all other morphisms between them are zero, and it seems that we have $$D \circ K \cong K \circ D \cong K \circ K \cong K$$.

In case it turns out that this monoidal category is too complicated, what about the category of monoids in it, i.e. the category of Tall-Wraith monoids in $$\mathbf{Mon}$$? Can we classify them? For $$\mathbf{Grp}$$ it is well-known by a result of Freyd.

• George Bergman characterizes representable endofunctors of monoids in 10.6 of math.berkeley.edu/~gbergman/245/3.2.pdf Dec 15, 2019 at 16:08
• Thanks! You can make this an answer, right? It is spelled out in great detail in Section 10.6. The comonoids are classified by E-systems. Dec 15, 2019 at 20:50
• Ok so another specific example is $\langle x,y : xy = 1 \rangle$ with $\nabla(x)=x_1 x_2$ and $\nabla(y)=y_2 y_1$. Dec 15, 2019 at 23:11
• your example co-represents the functor taking a monoid $M$ to the submonoid of $R\times L$ (where $R$ is the submonoid of right invertible elements and $L$ is the submonoid of left invertible elements) choosing all pairs $(x,y)$ with $xy=1$. Dec 16, 2019 at 19:21
• It's just a submonoid $M \times M^{op}$, also in alignment with Bergman's notation. Dec 18, 2019 at 8:42