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.

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    $\begingroup$ George Bergman characterizes representable endofunctors of monoids in 10.6 of math.berkeley.edu/~gbergman/245/3.2.pdf $\endgroup$ Dec 15, 2019 at 16:08
  • $\begingroup$ 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. $\endgroup$
    – HeinrichD
    Dec 15, 2019 at 20:50
  • $\begingroup$ Ok so another specific example is $\langle x,y : xy = 1 \rangle$ with $\nabla(x)=x_1 x_2$ and $\nabla(y)=y_2 y_1$. $\endgroup$
    – HeinrichD
    Dec 15, 2019 at 23:11
  • $\begingroup$ 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 $xy=1$. $\endgroup$ Dec 16, 2019 at 19:21
  • $\begingroup$ It's just a submonoid $M \times M^{op}$, also in alignment with Bergman's notation. $\endgroup$
    – HeinrichD
    Dec 18, 2019 at 8:42

1 Answer 1


George Bergman characterizes representable endofunctors of monoids in 10.6 of https://math.berkeley.edu/~gbergman/245/3.2.pdf by classifying the comonoids in the category of monoids.


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