# How is a monoidal category with one object a commutative monoid?

The nLab article for commutative monoid says:

Just as a monoid can be seen as a category with one object, a commutative monoid can be seen as a bicategory with one object and one morphism (or equivalently, a monoidal category with one object).

I am not sure why this is true.

Suppose there is a category $$C$$ with a single object, $$X$$. This defines a monoid $$M := hom(X, X)$$ with composition as its operation. If $$C$$ is monoidal, then all of the following are true about $$M$$:

1. There is a function $$\otimes : M \times M \to M$$
2. $$id \otimes id = id$$
3. $$(f_1 \otimes f_2)(g_1 \otimes g_2) = f_1 g_1 \otimes f_2 g_2$$
4. There are three invertible elements $$\alpha, \lambda, \rho \in M$$.
5. $$((f \otimes g) \otimes h) \alpha = \alpha(f \otimes (g \otimes h))$$.
6. $$(id \otimes f) \lambda = \lambda f$$
7. $$(f \otimes id) \rho = \rho f$$
8. $$\alpha (id \otimes \alpha^{-1}) = \alpha^{-1}(\alpha \otimes id)\alpha$$
9. $$\alpha(id \otimes \lambda) = \rho \otimes id$$

(Note that I am writing composition in diagrammatic order.)

The laws above are simplified versions of the laws for a general monoidal category. (2) and (3) state that $$\otimes$$ is a functor. (5), (6), and (7) are the naturality conditions for $$\alpha$$, $$\lambda$$, and $$\rho$$ respectively. And (8) and (9) are the "coherence conditions" that all monoidal categories must satisfy.

How can one prove from the nine properties above that $$M$$ is a commutative monoid? The Eckmann-Hilton argument seems promising, but that argument only applies if $$\otimes$$ is unital, and $$\otimes$$ is only unital up to conjugation by $$\lambda$$ and $$\rho$$.

• You might take inspiration from §1.3.1 in "Monoidal Categories and Topological Field Theory", by Turaev and Virezilier. A key point is that in this context, $\lambda_1 = \rho_1 : 1\otimes 1 \to 1$ are the same element. If there is only one object, then, the associator is redundant because it can be expressed in terms of $\lambda$. Mar 22, 2022 at 13:55

The endomorphism monoid of the unit of a monoidal category is commutative. In particular, a monoid has a monoidal structure as one-object category only if it is commutative. Moreover, in that case monoidal structures are given by $$f\otimes g=f\circ g$$, a strict associator, and an invertible element, considered as an automorphism $$m\colon I\otimes I=I\to I$$, corresponding to both unitors $$\lambda_I$$ and $$\rho_I$$ (the monoidal axioms follow immediately).
For the proof, coherence requires that $$\lambda_I=\rho_I\colon I\otimes I\to I$$ for the unit $$I$$ of a monoidal category (this used to be one of the original coherence axioms until shown redundant).
Combined with the fact that $$\lambda_X\colon I\otimes X\to X$$ and $$\rho_X\colon X\otimes I\to X$$ are natural, we have an isomorphism $$\lambda_I=\rho_I=m\colon I\otimes I\to I$$ such that $$m\circ I\otimes f=f\circ m=m\circ f\otimes I$$ for each endomorphism $$f\colon I\to I$$. Since $$f=(f\circ m)\circ m^{-1}$$, we get that $$f*g=m\circ f\otimes g\circ m^{-1}$$ is a binary operation on endomorphisms of $$I$$ with unit $$\mathrm{id}_I$$ that is homomorphic with respect to composition. Eckmann--Hilton therefore applies to conclude $$*$$ and $$\circ$$ coincide and are commutative. In particular, if $$I\otimes I=I$$, then $$f\otimes g=m^{-1}\circ(f\circ g)\circ m=f\circ g$$ because $$m$$ commutes with $$(f\circ g)$$.
Finally, coherence also requires that $$a_{I,I,I}\colon I\otimes I\otimes I\to I\otimes I\otimes I$$ be the identity since the latter factors as $$m^{-1}\otimes I\circ m^{-1}\circ m\circ m\otimes I\colon I\otimes I\otimes I\to I\otimes I\to I\to I\otimes I\to I\otimes I\otimes I$$.