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 $\rho$.

  • $\begingroup$ 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 $\lambda$. $\endgroup$
    – fosco
    Mar 22, 2022 at 13:55

1 Answer 1


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 $m^{-1}\otimes I\circ m^{-1}\circ m\circ m\otimes I\colon I\otimes I\otimes I\to I\otimes I\to I\to I\otimes I\to I\otimes I\otimes I$.

  • $\begingroup$ Thanks. I didn't realise that 𝜆𝐼 and 𝜌𝐼 were equal. $\endgroup$ Mar 23, 2022 at 23:19

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