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I'm studying symmetric monoidal categories and I have seen some authors saying that, due to the Eckmann-Hilton argument, given some symmetric category $C$, the category $Mon(Mon(C))$ of monoidal objects in the category of monoidal objects of $C$ is equivalent to the category $CoMon(C)$ of commutative monoids in $C$, i. e., we have

$$Mon(Mon(C)) \cong CoMon(C) $$

In Aguiar and Mahajan's book, "Monoidal Functors, Species and Hopf Algebras", Proposition 6.29 of page 180 states:

"A commutative monoid in $C$ gives rise to a double monoid, for which both monoid structures are identical. Conversely, let $A$ be a double monoid in $(C, •, •)$. Then the two products on $A$ coincide and are commutative."

I understood the proof, but it is not completely clear to me how this proposition allow us to conclude that there is an equivalence $Mon(Mon(C)) \cong CoMon(C) $.

Formally speaking, do we have an equivalence functor between $Mon(Mon(C))$ and $CoMon(C)$?

And there is more: since a commutative monoid in $C$ and a monoid in $Mon(C)$ are the same thing, why do we have an equivalence and not, in fact, an isomorphism?

Mayk Alves de Andrade
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1 Answers1

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Taking your quote from Aguiar, the functor $F:\text{Mon(Mon(C))}\to\text{CMon(C)}$is well defined by $(A,\bullet,\ast)\mapsto(A,\bullet)$.

On the other hand, the functor $G:\text{CMon(C)}\to\text{Mon(Mon(C))}$ is given by$(A,\bullet)\mapsto(A,\bullet,\bullet)$.

Easy to see that $FG$ and $GF$ are identity functors, hence these are isomorphisms.

Daniel Teixeira
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