# Is the inverse to a monoidal equivalence also monoidal?

Let ${\cal C,D}$ be two categories, and let $$F:{\cal C} \to {\cal D}, ~~~~~~~~~~~~~~~~~ G:{\cal D} \to {\cal C},$$ be an equivalence of categories. Let us now further assume that ${\cal C}$ can be endowed with a monoidal structure $\otimes$, and that ${\cal D}$ can also be endowed with a monoidal structure $\bullet$. Finally, let us assume, for each $X,Y \in C$, that we have isomorphisms $$J_{X,Y}: F(X \otimes Y) \to F(X) \bullet F(Y),$$ that give $F$ the structure of a monoidal functor.

Now it seems to me that this automatically gives $G$ the structure of a monoidal functor: Any two objects in ${\cal D}$ will be isomorphic to $F(X)$, and $F(Y)$, for some $X,Y \in {\cal C}$, and $$G(F(X) \bullet F(Y)) \simeq G(F(X \otimes Y)) \simeq X \otimes Y \simeq G(F(X)) \otimes G(F(Y)),$$ giving $G$ the structure of a monoidal functor. Thus, I would conclude that given a equivalence of monoidal categories as simple categories, such that one of the functors in the equivalence is a monoidal functor, then we always get an equivalence of monoidal categories.

Am I correct here, or have I missed some categorical subtlety?

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The $J_{X, Y}$ need to satisfy certain compatibility conditions which are laid out e.g. on Wikipedia: en.wikipedia.org/wiki/Monoidal_functor#Definition – Qiaochu Yuan Aug 16 '12 at 17:00
@QiaochuYuan: I believe the OP assumed these conditions when writing that the $J_{X,Y}$ "give $F$ the structure of a monoidal functor". – Brad Aug 16 '12 at 17:06
Yes, I assumed the compatibility conditions. Thanks for clarifying this though. – MikhailMatrix Aug 16 '12 at 17:12
Your argument looks good so far, but to be complete you should perhaps verify that these isomorphisms you produced satisfy the aforementioned compatibility conditions themselves. It would be pretty tedious, but I believe it works out. – Brad Aug 16 '12 at 17:19

This isn't a really problem though because if you have some object A, you know basically how to write it as F(X): just let X = G(A)! So we just have $$G(A \bullet B) \cong G(F(G(A))\bullet F(G(B))) \cong GF(G(A) \otimes G(B)) \cong G(A) \otimes G(B).$$