# Monoidal equivalence is: A monoidal functor that is also an equivalence of categories

Let $(\mathcal C, \otimes, I)$ and $(\mathcal D, \otimes', I')$ be two monoidal categories. And let $(F,\phi, \psi)$ and $(G,\phi', \psi')$ be a monoidal functors from $\mathcal C$ to $\mathcal D$.

Monoidal natural transformation is defined as: a natural transformation $\alpha : F\Rightarrow G$ such that it satisfies the following commutative equations:

$$\begin{array} ~F(X) \otimes'F(Y) & \stackrel{\alpha_X\otimes'\alpha_Y}{\longrightarrow} & G(X)\otimes'G(Y) \\ \downarrow{\phi} & & \downarrow{\phi'} \\ F(X\otimes Y) & \stackrel{\alpha_{A\otimes B}}{\longrightarrow} & G(X\otimes Y) \end{array}$$

$$\begin{array} ~I' & {=} & I' \\ \downarrow{\psi} & & \downarrow{\psi'} \\ F(I) & \stackrel{\alpha_I}{\longrightarrow} & G(I) \end{array}$$

Let's define monoidal equivalence as the following: Two monoidal categories $\mathcal C$ and $\mathcal D$ are monoidally equivalent if there are monoidal functors $F:\mathcal C\to \mathcal D$ and $G: \mathcal D \to \mathcal C$ such that $id_\mathcal C \Rightarrow GF$ and $FG\Rightarrow id_\mathcal D$ are monoidal natural isomorphisms.

Question: I would like to show that given two monoidal categories $\mathcal C$ and $\mathcal D$, an existence of a monoidal functor $F:\mathcal C \to \mathcal D$ which is an equivalence of $\mathcal C$ and $\mathcal D$ as categories is an equivalent definition of the above notion of monoidal equivalence.

One way is easy. If we have monoidal equivalence given by two monoidal functors $F: \mathcal C\to \mathcal D$ and $G:\mathcal D \to \mathcal C$ and two monoidal natural isomorphisms $\eta:id_\mathcal C \Rightarrow GF$ and $\epsilon: FG\Rightarrow id_\mathcal D$, then we do have the monoidal functor $F:\mathcal C\to \mathcal D$ which is an equivalence of $\mathcal C$ and $\mathcal D$ as categories (exhibited by considering $G$, $\eta$, $\epsilon$ as merely just an ordinary functor, and natural isomorphisms).

To prove the other way, consider we have a monoidal functor $F:\mathcal C \to \mathcal D$ which is an equivalence of $\mathcal C$ and $\mathcal D$, exhibited by a functor $G:\mathcal D\to \mathcal C$ and natural isomorphisms $\eta:id_\mathcal C \Rightarrow GF$ and $\epsilon: FG \Rightarrow id_\mathcal D$. I must prove that $G$ is not just a functor but it is also a monoidal functor. To see this:

$$G(X)\otimes G(Y) \cong GF(GX \otimes GY) \cong G(F(GX) \otimes' F(GY) ) \cong G(X\otimes' Y)$$ and $$I \cong GF(I) \cong G(I')$$

And these satisfy the axioms for monoidal functor. Hence $G$ is a monoidal functor.

I must also prove that $\eta$ and $\epsilon$ are not just natural isomorphisms, but they are monoidal natural isomorphisms.

To do this I have to show that the following diagram commutes:

$$\begin{array} ~FG(X) \otimes'FG(Y) & \stackrel{\epsilon_X\otimes'\epsilon_Y}{\longrightarrow} & X\otimes' Y \\ \downarrow{\phi} & & \downarrow{=} \\ F(GX\otimes' GY) & & X\otimes'Y \\ \downarrow{F(\phi')} & & \downarrow{=} \\ FG(X\otimes'Y) & \stackrel{\epsilon_{A\otimes B}}{\longrightarrow} & X\otimes'Y \end{array}$$

$$\begin{array} ~I' & {=} & I' \\ \downarrow{F(\psi')\circ\psi} & & \downarrow{=} \\ FG(I') & \stackrel{\epsilon_I}{\longrightarrow} & I' \end{array}$$

And likewise for $\eta$.

However I could not easily see how this work.

• I haven't really thought about the problem, but it might help to choose $G$ so that it's an adjoint equivalence. The adjunction identities might simplify the problem. – Hurkyl Dec 12 '17 at 12:19
• This statement is well-known but takes a long time to write down and uses naturality many, many times. Someone once told me that it was covered in the book of Etingof, however I cannot find it. I do have a written down proof, however it is part of an unpublished paper and I'm not sure whether I can put it online. It is straightforward but a lot of work to check all the details. – Mathematician 42 Dec 12 '17 at 12:24
• @Mathematician42 Is this statement actually true? – MathsMy Dec 12 '17 at 12:31
• @Mathematician42 It is indeed mentioned in Remark 2.4.10 in Etingof's book www-math.mit.edu/~etingof/egnobookfinal.pdf. But the proof wasn't given. – MathsMy Dec 12 '17 at 12:32
• Ah indeed. Now it is in some sense easy to show, but that doesn't mean it won't take you a long time. The important thing is that a tensor functor is more than just a functor, it also comes with some natural transformations $\phi_0$ and $\phi_2$ (see Kassel Quantum groups for this notation). There are obvious guesses for the $\phi_0$ and $\phi_2$ of $G$. These guesses will work. (but require work) – Mathematician 42 Dec 12 '17 at 12:39