# Every monoidal category is equivalent to a strict monoidal category

I'm reading the book "Quantum groups" by Kassel. I'm reading the proof that every tensor category is tensor equivalent with a strict tensor category. Here is the proof given. Below is more context.

The functor $$G: \mathcal{C}\to \mathcal{C}^{str}$$ is given by $$G(V) =(V)$$. However, I believe $$G$$ is NOT strict since $$G(U \otimes V) = (U \otimes V) \neq (U,V) = (U)*(V) = G(U)*G(V)$$

Is there a way to fix this proof? Maybe $$G$$ is not a strict tensor functor but still a tensor functor that gets the job done?

If not, can someone give a reference where this "finite sequence" approach to constructing the strict tensor category works?

Here is more context:

Let $$\mathcal{S}$$ be the class of finite sequences $$(V_1, \dots, V_k)$$ of objects in $$\mathcal{C}$$. We include the empty sequence $$\emptyset$$. The integer $$k$$ is by definition the length of the sequence. If $$S= (V_1, \dots, V_k)$$ and $$S' = (W_1, \dots, W_l)$$ are non-empty finite sequences, then we define $$S*S' := (V_1, \dots, V_k, W_1, \dots, W_l)$$. We also define $$S*\emptyset = S = \emptyset*S$$ for every finite sequence $$S \in \mathcal{S}$$.

We now associate to every $$S \in \mathcal{S}$$ an object $$F(S) \in \mathcal{C}$$. We give an inductive definition (recursion on the length of the finite sequence).

• $$F(\emptyset) = I$$

• $$F((V)) = V$$, $$\quad V \in \mathcal{Ob} \mathcal{C}$$

• $$F(S*(V)) = F(S) \otimes V$$, $$\quad S \in \mathcal{S}\setminus \{\emptyset\}, V \in \mathcal{Ob} \mathcal{C}$$

More generally, if $$(V_1, \dots, V_k) \in \mathcal{S}$$, we get $$F((V_1, \dots, V_k)) = (\dots((V_1 \otimes V_2) \otimes V_3) \otimes \dots )\otimes V_{k-1})\otimes V_k$$

We can now define the category $$\mathcal{C}^{str}$$. This is the category determined by:

• $$\mathcal{Ob} \mathcal{C}^{str}= \mathcal{S}$$

• $$Hom_{\mathcal{C}^{str}}(S,S') = Hom_\mathcal{C}(F(S), F(S')), \quad S,S' \in \mathcal{S}$$

• Composition of morphisms in $$\mathcal{C}^{str}$$ is simply the composition of the corresponding morphisms in $$\mathcal{C}$$.

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The functor $$G: \mathcal{C} \to \mathcal{C}^{str}$$ is defined by $$G(V) = (V)$$ and $$G(f) = f$$.

• I assume from what's said that the objects of $C^{\mathrm{str}}$ are finite sequences of objects of $C$ and that the functor $F$ multiplies a sequence of objects together using the tensor product of $C$. In this case I think the answer is just that $G$ is a strong monoidal functor. Jun 19, 2020 at 13:09
• @MikeShulman In the book, strict tensor functor $F$ is defined as a tensor functor $(F, \varphi_0, \varphi_2)$ where $\varphi_0$ and $\varphi_2$ are identities, which is not the case here? What transformations should I use then instead of identities?
– user159891
Jun 19, 2020 at 13:17
• Strong means the natural transformations are natural isomorphisms, that is strong is named because it's stronger than "lax" or "oplax." Strong does not mean strict! Strong is weaker than strict. I think Kassel just calls strong tensor functors "tensor functors." Jun 19, 2020 at 18:14
• @NoahSnyder And do you see how to prove that $G$ is a tensor functor then? What transformations should I use?
– user159891
Jun 19, 2020 at 18:24
• I agree that the monoidal functor $G$ is not strict. For the proof, it doesn't need to be strict. We only need a tensor equivalence, i.e. an equivalence of categories involving monoidal functors (in Kassel's terminology). The way Kassel defines a monoidal functor, it is what others (as Mike Shulman remarked), call a strong monoidal functor, i.e. the structural natural transformation $(G_2)_{U,V}\colon G(U)\otimes G(V)\to G(V\star U)$ consists of isomorphisms. Jun 19, 2020 at 18:56

Our goal is to find maps in $$\mathrm{Hom}_{\mathscr{C}^{str}}((U\otimes V), (U,V)) = \mathrm{Hom}_{\mathscr{C}}(U\otimes V, U\otimes V)$$ satisfying some identities. But there's a very natural element on the RHS which is $$\mathrm{id}_{U \otimes V}$$. Ah, and now we see why Kassel got confused, which is that these maps are written as identity maps, but they're not literally identity maps in $$\mathscr{C}^{str}$$ and so you don't literally have a strict functor.
• So if I understand correctly, we have that $(G, \varphi_0 = id_I, \varphi_2)$ is a tensor functor where $id_I \in Hom_{C^{str}}(\emptyset, (I))$ and $\varphi_2(U,V):= id_{U \otimes V} \in Hom_{C^{str}}((U,V),(U \otimes V))$?