# Does Tensor Categories section 2.6 have a mistake?

I'm reading the book Tensor Categories by Etingof, Gelaki, Nikshych, and Ostrik. In Section 2.6 they discuss the monoidal functors between two categories $$\mathcal{C}_{G_1}^{\omega_1}(A)$$ and $$\mathcal{C}_{G_2}^{\omega_2}(A)$$, where $$G_i$$ are finite groups, $$A$$ a fixed abelian group and $$\omega_i \in Z^3(G_i,A)$$.

The monoidal category $$\mathcal{C}_G^\omega$$ consists of the following datum:

• Objects are given by the fixed finite group $$G$$, i.e. $$\mathrm{ob}(\mathcal{C}_G^\omega) := G$$.
• Morphisms: $$\mathrm{Hom}(g,h) = \emptyset$$ if $$g\neq h$$ and $$\mathrm{Hom}(g,g) = A$$ for the fixed abelian group $$A$$. (In other words, as a category $$\mathcal{C}_G^\omega = \mathrm{B}A^{\amalg G}$$).
• Tensor products on objects is given by multiplication of $$G$$ and on morphisms by multiplication in $$A$$.
• Finally, $$\omega \in Z^3(G,A)$$ corresponds to the associators.

I think in addition to $$f:G_1\rightarrow G_2$$ (monoidal on objects) and $$\mu:G_1 \times G_1\rightarrow A$$ (monoidality isomorphism), a monoidal functor is also specified by an endomorphism $$g:A\to A$$ that determines how the functor maps morphisms. Also, $$\omega_1 = f^\ast \omega_2 \cdot d_3(\mu)$$ (see (2.31) in [EGNO]) should be changed into

$$g_{*}\omega_1=f^{*}\omega_2\cdot d_3(\mu)\,.$$

Therefore, a monoidal functor from $$\mathcal{C}_{G_1}^{\omega_1}$$ to $$\mathcal{C}_{G_2}^{\omega_2}$$ should correspond to a triple $$(f,g,\mu)$$, instead the mere pair $$(f,\mu)$$.

Who's correct? Me, the book, or neither?

• To make the question more accessible, I suggest to either add a link to the book (I think it was online, right?), or wrote down the definition of the categories, what (2.31) is saying and what brings you to your conclusion. Commented Nov 7, 2023 at 18:10

Yes, you are correct in that a monoidal functor $$F:\mathcal{C}_{G_1}^{\omega_1}\rightarrow \mathcal{C}_{G_2}^{\omega_2}$$ is determined by a triple $$(f,g,\mu)$$.
In EGNO they consider functors that act trivially on the coefficient group $$A$$, i.e. $$g = \mathrm{id}$$ in your example. They explain that in their corrections here.