$\newcommand{id}{\operatorname{id}}$
$\newcommand{ev}{\operatorname{ev}}$
$\newcommand{coev}{\operatorname{coev}}$
This question refers to Etingof, Gelaki, Nikshych, and Ostrik's book Tensor Categories.
Let $(\omega,c) \in Z^3_{ab}(G ; k^\times)$ be a normalized $3$-cocycle on a finite abelian group $G$, with $k$ a field of characteristic $0$. Let $\mathsf{Vec}_G^\omega$ be the monoidal category of finite dimensional $G$-graded vector spaces over $k$ with associator on simple objects $\delta_g,\delta_h,\delta_l$ defined by
$$a_{\delta_g,\delta_h,\delta_l} = \omega(g,h,l)\operatorname{id}_{\delta_{ghl}}$$
for every $g,h,l\in G$.
From now on let $X\in\mathsf{Vec}_G^\omega$ be an object in the isomorphism class of $\delta_g$.
On page 43 EGNO say that, in the case that $\omega$ is normalized (i.e. $\omega(g,1,h)=1$ for all $g,h\in G$ by the triangle axiom), we can take the coevaluation $\operatorname{coev}_X : \mathbf{1} \to X^*\otimes X$ to be the identity, and the evaluation $\operatorname{ev}_X : X^*\otimes X \to \mathbf{1}$ to be $\omega(g,g^{-1},g)^{-1}\operatorname{id}_{\mathbf{1}}$. Note the book does not say this, but the errata gives a correction.
In checking that these morphisms (left) dualize simple objects, I verified that these morphisms make the composition
$$X \xrightarrow{\coev_X\otimes \id_X}(X\otimes X^*)\otimes X \xrightarrow{a_{X,X^*,X}}X\otimes (X^*\otimes X) \xrightarrow{\id_X \otimes \ev_X} X$$
the identity morphism, but I'm having trouble seeing why the dual
$$X^* \xrightarrow{\id_{X^*}\otimes \coev_X}X^*\otimes (X\otimes X^*) \xrightarrow{a^{-1}_{X^*,X,X^*}}(X^*\otimes X)\otimes X^* \xrightarrow{\ev_X \otimes \id_{X^*}} X^*$$
should be the identity.
Checking the definitions, we end with a factor $\omega(g^{-1},g,g^{-1})^{-1}\omega(g,g^{-1},g)^{-1}$. How can I realize this as 1? Or have I made a mistake in my work?
Best Answer
$\newcommand{id}{\operatorname{id}}$
I believe I've resolved my problem.
The 3-cocycle condition says that $$\omega(g_1g_2,g_3,g_4)\omega(g_1,g_2,g_3g_4)=\omega(g_1,g_2,g_3)\omega(g_1,g_2g_3,g_4)\omega(g_2,g_3,g_4)$$ for all $g_i\in G$. Set $g_1=g^{-1},g_2=g, g_3=g^{-1},g_4=g$. The cocycle condition says that $$\omega(1,g^{-1},g)\omega(g^{-1},g,1)=\omega(g^{-1},g,g^{-1})\omega(g^{-1},1,g)\omega(g,g^{-1},g).$$ Since $\omega$ is normalized, $\omega(g^{-1},1,g)=1$, so the right hand side is what we want to compute. If we show the left hand side is $1$, we will be done.
The first commutative diagram of EGNO Proposition 2.2.4 says $l_X \otimes \id_Y = l_{X\otimes Y} \circ a_{\mathbf{1},X,Y}$ for all objects $X,Y$, and the second commutative diagram says $r_{X\otimes Y} = (\id_X\otimes r_Y)\circ a_{X,Y,\mathbf{1}}$ for all objects $X,Y$. Replacing $Y$ with $X$ and $X$ with $X^*$ in the proposition, these equalities give relations \begin{align} r_{X^*\otimes X} &=\omega(g^{-1},g,1) \id_{X^*}\otimes r_{X} \\ l_{X^*}\otimes\id_{X} &=\omega(1,g^{-1},g)l_{X^*\otimes X}, \end{align} but EGNO Exercise 2.3.9 says that $l_Z=r_Z=\id_Z$ for all objects $Z$ if and only if $\omega$ is normalized. So all morphisms in the above relations pick up factors of $1$. Hence $\omega(g^{-1},g,1)\omega(1,g^{-1},g)=1$.
The same argument goes through by symmetry by swapping roles $g^{-1}\leftrightarrow g$ and $X^* \leftrightarrow X$.