TITLE: A tensor category need not be isomorphic to a strict tensor category
QUESTION [5 upvotes]: This question was originally posted on MSE, but got no answer even after putting a bounty on it, so I'll try here.

I'm reading the book "Tensor categories" by Etingof, Gelaki, Nikshych, and Ostrik. In remark 2.8.6 (posted below), it is claimed that the category $\mathcal{C}_G^\omega$ (defined in example 2.3.8, also below) is not isomorphic to a strict tensor category whenever $\omega$ is not cohomologically trivial. Can someone explain this a bit more?
If someone knows another example of a tensor category that is not isomorphic to a strict tensor category, I'm also interested in that.

REPLY [6 votes]: First consider the category $\mathcal{C}_G$ with its bifunctor $\otimes$ and unit. How many ways are there to enhance this to a monoidal category structure? The missing data are precisely the associators. The associators will be a choice of map
$$a_{\delta_g, \delta_h, \delta_m}: (\delta_g \otimes \delta_h) \otimes \delta_m \to \delta_g \otimes (\delta_h \otimes \delta_m)$$
for each $g,h,m \in G$. However in $\mathcal{C}_G$, the hom spaces are identified with $A$. Thus the associator amounts to an assignment of elements $\omega(g,h,m) \in A$ for each triple $g,h,m \in G$. Now the associator is required to satisfy some equations, such as the pentagon equation. Written in terms of the $\omega$, the pentagon equation is equivalent to saying that $\omega$ satisfies the cocycle condition, which we might write as $\delta  \omega= 0$. (It should actually be a normalized cocycle because of the unit triangle equations, but lets ignore that).
When are two of these structures isomorphic? For that we need a tensor isomorphism. Assume for simplicity that it is the identity on objects and morphisms and preserves the $\otimes$ bifunctor strictly. This is not really a restriction as we could use the isomorphism to make this identification. There is still another piece of data, which is a collection of isomorphisms:
$$f_{\delta_g, \delta_h}: F(\delta_g) \otimes F(\delta_h) \to F(\delta_g \otimes \delta_h)$$
for each pair $g,h \in G$. As before we can identify this with a collection of elements $c(f,g) \in A$ for each pair $g,h \in G$. These are required to satisfy a hexagon identity which involves the two associators. When written in terms of these elements, this becomes:
$$\omega_1 - \omega_2 = \delta c$$
So that the two cocycles $\omega_1$ and $\omega_2$ differ by a cochain. In summary if two of these monoidal structures are related by an isomorphism of tensor categories, the two cocycles must be cohomologous.
Note: to be strict, means that the corresponding cocycle is the trivial cocycle. Putting this together we see that to be isomorphic to a strict tensor category requires that the underlying cocycle be cohomologically trivial.