In [arxiv:0909.3140](https://arxiv.org/abs/0909.3140) the $G$-extensions of a fusion category $\mathcal{D}$ are classified via monoidal 2-functors $G \to \underline{\underline{BrPic}}(\mathcal{D})$. A crucial part of the classification is Theorem 7.7 that shows that equivalence classes of such functors are in 1-1 correspondence with equivalence classes of $G$-extensions of $\mathcal{D}$. However in the proof they never show that the tensor categories they obtain from such functors are actually rigid. I was also unable to find this statement anywhere else in the paper, it is not even mentioned. Is there some obvious proof of this that I am missing or are they in fact classifying not necessarily rigid $G$-extensions? (The latter would however contradict Definition 2.1, which defines a G-extension to be in particular a fusion category, hence rigid.)



Why are $G$-Extensions of fusion categories rigid, when constructed via monoidal 2-functors $G \to \underline{\underline{BrPic}}(\mathcal{D})$?
Great question! Yes, this is missing from the original paper and is certainly needed for the main result. It is true though. There's two different proofs currently in the literature, [Deshpande and Mukhopadhyay](https://arxiv.org/pdf/1909.10799.pdf) Corollary 2.11 and [Davydov and Nikshych](https://arxiv.org/pdf/2006.08022.pdf) proof of Theorem 8.5.


You can also prove it using an argument I learned from Theo Johnson-Freyd, namely given an object $X$ in $C\_g$ you can realize $C\_g$ as $A$-mod in $C\_e$ with $X$ corresponding to $A$ itself, then $C\_{g^{-1}}$ can be identified with mod-$A$ and the dual object will be $A$ again and the evaluation and coevaluation maps are given by multiplication and the unit of the algebra structure on $A$.


As to your last two sentences, rigidity is pretty essentially to the whole setup, otherwise the graded parts don't need to be invertible and then you leave the world of homotopy theory. For example, take the semisimple category with two objects 1 and X with tensor product defined by $X \otimes X = 0$ (aka the universal counterexample to all questions involving rigidity), this is $\mathbb{Z}/2\mathbb{Z}$ graded but the odd part is not invertible as a bimodule over the even part.


