In my answer, [EGNO] refers to Etingof, Gelaki, Nikshych, Ostrik: "Tensor Categories", AMS 2015, freely available on Etingof's homepage http://www-math.mit.edu/~etingof/books.html (note the copyright).
- If $C$ is simply a spherical fusion category, how is it additive? That is, why can you consider its tensor product as defined above?
By definition, a fusion category is a $\mathbb k$-linear abelian category and thus is, in particular, additive (see [EGNO, Definition 4.1.1]). The tensor products between the Hom spaces are therefore tensor products of $\mathbb k$-vector spaces.
- What does the expression $\bigoplus X_i \boxtimes Y_i, X_i \in Obj(C_1), Y_i \in Obj(C_2)$ stand for? That is, what does the symbol $\boxtimes$ refer to?
For $\mathbb k$-linear abelian categories $\mathcal C$, $\mathcal D$, the $\mathbb k$-linear abelian category $\mathcal C \boxtimes \mathcal D$ denotes their Deligne tensor product. The definition is given in terms of a universal property that is similar to the tensor product of other algebraic structures.
$\mathcal C \boxtimes \mathcal D$ is the $\mathbb k$-linear abelian category that is universal for functors from the category of $\mathbb k$-linear abelian categories to the category of right exact bilinear bifunctors out of $\mathcal C \times \mathcal D$, in the following sense: There is a right exact bifunctor $F_\boxtimes: \mathcal C \times \mathcal D \to \mathcal C \boxtimes \mathcal D$ such that any other right exact bifunctor $\mathcal C \times \mathcal D \to \mathcal A$ factors uniquely (in the appropriate sense) through $F_\boxtimes$ (see https://ncatlab.org/nlab/show/Deligne+tensor+product+of+abelian+categories or [EGNO, Definition 1.11.1] for details).
Accordingly, the object $X\boxtimes Y$ is the unique object of $\mathcal C \boxtimes \mathcal D$ that is associated to $(X,Y)\in \mathcal C \times \mathcal D$, i.e. $X\boxtimes Y \equiv F_\boxtimes (X,Y)$. The $\mathbb k$-linear abelian structure allows to take sums $\bigoplus_i X_i \boxtimes Y_i$ of such objects.
- How does the pivotal structure give these functorial isomorphisms? I tried reading the reference given in the paper, namely chapter 5.3 (which deals with Moore-Seiberg data) of Lectures on tensor categories and modular functor by Bakalov and Kirillov. However, I couldn't find a proof of that claim.
The isomorphisms $\langle V_1,\ldots, V_n \rangle \overset{\cong}{\to} \langle V_n,V_1,\ldots, V_{n-1} \rangle$ are obtained in the simplest way: Suitably employ the coevaluation and evaluation on the object $V_n$ to map a morphism $1\to V_1 \otimes\ldots\otimes V_n$ to a morphism $1\to V_n \otimes V_1 \otimes\ldots\otimes V_{n-1}$. I suggest you use graphical calculus (string diagrams) to represent what is going on algebraically: First use a coevaluation to bend the string labeled $V_n$ down, and then an evaluation to bring that string all around the other factors $V_1 \otimes\ldots\otimes V_{n-1}$ to the first position. What you get is a morphism in $\langle V_n,V_1,\ldots, V_{n-1} \rangle$. Reversing these operations is the inverse, thanks to the "snake identity" ("zig-zag moves") satisfied by the duality morphisms. Here the pivotal structure has to be used, otherwise one does not get the necessary identification between left and right duals.
Best Answer
The notation is perhaps a bit misleading. Here's what actually happens. Let $\{X_i\}$ be a complete collection of simple objects, and consider a basis $\psi(i)$ of $\mathrm{Hom} (1 , V_n^\ast \otimes \ldots \otimes V_1^\ast \otimes X_i)$ for every $i$.
Then by Lemma 1.8 in the notes, one has the diagram equality as shown in the proof, but with $\varphi,\varphi^\ast$ replaced by $\psi,\psi^\ast$ for better distinction. Note that the equation holds for every disconnected subgraphs $A$, $B$, which do not play a role in the argument.
As explained in the proof, the sum collapses to a single term corresponding to $X_i = 1$, and that term involves a basis of $\mathrm{Hom} (1 , V_n^\ast \otimes \ldots \otimes V_1^\ast \otimes 1)$ and its dual basis. This is isomorphic to a basis $\varphi$ of $\mathrm{Hom} (1 , V_n^\ast \otimes \ldots \otimes V_1^\ast)$ and its dual, and so we arrive on the left-hand side of the desired equation in the lemma.