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Yes. This follows from two simple facts: If $F:C \simeq D$ is an equivalence of categories, and $D$ has a braided monoidal structure, then there exists a braided monoidal structure on $C$ and an enh …

That equation is not correct. You should be suspicious because the definition of the $\mathcal{C}$-module category structure on $\mathcal{M} \boxtimes \mathcal{N}$ doesn't use the $\mathcal{C}$-module …

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 associat …

I don't understand what you mean about the "directed sphere" so will focus on the other questions. The free Picard $n$-category on one object has a description as a bordism $n$-category. Specifically …

The Reshetikhin-Turaev construction take as input a Modular Tensor Category (MTC) and spits out a 3D TQFT. I've been told that the other main construction of 3D TQFTs, the Turaev-Viro State sum constr …

Here is one set of data that will be sufficient. To get the monoidal structure you don't actually need a (monoidal) functor $C \to C \boxtimes C$. It is sufficient to have a bimodule category M from C …

The answer is no in general. Here is a counter example. Let us work over a ground field $k$, and let $ H = \oplus_n k$ be the direct sum of $n$ copies of $k$, with $n \geq 2$. This is a commutative, c …

So one way to categorify the natural numbers is to replace them with vector spaces. Then the dimension of the vector space reproduces the natural number. More generally you can categorify integers to …

It is indeed true that any strong braided monoidal functor from the braid category to a strict braided monoidal category is equivalent to a strict braided monoidal functor, however that comes out of t …

The categories of (finite dimensional) vector spaces and super vector spaces are rigid and symmetric monoidal and so admit ribbon structure where the twist is trivial. The forgetful functor from super …