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Any unitary fusion category has a canonical spherical structure. See Example 2.12 of this paper and the references therein. It also follows from a more general result Prop 8.23 in this paper. So you …

As Oscar has explained in comments, with the most common definitions it's just not true that $M \sqcup N$ is exactly the same as $N \sqcup M$. But even if you were working with some version of the ca …

As Charlie pointed out in comments Balsam and Kirillov were working on this, and since his comment they posted a preprint to the arxiv: http://arxiv.org/abs/1004.1533 For another point of view see Ke …

Kevin's parenthetical about needing things to be sufficiently finite and semisimple suggests that thought of another way the answer is "no." In particular, there are known non-semisimple TQFTs. I kn …

Let's first think about 1-dimensional TQFTs. As is well-known, these correspond exactly to finite dimensional vector spaces as follows. The positive point is assigned to some vector space V, the nega …

I think that the answer is "yes" if by TQFT you mean one that extends all the way down to 1-manifolds. The MTC is the thing assigned to a circle by this extended TQFT. There's also lurking somewhere …

I'm not comfortable enough with the $\infty$ setting to give a full answer, but for ordinary algebras the condition you're looking for here should be that of a symmetric Frobenius algebra (i.e. the Fr …

My initial answer was wrong, here's the correct version plus a reference: Turaev-Turner New generating morphisms: The Mobius strip $\emptyset \rightarrow S^1$ and the "orientation reversing" diffemor …

This is addressed in Remark 2.4.30 of Jacob's paper. The PL case has a very nice description but the topological case does not. In particular, there's no difference between framed bordisms in the PL …

In the non-unitary setting ENO proved that if $Z(C)$ and $Z(D)$ are equivalent as braided tensor categories, then C and D are Morita equivalent. This is Theorem 3.1 of this paper. Note that they sa …

Let me try to answer your questions at least in part. My apologies for references I've missed. For an overview of the ideas without references, you might enjoy Pavel Safranov's talks at the intro co …