For $U\_q(\frak{g})$ the Drinfeld--Jimbo quantum group, its category of representations is equivalent to the category of representations of $U(\frak{g})$, or equivalently the category of Lie algebra representations of $\frak{g}$. Both categories have an obvious monoidal structure, what is not obvious is if this is an equivalence of monoidal categories.


Edit: As Phil's comment below, this is not a monodial equivalence. As the linked answer says, the problem is that the associators in both cases are different. What is the easiest way to see that this is so? The discussion about 6j symbols and coordinates on stacks is unfortunately lost on me. In fact, for modules over a Hopf algebra the assoicators look easy to the naive untrained eye, why is it about the assoiators on either category that is non-trivial?



Monoidal Equivalence for Drinfeld--Jimbo Quantum Groups
For simplicity let’s just do the $\mathfrak{sl}(2)$ case.
Let X be the 2-dimensional natural representation. From the fusion rules, $\mathrm{Hom}(X \otimes X,1)$ and $\mathrm{Hom}(1, X \otimes X)$ are one-dimensional. Choose a map in each normalized such that the zig-zag is the identity:
$$X = X \otimes 1 \rightarrow X \otimes (X \otimes X) \rightarrow (X \otimes X) \otimes X \rightarrow 1 \otimes X = X.$$
Note that this fixes the maps up to rescaling them in opposite ways. Now compute the quantum dimension (aka the circle):
$$1 \rightarrow X \otimes X \rightarrow 1.$$
This is a numerical invariant of such a monoidal category and a calculation shows that it is $q+q^{-1}$. Thus the monoidal categories are not equivalent except for replacing q by its inverse.


In the classical case these maps are the determinant map and the copairing associated to the determinant map. I.e. since $\det(e\_1, e\_1) = 0$, $\det(e\_1, e\_2) = 1$, $\det(e\_2, e\_1) = -1$, $\det(e\_2, e\_2) = 0$, the copairing sends 1 to $e\_1 \otimes e\_2 - e\_2 \otimes e\_1$. Thus the value of the circle is $$\det(e\_1 \otimes e\_2 - e\_2 \otimes e\_1) = 1+1 = 2.$$


(Some side notes for experts. 


* If $X$ were not self-dual you'd have to replace some of the $X$'s above with $X^{\*}$ which results in some trickiness, the upshot of which is you either need a pivotal structure, or (following ENO) you compute $\mathrm{qdim}(X)\mathrm{qdim}(X^{\*})$ to get a monoidal invariant. This wouldn't quite let you distinguish everything because of switching q and -q.
* When $X$ is self-dual something very tricksy is going on with the signs here... I will update this bullet when I understand it better.)


