TITLE: What does "quantization is not a functor" really mean?
QUESTION [41 upvotes]: The answers to this question do a good job of exploring, at a heuristic level, what "quantization" should be.  From my perspective, quantization involves replacing a (commutative) Poisson algebra by some related noncommutative associative algebra.  Poisson algebras arise naturally especially as algebras of functions in geometry and physics.  Noncommutative algebras arise naturally as  algebras of operators on linear spaces.
I've often heard it said that "quantization is not a functor".  I'm wondering what a precise statement of this is.
For example, I could imagine statements of the following form.

There is no functor from the category of Poisson manifolds (and Poisson maps?) to the (opposite of the) category of associative algebras satisfying some nice property.
There is no functor from the category of symplectic manifolds (and Poisson maps?) to the (opposite of the) category of associative algebras satisfying some nice property.
Recall that for any smooth manifold, its cotangent bundle is naturally symplectic.  There is no functor from the category of smooth manifolds to the category of associative algebras that quantizes the cotangent bundle.
Recall that the dual to the universal enveloping algebra of a Lie bialgebra is naturally Poisson Hopf.  There is no functor from the category of Lie bialgebras to the category of Hopf algebras satisfying some nice property.

Actually, 4. is false.  Indeed, Etingof and Khazdan constructed a functor from bialgebras to Hopf algebras satisfying a host of properties, and Enriquez classified all the ones with nice properties.  Note that Kontsevich does give a quantization of any Poisson manifold, but perhaps his isn't functorial?

REPLY [23 votes]: This is meant to explain a bit more some of the things that were already mentioned before. 
Quantization of Lie bialgebras is indeed a functor, as was shown in my work with Kazhdan. However, the Konstevich or Fedosov deformation quantization are not functors in this sense. In fact, there is not even a functorial deformation quantization of the symplectic plane, which is sometimes referred to as "Groenewald-van Hove theorem", mentioned in one of the previous answers. The essence of this theorem is that there is no way to quantize the symplectic plane preserving all its symmetries, i.e. symplectic diffeomorphisms. Speaking algebraically, the Lie algebra of infinitesimal symmetries of the symplectic plane is $L_0=C^\infty(\Bbb R^2)/\Bbb R$, and the Lie algebra of symmetries of the (say, Moyal) quantization is the quotient of the quantized algebra by its center, $L=C^\infty(\Bbb R^2)[[h]]_\ast/\Bbb R[[h]]$, with bracket $[a,b]:=(a\ast b-b\ast a)/h$. Now, $L$ is a flat deformation of $L_0$, but the key point is that it is ${\bf nontrivial}$. This nontriviality (which is the gist of the Groenewald-van Hove theorem) is the obstruction to existence of a quantization functor. 
However, as was mentioned in a previous answer, a Drinfeld associator does, according to Tamarkin, give a map between sets of isomorphism classes of Poisson structures and star products (which is not a functor, however, since we have modded out by automorphisms).