Q:

Why "Classification" of 4 manifolds is NOT possible?

I know classification of 2 manifolds and geometrization for 3 manifolds.
Why for dimension great or equal to 4, this task become impossible?
edit: Or should I ask "why geometrization won't be possible for 4 or higher dimension?"

A:

I'm guessing that you heard this from someone whose reasoning goes "Every finite presentation of a group can be made to give the $\pi_1$ of a smooth 4-manifold. If we could put any 4-manifold into the Magic List of All, then we could recognize presentations of the trivial group. But no algorithm can do that."
Often people worry about classifications of simply connected manifolds, and don't have to deal with this. (Of course in three dimensions this becomes Perelman's theorem.)

A:

As pointed out in a comment by Richard Kent to Allen Knutson's answer,
the problem is a bit more subtle than it may appear. In order to
prove that the homeomorphism problem for compact 4-manifolds, say in
the topological category, is recursively unsolvable, it is not enough
to know that (1) every finitely presented group can be realized as
the fundamental group of some compact 4-manifold, and (2) the isomorphism problem for finitely presented groups is recursively unsolvable.
Instead, what you do is give a construction which to any finite presentation $< S | P >$ of a group associates a 4-manifold $M(S,P)$ in such a way that $\pi_1(M(S,P))$ is isomorphic to the group defined by
the presentation $< S | P >$, and moreover two such manifolds are homeomorphic if and only if they have isomorphic fundamental groups.
Then you have constructed a class of 4-manifolds for which the homeomorphism problem is equivalent to the isomorphism problem for
finitely presented groups, and therefore unsolvable.
About "geometrization for manifolds of dimension 4 or higher", well
as far as I know there is no theorem which says it is impossible. It
depends on what you mean by `geometric structure', and what you
want those structures to do for you.

