TITLE: Classification problem for non-compact manifolds
QUESTION [46 upvotes]: Background
It is well-known that the compact two-dimensional manifolds are completely classified (by their orientability and their Euler characteristic).
I'm also under the impression that there is also a classification for compact three-dimensional manifolds coming from the proof of the Geometrization Conjecture and related work.
Unfortunately for $n\ge4$ no similar classification is possible because it can be shown that it is at least as hard as the word problem for groups. Thus for higher-dimensional manifolds we instead focus on classifying all the simply-connected compact manifolds.
My question
Why in these "classification problems" are we only considering compact manifolds? Is there an easy reason why we restrict ourselves to the classification of compact manifolds? Does a classification of general (not necessarily compact) manifolds follow easily from a classification of compact manifolds?

REPLY [6 votes]: To illustrate both sides of all of the above comments (non-compact builds on compact, vs. non-compact is much harder than compact), I point out that ALL connected, separable, metric 2-manifolds (with or without boundary) meaning all connected, separable, metric spaces in which every point has an open neighborhood homeomorphic to the closed 2-disk have been classified.  See Brown, Edward M.; Messer, Robert: The classification of two-dimensional manifolds.  Trans. Amer. Math. Soc. 255 (1979), 377–402.  MR0542887 
The point is that the analysis is rather delicate and has some surprises, showing that the non-compact case does not follow easily from the compact.  Given the examples in dimension 3 by McMillan mentioned by algori, one of the surprises is that it could be done at all.