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?
Best Answer
I have no access to Jstor right now, so I rely on my memory and, well, you may want to check all this.
Ian Richards theorem says that non-compact surfaces (without boundary) are classified by their orientablility, their genus (possibly infinite) and a triple of spaces, each one embedded in the preceding, that are:
the space of its ends,
the space of its ends with genus,
the space of its unorientable ends.
The space of ends is constructed by taking an increasing sequence of compact subset that cover a topological space $T$, and looking at the connected components of their complements. An end of $T$ is an infinite, decreasing sequence of such connected components. The point is that you can do that in a way that does not substancially depend on the sequence of compact you chose.
For example, $\mathbb{R}^n$ has only one end provided $n\ge2$ (look at the sequence of balls of integer radius and centered at some point), while $\mathbb{R}$ has two ends. The space of ends of a regular tree is a Cantor set.
An end is said to have genus if the connected components that define it all have genus (they never reduce to annuli). An end is said to be unorientable if the connected components that define it all are unorientable.
Now, consider the surface $S^n$ ($n=1,2$ or $3$ defined as the boundary of a tubular neighborhood of the usual embedding of the usual Cayley graph of $\mathbb{Z}^n$ into $\mathbb{R}^3$ (for $n=1$ you get a cylinder; for $n=2$ some sort of grid; for $n=3$ it is sometimes called a jungle gym). $S^2$ and $S^3$ are the surfaces described by Richard Kent in his third paragraph. These two surfaces have exactly one end which is orientable but has genus. Therefore they all are homeomorphic. This is a pretty incredible result in my opinion. The most simple presentation of this surface is called the Loch-Ness monster: it is constructed by adding to a plane a sequence of handles placed in a row.