There is a full classification of open surfaces (noncompact surfaces without boundary); see here. This is perhaps a bit more daunting than the classification of compact surfaces, as the latter needs only two simple invariants: the number of boundary components and the Euler characteristic. The added complexity in the noncompact case should not be surprising, as one can have truly wild-looking spaces (imagine making an infinite binary tree and for each edge, inserting a copy of the twice-punctured torus). But it does provide the following nice fact: the only simply connected surfaces without boundary are the sphere and plane. (If you allow boundary, you in addition gain the closed disc and any subset of the closed disc obtained by deleting closed subsets of the boundary.) We don’t have so nice a classification in higher dimensions; there are uncountably many pairwise non-homeomorphic noncompact, contractible 3-manifolds. (However, if you’re “simply connected at infinity” and contractible, then in dimensions at least 5, you’re homeomorphic to .)

The purpose of this post is to provide two proofs of that, one unnecessarily difficult and the other elementary.

Given an open subset of the plane, somehow the most obvious approach is to apply the Riemann mapping theorem, which provides a conformal map to the disc. To generalize this to other surfaces, we need to invoke something for more general Riemann surfaces: the uniformization theorem. This is annoyingly difficult, since you need to start by having a complex structure on your open surface. To do this, start by finding an almost complex structure on it; this is a smoothly varying automorphism such that . It’s straightforward to see than an orientation and Riemannian metric determine such an automorphism: let be the unique tangent vector orthogonal to with a positively oriented basis. Now invoke the Newlander–Nirenberg theorem that a manifold carries an honest complex structure inducing the almost complex structure if and only if its Nijenhuis tensor vanishes, and one can check by hand that this is always true for surfaces (rather miraculously; this is wildly false in higher dimensions). This is known in dimension 2 as “the existence of isothermal coordinates”. This gives an honest complex structure on our open manifold, and the uniformization theorem then literally gives us what we want. If you want to take this approach, you probably want both a proof of Newlander-Nirenberg in two dimensions and the uniformization theorem; the first is reasonably classical (due originally to Gauss in the real analytic case), and can be solved by solving what’s known as the Beltrami equation, and my favorite proof of the second (since we’re already doing PDE) is in Mazzeo-Taylor, Curvature and Uniformization (thanks to Willie Wong on StackExchange for pointing this paper out to me here).

Here’s a less preposterous approach (in that it involves no more than basic algebraic topology); in flavor, it is very similar to the link in the first sentence. First, note that every open smooth manifold is the increasing union of a sequence of compact submanifolds. To prove this, you should first prove the existence of a proper smooth function ; then pick a sequence of regular values , and let . One unfortunate fact is that these might always be disconnected. (Imagine a disc with lowest point at height centered at , and a height-increasing “bridge” from one to the next, sitting inside , and extend these discs more-or-less vertical so that this is a closed submanifold of ; this is diffeomorphic to a plane, but if you set to be the above height function, and pick , we run into this problem. To resolve this, we just force the first stage to be connected: if is eg two components for convenience of discussion, pick a path from the boundary of one to the other, and take a tubular neighborhood of it; rounding out the corners, union this bridge is a connected compact manifold. Using the same idea, we may further demand that all of the are connected.

Henceforth for convenience all homology will be taken with coefficients.

Fact 1: If is a -dimensional noncompact manifold and a -dimensional compact boundary component, then is injective. This follows from the relative long exact sequence, because , because is a noncompact manifold. A similar proof shows that if (now possibly compact) has at least two compact boundary components, then has positive rank.

Fact 2: All of the have genus zero, and hence (this is just what it means for a compact surface with boundary to be genus zero!) are given as a sphere minus the interiors of finitely many closed discs in the sphere. To see this, apply Mayer-Vietoris to ; on the way, you should prove that the genus zero assumption is equivalent to being surjective.

Fact 3: The complement of has all but one component compact. If it didn’t, Apply Fact 1 and Mayer-Vietoris to derive a contradiction.

Fact 2 implies that these compact components must be discs by the classification of compact surfaces. So define to be the union of and these discs, so is a disc for all . Thus is an annulus. A union of successive annuli is diffeomorphic to , which is diffeomorphic to a punctured disc; we conclude that is diffeomorphic to gluing this collar to the boundary of the disc , which is indeed a plane.