Is the hyperbolic plane the only simply connected hyperbolic 2-manifold? Let $S$ be a simply connected Riemannian 2-manifold with everywhere negative curvature.  Is $S$ necessarily diffeomorphic to $\mathbb{R}^2$?
 A: Yes, by Uniformization theorem
A: Yes, the geometric classification of surfaces tells us that a simply connected Riemannian surface $S$ must be (up to diffeomorphism) the sphere $S^2$, the complex plane $\mathbf{C}$, or the hyperbolic plane $\mathbf{H}$. Given that $\mathbf{H}$ is the only one of these with negative curvature, $S$ must be the hyperbolic plane.
A: Just to be ultra clear about some issues related to the original question, there are plenty of simply connected surfaces of constant curvature K = 0 or K = -1 that are not isometric to either R2 (K = 0) or H2 (K = -1). Some of these are each the interior of a simple closed curve drawn on either R2 or H2. Some others are each one side of an infinite simple arc drawn on R2 or H2 that goes off to infinity in both directions. The rest can get extremely weird as one of the complementary regions, in R2 or H2, of one of some very strange closed subsets. In fact, all of them can be described as the universal Riemannian covering space of the complement of an arbitrary closed subset of R2 or of H2. These are all (as Andrew Hwang has pointed out) not complete. 
The Uniformization Theorem shows that each of these simply connected subsets of constant curvature K = 0 or K = -1 is, however, conformally equivalent to either R2 or H2. (Given an arbitrary Riemannian metric on a simply connected non-compact surface, it is a difficult problem to decide whether it is conformally equivalent to R2 or to H2.)
