Let $U\subset \Bbb R^2$ be open, bounded, and simply connected. Is it true that $U$ is smoothly contractible to a point?
By the Whitney Approximation Theorem, it suffices to find a continuous $H :U\times [0,1]\to U$ s.t. $H(x,0)=x$ and $H(x,1)=x_0\in U$. If I could assume the Riemann mapping theorem, this would be trivial. Just send $U$ to the unit disk and contract to the origin. Unfortunately, this is a question from Spivak's Differential Geometry books, and so far there has been no mention of the Riemann mapping theorem, so I believe I should not use it in the proof.
What has been covered so far is some basic de Rham cohomology, but all I know in this connection is that $H^1(M)=0$ for a simply connected manifold. I do not have any sort of converse result where the cohomology group tells me anything about the underlying space. Am I barking up the wrong tree? Is the proof along another direction?