- Suppose \(U\) and V are conformally equivalent. Prove that if U is simply connected, then so is V . Note that this conclusion remains valid if we merely assume that there exists a continuous bijection between U and \(V .\)
Solution
Let \(\Phi:U\to V\) be a conformal equivalence. Since \(\Phi\) and \(\Phi^{-1}\) are holomorphic, they are continuous, so \(\Phi\) is a homeomorphism.
Assume \(U\) is simply connected. Let \(\gamma:[0,1]\to V\) be a closed curve. Then \(\Phi^{-1}\circ\gamma\) is a closed curve in \(U\), so there is a homotopy \[ H:[0,1]^2\to U \] from \(\Phi^{-1}\circ\gamma\) to a constant curve. Composing with \(\Phi\) gives a homotopy \(\Phi\circ H\) in \(V\) from \(\gamma\) to a constant curve. Hence every closed curve in \(V\) is null-homotopic, so \(V\) is simply connected.
More generally, suppose only that \(\Phi:U\to V\) is a continuous bijection. Since \(U,V\) are open subsets of \(\mathbb R^2\), invariance of domain implies that the continuous injective map \(\Phi\) is open. Therefore \(\Phi^{-1}\) is continuous, so \(\Phi\) is again a homeomorphism, and the same homotopy argument applies.