- (15 points) Show that there is no one-to-one holomorphic mapping of the open annulus \(\{ z : 1 < | z | < 2 \}\) onto the punctured unit disc \(\{ z : 0 < | z | < 1 \}\) . (HINT: consider the inverse mapping)
- (5 points) Give an example of a one-to-one \(C ^ { \infty }\) mapping of the open annulus \(\{ z : 1 < | z | < 2 \}\) onto the punctured unit disc \(\{ z : 0 < | z | < 1 \}\)
Solution
1. Part (a): a biholomorphism would give a bounded inverse on the punctured disk.
Suppose, for contradiction, that \[ f:A=\{1<|z|<2\}\longrightarrow D^*=\{0<|w|<1\} \] is one-to-one, holomorphic, and onto. Since a one-to-one holomorphic map has nonzero derivative, its inverse \[ g=f^{-1}:D^*\to A \] is holomorphic. It satisfies \(1<|g(w)|<2\), so \(g\) is bounded near the puncture. The removable-singularity theorem extends it to a holomorphic map \(G:D\to\mathbb C\).
Continuity and the inequalities on \(D^*\) give \[ 1\le |G(0)|\le2. \] If \(1<|G(0)|<2\), then \(G(0)\in A\). Since \(f\circ g\) is the identity on \(D^*\), continuity of \(f\) at \(G(0)\) gives \[ f(G(0))=\lim_{w\to0}f(g(w))=\lim_{w\to0}w=0, \] contradicting \(f(A)\subset D^*\).
If \(|G(0)|=2\), then \(|G|\le2\) on \(D\) and \(G\) attains its maximum modulus at the interior point \(0\), so the maximum modulus principle makes \(G\) constant, contradicting that \(g\) is the inverse of a bijection. If \(|G(0)|=1\), then \(G\) is nowhere zero and \(|1/G|\le1\) on \(D\), with equality at \(0\); the same principle applied to \(1/G\) again makes \(G\) constant. All cases are impossible. Thus no such holomorphic bijection exists.
2. Part (b): a radial rescaling is a smooth bijection.
Define \[ F(z)=\left(1-\frac1{|z|}\right)z, \qquad 1<|z|<2. \] Writing \(z=re^{i\theta}\) gives \[ F(re^{i\theta})=(r-1)e^{i\theta}. \] Hence \(0<|F(z)|<1\), and the map preserves the angular coordinate while sending \(r\in(1,2)\) bijectively to \(r-1\in(0,1)\). Its inverse is \[ F^{-1}(w)=\left(1+\frac1{|w|}\right)w, \qquad 0<|w|<1. \] Both formulas are \(C^\infty\) on their respective domains because the modulus function is smooth away from zero. Thus \(F\) is a one-to-one \(C^\infty\) mapping of the annulus onto the punctured unit disk, as required.