Analysis Problem of the Day 83

Today’s problem appeared as Problem 10 on the UCLA Spring 2018 Analysis Qual:

Problem 10. Let $\widehat{\mathbb{C}}$ be the Riemann sphere and let $\Omega = \widehat{\mathbb{C}} \setminus \{0,1\}.$ Let $f: \Omega \to \Omega$ be a holomorphic function.

a) Prove that if $f$ is injective, then $f(\Omega)=\Omega.$

b) Make a list of all such injective functions $f.$

Solution: By means of the Möbius transformation $h(z):=\frac{z}{z-1} = 1+\frac{1}{z-1}$ (which is an automorphism of the Riemann sphere) such that $h(0)=0, h(1)=\infty,$ and its inverse $h^{-1}(z) = h(z) = 1+\frac{1}{z-1},$ we may suppose that $\Omega = \widehat{\mathbb{C}} \setminus \{0,\infty\} = \mathbb{C} \setminus \{0\}$ since $f:= h \circ \widetilde{f} \circ h^{-1}: \mathbb{C} \setminus \{0\} \to \mathbb{C} \setminus \{0\}$ is injective. Thus, $0$ is an isolated singularity of $f.$ Then, note that by the Great Picard theorem, $0$ cannot be an essential singularity, as otherwise $f$ would fail to be injective in a neighborhood of 0. Similarly, if $0$ is a removable discontinuity, then we may extend $f$ to a nonconstant entire function, which by the Little Picard theorem attains all except at most one value infinitely many times. Thus, $0$ is a pole of $f,$ i.e. if $U$ is a neighborhood of $0,$ then $f(U)$ is a neighborhood of $\infty.$ Since $f$ is injective, it follows that $f$ cannot have a pole at $\infty,$ and by Great Picard and injectivity yet again it cannot have an essential singularity at $\infty,$ i.e. $\infty$ is a removable singularity. This means that $g(z):=f(\frac{1}{z})$ is an entire injective function with a pole at $\infty,$ and since entire functions with a pole at $\infty$ are precisely the polynomials, from injectivity it follows that $g$ is precisely a nondegenerate linear function $g(z) = az+b, a \not =0.$ It follows that $f(z) = \frac{a}{z}+b, a,b \in \mathbb{C}, a \not =0$ are all such injective functions, so that $\widetilde{f}$ takes the form

$\widetilde{f}(z) = (h^{-1} \circ f \circ h)(z) = 1+\frac{1}{\frac{a}{1+\frac{1}{z-1}}+b-1} = 1+\frac{z}{(a+b-1)z-a}:= 1+\frac{z}{cz-d}$

for $c \not =0.$ Since $f$ is a homeomorphism of $\mathbb{C} \setminus \{0\}$ and $h$ is an automorphism of the Riemann sphere, we thus conclude both (a) and (b).

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