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Let $D \subset \mathbb{C}$ be a domain and let $a \in D$. Suppose $f: D \smallsetminus \{a\} \to \mathbb{C}$ is analytic and that $a$ is an essential singularity of $f$. Show that $f$ cannot be univalent (= injective) in any neighborhood of $a$.

This is a trivial consequence of the Picard theorem. But I don't know if there is any elementary approach.

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Please edit your post so that the "title" is a succinct description of your question, and the body of your question is contained entirely in the body text. Don't split the question text partly between the title and the main part. And please punctuate correctly. Also, if you enclose mathematics expressions in dollar signs $, you can make use of our [built-in LaTeX mathematics display]( meta.math.stackexchange.com/questions/1773/…). –  Willie Wong Jul 23 '11 at 14:20
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Casorati-Weierstrass –  t.b. Jul 23 '11 at 14:33
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By the way: proving this via Picard is cracking a nut with a sledgehammer. Now that you know how the result is called you should be able to locate it in any book on complex analysis. –  t.b. Jul 23 '11 at 14:46
    
Ah, looks like @Theo cleaned up the question for you. –  Willie Wong Jul 23 '11 at 17:30
    
The case where the essential singularity is at infinity was covered in answers to two previous questions: math.stackexchange.com/questions/39479/… and math.stackexchange.com/questions/29758/entire-1-1-function/…. (In each of those it was assumed that the function is entire, but the idea is the same as in Theo's answer here.) –  Jonas Meyer Jul 28 '11 at 7:47

1 Answer 1

This is a simple consequence of the Casorati–Weierstrass theorem:

If $f: D \smallsetminus \{a\} \to \mathbb{C}$ is holomorphic and $a$ is an essential singularity of $f$ then $f(D \smallsetminus \{a\})$ is dense in $\mathbb{C}$.

One proof of the Casorati–Weierstrass theorem is via Riemann's theorem on removable singularities and is given on the Wikipedia page I linked to above. It's very simple:

If $f(D \smallsetminus \{a\})$ is not dense then by definition of density we can find $w_{0} \in \mathbb{C}$ and $\varepsilon \gt 0$ such that $\{w\,:\,|w - w_0| \lt \varepsilon\} \cap f(D\smallsetminus\{a\}) = \emptyset$. But then $g(z) = \dfrac{1}{w_0 - f(z)}$ is bounded and analytic on $D \smallsetminus \{a\}$ and the straightforward Riemann theorem on removable singularities allows us to extend $g$ to an analytic function $\tilde{g}: D \to \mathbb{C}$. The function $\tilde{f}(z) = w_0 - \dfrac{1}{\tilde{g}(z)}$ coincides with $f$ on $D \smallsetminus \{a\}$ and a quick check shows that $a$ must either be a removable singularity or a pole of $f$ (depending on $\tilde{g}(a) \neq 0$ or $\tilde{g}(a) = 0$). Either way this contradicts the hypothesis that $a$ is an essential singularity of $f$.


To prove that $f$ cannot be injective on any (pointed) neighborhood $W$ of the essential singularity $a$, let $x \neq a$ be any point of $W$. Choose $0 \lt \varepsilon \lt \frac{|x-a|}{2}$ so small that both $U = \{z\,:\,0 \lt |z-a| \lt \varepsilon\}$ and $V = \{z\,:\,0 \leq |z-x| \lt \varepsilon\}$ are contained in $W$. Observe that $U$ and $V$ are disjoint and open.

  1. By Casorati-Weierstrass, $f(U) \subset \mathbb{C}$ is dense (since $a$ is also an essential singularity of $f\,|_{U}$) and by the open mapping theorem $f(V)$ is open (since $f$ is not constant). Therefore $f(U) \cap f(V) \neq \emptyset$.

  2. If $f$ were injective then $f(U) \cap f(V) = \emptyset$ by disjointness of $U$ and $V$, contradicting $1.$

This shows that $f$ cannot be injective (univalent) on any neighborhood $W$ of the essential singularity $a$.

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I think: Univalent = One-to-one. –  GEdgar Jul 23 '11 at 18:56
    
@GEdgar: You're right. I'm always confusing univalent and schlicht, thanks. –  t.b. Jul 23 '11 at 19:28

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