This is an exercise for my complex analysis course. I have some ideas about this exercise, but I am not sure if I am correct.

Use the Inverse Function Theorem to show that if $f: A \subset \mathbb{C} \rightarrow \mathbb{C}$ is analytic and $f'(z) \neq 0$ for all $z \in A$, then $f$ maps open sets in $A$ to open sets.

By the Inverse Function Theorem, there exists a neighborhood $U$ of $z$ and a neighborhood $V$ of $f(z)$ such that $f: U \rightarrow V$ is a bijection.

Let $S \subset A$ be open. Then, for any $z_1 \in S$, we can find a neighborhood $U_1 \subset S$ of $z_1$. So, we know that there exists an open set $V_1$ of $f(z)$ such that $f(U_1) = V_1$. Can we say that $f(S) = \bigcup V_i$, the union of infinitely many open sets, which is open?

  • 1
    $\begingroup$ Yes, the proof looks correct. $\endgroup$ – Mark Feb 6 at 19:55

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