# Show that $f$ is an open mapping (using Inverse Function Theorem)

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?

• Yes, the proof looks correct. – Mark Feb 6 at 19:55