Open mapping theorem

Statement

Let $D \subseteq \mathbb{C}$ be a domain and let $f : D \to \mathbb{C}$ be holomorphic on $D$. Assume that $f$ is not constant on $D$. Then $f$ is an open map: for each open set $V \subseteq D$, the image $f(V)$ is an open subset of $\mathbb{C}$.

Remarks

The maximum modulus principle follows. If $|f|$ had a local maximum at $z_0$, then $f$ would map a small disc around $z_0$ onto a set that contains no disc around $f(z_0)$. The connectedness of $D$ matters only to exclude functions that are constant on one part of $D$. There is no real analogue: $x \mapsto x^2$ maps $(-1, 1)$ onto $[0, 1)$, which is not open in $\mathbb{R}$. In contrast, $z \mapsto z^2$ maps the unit disc onto the unit disc.

Explanations

L1 human

A nonconstant holomorphic function never squeezes an open set into a curve or a point. Around each point of the image, the image also contains a small disc.