Difficulty in understanding the proof of open mapping theorem and maximum modulus principle

Question

It was taken from Complex Analysis (Princeton Lectures in Analysis, No. 2) by Elias M. Stein and Rami Shakarchi page 92.

This is the rough idea of the proof of open mapping theorem:

Theorem: If $f$ is non-constant holomorphic function in a region $\Omega$, then $f$ is open.

Let $w_0$ belong to the image of $f$, say $w_0=f(z_0)$. We must prove that all points $w$ near $w_0$ also belongs to the image of f. Define $g(z) = f(z) - w$. Choose $\delta \gt 0 $ s.t. the disc $|z-z_0| \lt \delta$ is contained the the domain $\Omega$. Then at the end it proves that $g$ has a zero inside the circle $|z-z_0| = \delta$.

Why is it the end of the proof? Can someone elaborate a little bit more?

And this is the rough idea of the proof of maximum modulus principle:

Theorem: If $f$ is non-constant holomorphic function in a region $\Omega$, then $f$ cannot attain a maximum in $\Omega$.

Suppose that $f$ did attain a maximum at $z_0$. Since $f$ is holomorphic it is an open mapping, and therefore, if $D \subset \Omega $ is a small disc centered at $z_0$, its image $f(D)$ is open and contains $f(z_0)$. This proves that there are points in $z \in D$ such that $|f(z)| \gt |f(z_0)|$, a contradiction.

How comes there are points in $z \in D$ such that $|f(z)| \gt |f(z_0)|$? Did I miss some basic stuffs?

Answer 1

If $g$ has a zero in the disc $D=\{z:|z-z_0|<\delta\}$, then there is $z\in D$ with $f(z)=w$. This means that if $z$ is "near" to $w_0$, say in an $\varepsilon$-neighbourhood $U=\{w:|w-w_0|<\varepsilon\}$ then $w\in f(D)$. This means that $U\subseteq f(D)\subseteq f(\Omega)$. Therefore $f(\Omega)$ is a neighbourhood of each of its points, and so is open.

For the last question, split into cases $f(z_0)=0$ and $f(z_0)\ne0$. For the first case, you only need that $f$ takes nonzero values near $z_0$. In the second case you have $f(z_0)(1+\varepsilon)\in f(D)$ for small positive $\varepsilon$.