23 - Open Mapping Theorem

Open Mapping Theorem

Suppose $f:U\to \mathbb{C}$ is holomorphic and non-constant on a domain $U$ then

For any open set $V\subset U$ then set $f(V)$ is also open

Proof

Suppose $w_0\in f(V)$

Consider $z_0$ such that $f(z_0)=w_0$ so
Let $g(z):=f(z)-w_0$ such that it has a zero at $z_0$

Since $g$ is non-constant then the zero is isolated

Thus there exists some $r>0$ such that $g(z)$ has no other zeros inside $\overline{B}(z_0,r)\subset U$

Since $\partial B(z_0,r)$ is compact then there is $\delta >0$ such that

$$|g(z)|\ge \delta >0\text{ on }\partial B(z_0,r)$$

Consider any $w$ such that $|w-w_0|<\delta$ then

$$|w-w_0|<\delta \le |g|\text{ on }\partial B(z_0,r)$$

Hence by Rouché’s Theorem then

$$g(z)\text{ and }g(z)+w_0-w=f(w)-w$$

have the same number of zeros (counting multiplicities) inside $B(z_0,r)$

Since $g(z)$ has a zero at $z_0$ of multiplicity of at least one then
Equation $f=w$ has at least one solution inside $B(z_0,r)$

Thus

$$B(w_0,\delta )\subset f(B(z_0,r))$$

and $f(U)$ is open