Notes - Complex Analysis MT23, Open mapping theorem

Suppose:

• $U$ domain
• $f : U \to \mathbb C$ holomorphic, non-constant
• $V \subseteq U$ open

Then:

• $f(V)$ is also open.

Overall idea: Pick some point $w _ 0$ in the image which we want to find an open ball around, and construct a new function $g$ by shifting so that this function has a zero at $w _ 0$. Then since $g$ is in particular continuous, we can find some $\delta$ such that $ \vert g(z) \vert \le \delta$ on some circle around $g$. But then by considering all $w$ such that $B(w, w _ 0) < \delta$, $ \vert g(z) \vert \le \vert w - w _ 0 \vert $ and so $g$ and $g + (w _ 0 - w) = f - w$ have the same number of zeroes inside this circle. So then we’ve found a little ball around $w _ 0$ such that for any $w$ in this ball, we can find another point in the domain that maps to $w$, i.e. it is in the image.s

Fix some $w _ 0 \in f(V)$. We aim to find some $\delta > 0$ such that $B(w _ 0, \delta) \subseteq f(V)$.

Pick some $z _ 0 \in V$ such that $f(z _ 0) = w _ 0$. Then consider the function

\[g(z) = f(z) - w_0\]

This has a zero at $z _ 0$, which is isolated since $f$ is non-constant. Hence $\exists r > 0$ such that $g(z) \ne 0$ on $\bar B(z _ 0, r) \setminus \{z _ 0 \} \subseteq U$.

Since $\partial B(z _ 0, r)$ is compact, $\exists \delta > 0$ such that

\[|g(z)| \ge \delta\]

on $\partial B(z _ 0, r)$. But then for $w$ such that $ \vert w - w _ 0 \vert < \delta$, it follows

\[|g(z)| > |w-w_0|\]

on $\partial B(z _ 0, r)$. Then by Rouche’s theorem, since $g(z)$ has a zero in $B(z _ 0, r)$, so does

\[h(z) := g(z) + (w_0 - w) = (f(z) - w_0) + w_0 - w = f(z) - w\]

In other words, $f(z) = w$ for some $z \in B(z _ 0, r)$. Thus

\[B(w_0, \delta) \subseteq f(B(z_0, r)) \subseteq f(V)\]

Hence $f(V)$ is open.

Suppose:

• $f : U \to \mathbb C$ injective, holomorphic
• $f’(z) \ne 0$ for all $z \in U$ (this condition is actually implies by the fact $f$ is injective and holomorphic)

Then:

• $\frac{\text d}{\text d z} f^{-1} (w) = \frac{1}{f’(f^{-1}(w))}$

Suppose that $\exists z _ 0$ such that $f’(z _ 0) = 0$. Then

\[f(z) = f(z_0) + (z-z_0)^k g(z)\]

where $k \ge 2$, $g$ holomorphic and $g(z _ 0) \ne 0$ (this is just using the fact that $f(z) - f’(z)$ has a zero of degree at least $2$ at $z _ 0$, where the degree is at least $2$ since $f’(z _ 0) = 0$.

Then we know $\exists \delta$ such that on $B(z _ 0, \delta)$, $g(z) \ne 0$. Hence we can find a holomorphic branch of $z \mapsto z^{1/k}$ on such a disc, so that defining

\[\phi(z) = (z - z_0) \cdot g^{1/k} (z)\]

is a holomorphic function where

\[f(z) = f(z_0) + \phi^k(z)\]

But $\phi(B(z _ 0, \delta)) \supseteq B(0, r)$ for some $r$ since $B(z _ 0, \delta)$ contains $z _ 0$ and $\phi(z _ 0) = 0$.

Since $\phi$ is holomorphic, it maps open sets to open sets, and so the image must contain some neighbourhood of $0$. Thus $\exists a, b \in B(z _ 0, \delta)$ s.t. $\phi(a) = \varepsilon$, $\phi(b) = \varepsilon \exp(2\pi i/ k)$. But then $f(a) = f(z _ 0) = \varepsilon^k = f(b)$, which contradicts the injectivity of $f$.

Proof sketch: First note that $g$ must be continuous by the open mapping theorem. Write as

\[\lim_{w \to w_0} \frac{g(w) - g(w_0)}{w - w_0}\]

and then use standard trick of rewriting $w$ as $f(z)$ in the denominator. For the