Open mapping theorem (complex analysis)

In complex analysis, the open mapping theorem states that if U is a connected open subset of the complex plane C and f : U → C is a non-constant holomorphic function, then f is an open map (i.e. it sends open subsets of U to open subsets of C).

The open mapping theorem points to the sharp difference between holomorphy and real-differentiability. On the real line, for example, the differentiable function f(x) = x² is not an open map, as the image of the open interval (-1,1) is the half-open interval [0,1).

The theorem for example implies that a non-constant holomorphic function cannot map an open disk onto a portion of a line.

Proof

Blue dots represent zeros of g(z). Black spikes represent poles. The boundary of the open set U is given by the dashed line. Note that all poles are exterior to the open set. The smaller red circle is the set B constructed in the proof.

Assume f:U→C is a non-constant holomorphic function and $U$ is a connected open subset of the complex plane. We have to show that every point in $f(U)$ is an interior point of $f(U)$, i.e. that every point in $f(U)$ is contained in a disk which is contained in $f(U)$.

Consider an arbitrary $z_{0}$ in $U$. Since U is open, we can find $d>0$ such that the closed disk $B$ around z₀ with radius d is fully contained in U. Since U is connected and f is not constant on U, we then know that f is not constant on B. Consider the image point, $w_{0}=f(z_{0})$. Then $f(z_{0})-w_{0}=0$, making $z_{0}$ a root of the function $g(z)=f(z)-w_{0}$.

We know that g(z) is not constant, and by further decreasing d, we can assure that g(z) has only a single root in B. (The roots of holomorphic non-constant functions are isolated.) Let e be the minimum of |g(z)| for z on the boundary of B, a positive number. (The boundary of B is a circle and hence a compact set, and |(g(z)| is a continuous function, so the extreme value theorem guarantees the existence of this minimum.) Denote by $D$ the disk around $w_{0}$ with radius $e$. By Rouché's theorem, the function $g(z)=f(z)-w_{0}$ will have the same number of roots in B as $f(z)-w$ for any $w$ within a distance $e$ of $w_{0}$. Thus, for every $w$ in $D$, there exists one (and only one) $z_{1}$ in $B$ so that $f(z_{1})=w$. This means that the disk D is contained in f(B), which is a subset of $f(U)$.