Lecture 24: the open mapping theorem

Complex analysis, lecture 2

(April 24, 2026)

Our first goal today is to prove the open mapping theorem. First, we need to review some topology.

Topology is not a prerequisite for this class; but fortunately we’ve already seen most of the notions we need, most importantly open and closed sets. In general topology, the main notion is a “topological space,” which is a set $X$ together with some labeling of which subsets of $X$ are open (or closed, neither, or both), satisfying certain axioms. For our purposes, we can always think of $X$ as some subset of $\mathbb{C}$ , so the open (and closed) sets will be as usual.

Here is the topological definition of continuity: let $X$ and $Y$ be topological spaces, and $f:X\to Y$ a function. Then $f$ is continuous if and only if for every open set $U\subset Y$ , the preimage $f^{-1}(U)$ is open in $X$ .

This is not obviously equivalent to our familiar $\epsilon$ - $\delta$ definition! To translate between them, think of the set $U$ as a small ball of radius $\epsilon$ around a point $y\in Y$ . We may as well assume $y$ is in the image of $f$ ; otherwise either the preimage is empty, hence open, or we could choose a different point, so write $y=f(x)$ . Then the condition is that the preimage of $U$ is an open set containing $x$ , i.e. it contains a ball of some sufficiently small radius $\delta$ centered at $x$ : that is, points within $\delta$ of $x$ are mapped to points within $\epsilon$ of $y=f(x)$ . This is our more familiar definition of continuity. One can also reverse the logic to see that the two really are equivalent.

It is perhaps strange that the definition relies on the inverse mapping $f^{-1}$ , rather than the function $f$ itself. Here we’re taking the preimage on sets, so $f^{-1}(U)$ is always well-defined (as the set of points $x\in X$ such that $f(x)\in U$ ) even when $f$ is not a bijection; but it still seems perhaps surprising. A more naive guess would be that the following property is the fundamental one: say that $f:X\to Y$ is an open mapping, or just open, if for every open set $V\subset X$ , the image $f(V)$ is an open set.

This is also an important property, but it turns out not to match what we mean by continuity. For example, consider $X=Y=\mathbb{R}$ , and $f:X\to Y$ given by $f(x)=x^{2}$ . This is a continuous function; it’s even differentiable (and in fact real-analytic, being a polynomial). But it is not an open mapping: the open set $(-1,1)$ is mapped to $f((-1,1))=\{x^{2}|-1<x<1\}=[0,1)$ , which is not an open set.

(It is also possible to find examples of open maps $\mathbb{R}\to\mathbb{R}$ which are not continuous! These are a little trickier to write down, so I’ll skip them for now; take it as an informal exercise to find one, if you like.)

Over the complex numbers, one can find similar examples of continuous functions which are not open (and vice versa). But if we go so far as to require analyticity, there are no such examples:

Theorem (Open mapping theorem).

Let $D\subseteq\mathbb{C}$ be a domain and $f:D\to\mathbb{C}$ a non-constant analytic function. Then $D$ is an open mapping.

If not for our discussion above, this would be a very innocent-looking theorem: it hardly seems surprising that open sets should map to open sets. But as we saw above, it fails over the real numbers! Perhaps more surprisingly from a certain point of view, it also fails in higher dimensions: if we were studying, say, analytic maps $\mathbb{C}^{2}\to\mathbb{C}^{2}$ —as we might $\mathbb{R}^{2}\to\mathbb{R}^{2}$ —it also fails.¹¹1This might inspire you to ask if we can come up with some new space $\mathbb{H}$ of “numbers” which are isomorphic to $\mathbb{C}^{2}$ the same way $\mathbb{C}$ is isomorphic to $\mathbb{R}^{2}$ , for which there is a better notion of “analytic” maps $\mathbb{H}\to\mathbb{H}$ which does satisfy the open mapping theorem and which gives a whole new field of “ $\mathbb{H}$ -analysis.” The answer is yes in a very particular sense: one has to significantly broaden what a “number” should mean, but there is such an object $\mathbb{H}$ , called the quaternions, with these properties. The theory of quaternionic analysis has fewer direct applications than complex analysis and is newer, but can be developed; however one can’t continue in this way (i.e. keep replacing e.g. $\mathbb{H}^{2}$ by something else, or more copies of $\mathbb{C}$ , etc.) indefinitely. More typically, the study of quaternions is subsumed into the theory of Lie algebras and related fields.

Proof.

Let $V\subset D$ be an open set. We want to show that $f(V)$ is open: that is, for every $w_{0}\in f(V)$ , we can find some $\epsilon>0$ with all points $w$ such that $|w-w_{0}|<\epsilon$ contained in $f(V)$ .

Since $w_{0}\in f(V)$ , we can find $z_{0}\in V$ with $f(z_{0})=w_{0}$ . Since $V$ is open, we can find some $\delta>0$ such that all $z$ within $\delta$ of $z_{0}$ are also in $V$ .

Define $g(z)=f(z)-w_{0}=f(z)-f(z_{0})$ . This has a zero at $z_{0}$ ; and since $f$ is non-constant, $g$ is not the constant zero function, so $g$ has isolated zeros. Therefore by choosing $\delta$ small enough, we can guarantee that $g(z)\neq 0$ for any $0<|z-z_{0}|<\delta$ .

In particular, on the boundary $\{z:|z-z_{0}|=\delta\}$ , $g$ is always nonzero. Since this is a circle (of radius $\delta$ ), it is a compact set, so the continuous real functions $-|g(z)|$ attains its maximum; since $-|g(z)|$ is always a non-positive real number and, by the above, is always nonzero on this circle, this maximum must be strictly less than $0$ , i.e. there exists some $\epsilon>0$ such that $-|g(z)|\leq-\epsilon$ for all $z$ on the boundary, or equivalently $|g(z)|\leq\epsilon$ .

Now, let $w$ be any point with $|w-w_{0}|<\epsilon$ . We claim that $g(z)=f(z)-w_{0}$ and $h(z)=f(z)-w$ have the same number of zeros within $\{z:|z-z_{0}|<\delta\}$ . Our method is to apply Rouché’s theorem: by the above, $|g(z)|\geq\epsilon$ , while $|w_{0}-w|<\epsilon$ , i.e. $|g(z)|>|w_{0}-w|$ ; so by Rouché’s theorem $g(z)$ and $g(z)+w_{0}-w=f(z)-w$ have the same number of zeros in this region.

We found above that in this region $g$ has exactly one zero, namely at $z=z_{0}$ . Therefore there is also one solution to $f(z)-w=0$ , i.e. there is a unique $z$ within $\delta$ of $z_{0}$ such that $w=f(z)$ . Hence any $w$ within $\epsilon$ of $w_{0}$ is in the image $f(V)$ , i.e. there is a neighborhood of $w_{0}$ inside $f(V)$ . Since $w_{0}$ was an arbitrary point of $f(V)$ , this shows that in fact $f(V)$ is an open set. ∎

Since the continuous image of a connected set is connected, it follows that for any domain $D\subseteq\mathbb{C}$ and analytic map $f:D\to\mathbb{C}$ , the image $f(D)$ of the whole domain $D$ is itself a domain, and likewise for any subdomains.

This lets us immediately recover some earlier results. For example, recall that if $f:D\to\mathbb{C}$ is an analytic function with real image, we saw that it has to be constant using the Cauchy–Riemann equations. We can now see this directly using the open mapping theorem: in this language, we are given that $f(D)\subset\mathbb{R}$ ; but $\mathbb{R}$ is not a domain, and does not contain any domains, since it is one-dimensional and hence can’t contain any nonempty open sets (which must include two-dimensional neighborhoods). Hence the only way in which this is possible, given the open mapping theorem, is if $f$ is constant.

In fact, this lets us prove much more general results: for example, if $f:D\to\mathbb{C}$ has image contained in the curve $\{z=x+iy:y=x^{2}\}$ , then it must be constant. Indeed, this curve again does not contain any nonempty open sets, so the only way an analytic function can have image contained in it is if it’s constant. In principle, we could also show this using the Cauchy–Riemann equations, but it would be much more difficult! The same principle applies to any curve: we can think of this as saying that a non-constant analytic function preserves dimension, so a two-dimensional space can’t be sent to a one-dimensional one. (Indeed, any “two-dimensional space” should contain some nonempty open set, while a “one-dimensional space” cannot, though we avoid making this precise.)

A perhaps less immediate application is as follows.

Proposition.

Let $f:D\to\mathbb{C}$ be an analytic function. Let $E=f(D)\subseteq\mathbb{C}$ be the image of $D$ , so we can think of $f$ as a map $D\to E$ . Suppose that $f:D\to E$ is a bijection. Then its inverse $f^{-1}:E\to D$ is again analytic.

Note that by the above, $E$ is itself a domain, so this makes sense.

This might again seem obvious, but let’s be cautious: for example, it is possible to have continuous bijections $f:X\to Y$ which therefore have an inverse function $f^{-1}:Y\to X$ which turns out not to be continuous! For example, let $S^{1}=\{z:|z|=1\}$ , the unit circle, and consider the function $f:[0,1)\to S^{1}$ sending $x\mapsto e^{2\pi ix}$ . This is a continuous function; it is one-to-one and onto, so it is a bijection, with inverse $z\mapsto\frac{1}{2\pi}\arg(z)$ where we take $0\leq\arg(z)<2\pi$ ; but this inverse is not continuous at $z=1$ , since the argument jumps from $2\pi$ down to $0$ .

We could also state the above theorem in the real setting—so everything is not only continuous but real-analytic—and it again fails: for example, $f(x)=x^{3}$ is an analytic bijection $\mathbb{R}\to\mathbb{R}$ , but its inverse $f^{-1}(x)=\sqrt[3]{x}$ is not analytic: it is not differentiable at $0$ . So some nontrivial complex analysis is needed for the above theorem.

Proof.

Note that since $f$ is analytic, it is continuous. We first show that its inverse (which exists since $f$ is a bijection) is also continuous.

The claim that $f^{-1}$ is continuous is equivalent to the claim that for every open set $U\subset D$ , its preimage $(f^{-1})^{-1}(U)$ under $f^{-1}$ , equivalently its image $f(U)$ under $f$ , is open. Since $f$ is a bijection to a domain, it is not constant, and since it is analytic it is an open map by the open mapping theorem; that is, $f(U)$ is indeed an open set. So the first claim is proved.

Now, recall that we actually have a formula for the derivative of an inverse function of an analytic function: $(f^{-1})^{\prime}(z)=\frac{1}{f^{\prime}(f^{-1}(z))}$ . So the derivative is automatically well-defined, and indeed continuous, so long as $f^{\prime}(f^{-1}(z))$ is nonzero.

Let $S$ be the set of zeros of $f^{\prime}$ . (Note that since $f^{\prime}$ is analytic, every point in $S$ is isolated.) The zeros of $f^{\prime}(f^{-1}(z))$ are points $z$ such that $f^{-1}(z)\in S$ , i.e. $z\in f(S)$ . So we know that $f$ is analytic away from some isolated set of points $f(S)$ .

Therefore each of these points is an isolated singularity, which we now know how to classify: each is either an essential singularity, a pole, or a removable singularity. But actually we also know that $f^{-1}$ , which is analytic on $E\setminus f(S)$ , extends to a continuous function on all of $E$ : in particular, for any $z_{0}\in f(S)$ , the limit $\lim_{z\to z_{0}}f^{-1}$ exists. Therefore the singularity at $z_{0}$ must be removable, which means that $f^{-1}$ actually extends to an analytic function at that point. Since this is true for every point in $f(S)$ , we find that $f^{-1}$ is actually analytic on all of $E$ , proving the claim. ∎

Note that as a corollary of the proof above, we obtained the following fact: if $f:D\to E$ is an analytic bijection, then $f^{\prime}$ is nowhere zero.

We conclude with a more quantitative version of the open mapping theorem. First let’s rephrase it slightly. We know that a non-constant analytic function $f:D\to\mathbb{C}$ sends open sets to open sets. The key property of open sets in the complex plane is that around every point $z_{0}$ , they contain some sufficiently small disk $D(z_{0},r)$ ; so another way to state the theorem would be to say that for every $z_{0}\in D$ , the image of the disk $D(z_{0},r)$ of radius $r$ , i.e. $f(D(z_{0},r))$ , contains a disk $D(f(z_{0}),R)$ of some radius $R$ . Bloch’s theorem tells us that we can make precise the relationship between $r$ and $R$ :

Theorem (Bloch’s theorem).

There exists a constant $c>0$ such that for any non-constant analytic function $f:D\to\mathbb{C}$ , for any $z_{0}\in D$ and $r>0$ with $D(z_{0},r)\subset D$ , the image $f(D(z_{0},r))$ contains the disk $D(f(z_{0}),R)$ where $R=c\cdot r\cdot|f^{\prime}(z_{0})|$ . Further, when restricted to $D(z_{0},r)$ , $f$ becomes a bijection onto its image, so there is a right inverse $D(f(z_{0}),R)\to D(z_{0},r)$ .

In fact, $c\geq\frac{1}{72}$ .

Note that the nonvanishing of $f^{\prime}$ is crucial in order for this theorem to have any content!

We won’t prove Bloch’s theorem—the proof is not too hard, but is a little long. Let’s mention though that although we know thanks to Bloch that the constant $c$ must be at least $\frac{1}{72}\approx 0.01389$ , this is not the best possible bound. The true largest possible value for $c$ is unknown; it is known that $c\geq\frac{\sqrt{3}}{4}\approx 0.43301$ , and conjectured that the true value is $\sqrt{\frac{\sqrt{3}-1}{2}}\cdot\frac{\Gamma(1/3)\Gamma(11/12)}{\Gamma(1/4)}% \approx 0.47186$ .