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One way of stating the Open Mapping Theorem is that non-constant holomorphic functions map open sets to open sets. I am kind of confused why this is remarkable. Isn't this a characteristic property of continuous functions anyways?

John
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    (Commenting instead of answering since I think this is a duplicate:) Absolutely not! Continuity is the opposite direction: the preimage of an open set with respect to a continuous function is open. For instance, consider (in $\mathbb{R}$) the $\sin$ function: this is continuous, but we have $$\sin[(0,17)]=[-1,1]$$ (remember $f[A]={f(a):a\in A}$) which is not open. The open mapping theorem is actually extremely powerful: in $\mathbb{R}$-land, most(?) of the continuous functions we're used to are not open (a function is open iff it sends open sets to open sets). – Noah Schweber May 12 '22 at 17:57
  • Or even just a constant function – MPW May 12 '22 at 17:59
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    @MPW I wanted to pick a function whose "complex version" satisfies the hypotheses for the open mapping theorem, though. – Noah Schweber May 12 '22 at 18:00
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    Constant functions actually have to be excluded in this statement of the Open Mapping Theorem anyway. So it's a good idea to show the plethora of non-constant continuous functions which are not open. Like $f(z) = \lvert z \rvert$. – Torsten Schoeneberg May 12 '22 at 18:02
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    Just among the related question the AI shows up for me here, https://math.stackexchange.com/q/3932438/96384 looks like a near-duplicate, https://math.stackexchange.com/q/124846/96384 looks like a nice "explanation", and https://math.stackexchange.com/q/1734141/96384 gives an easy application which shows how much this differs from plain old real calculus, let alone just any continuous functions. – Torsten Schoeneberg May 12 '22 at 18:10

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