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I have been reading about the open mapping theorem for non constant holomorphic functions, all the proofs involve theorems of complex analysis (like Rouche, argument principle). What if we just applied the inverse function theorem to obtain that holomorphic functions are local diffeomorphism and therefore open, is this valid?

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    $\begingroup$ $f(z)=z^2$ is non-constant, and entire, but the inverse function theorem doesn’t apply around the origin. $\endgroup$
    – peek-a-boo
    Commented Jan 18, 2023 at 15:29
  • $\begingroup$ but i should say that it is possible to use the IFT, just that one has to be more sophisticated. The above function is indeed open, since it is a pointwise product of (homeomorphisms and hence) open maps. I’ll elaborate more if I have time later $\endgroup$
    – peek-a-boo
    Commented Jan 18, 2023 at 15:38
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    $\begingroup$ @peek-a-boo $z \mapsto z$ and $z \mapsto \bar{z}$ are both open but their pointwise product $z \mapsto |z|^2$ is not open. $\endgroup$ Commented Jan 18, 2023 at 22:16
  • $\begingroup$ @StevenGubkin very good point, I had convinced myself that if $f,g:X\to Y$ are open maps, then so is $(f,g):X\to Y\times Y$ (confused it with $f\times g:X\times X\to Y\times Y$, $(x_1,x_2)\mapsto (f(x_1),g(x_2))$). Good thing the proof below doesn't use my "fact". $\endgroup$
    – peek-a-boo
    Commented Jan 18, 2023 at 23:02

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Yes, you can prove it using the Inverse Function Theorem:

Consider a non-constant holomorphic function $f:\Omega\longrightarrow \mathbb{C}$, where $\Omega$ is a connected open set. Let $z_0\in \Omega$.

Since $f$ is not constant, the zero $z_0$ from the function $f-f(z_0)$ can be factored out. That is, there exists a function $g$ holomorphic and non-zero in some disk $D(z_0,r)$ such that $$f(z)-f(z_0)=(z-z_0)^ng(z),$$ where $n$ is the multiplicity of $z_0$.

Now, since $g$ is non-zero, it has an holomorphic logarithm: there exists a function $h\in \mathcal{H}(D(z_0,r))$ such that $$g=e^h.$$

Therefore, $$f(z)-f(z_0)=(z-z_0)^ne^{h(z)}=\left((z-z_0)e^{\frac{h(z)}{n}}\right)^n.$$

So $f$ can be expressed locally as $$f=f(z_0)+F^n,$$ where $F:=(z-z_0)e^{h/n}$. You can check that $F'$ doesn't annihilate in $z_0$. Thus, $F$ is locally open by the Inverse Function Theorem. Furthermore, $F^n$ is also open since it is the composition of two open functions, $F$ and $z\mapsto z^n$. Finally, adding the constant $f(z_0)$ doesn't change anything, so $f$ is open.

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    $\begingroup$ It should be noted that $\Omega$ needs to be connected for obvious reasons (and this is invoked in your third line when you say “the zero $z_0$ from $f-f(z_0)$ can be factored out”). $\endgroup$
    – peek-a-boo
    Commented Jan 18, 2023 at 21:02
  • $\begingroup$ True! Thanks for noticing. $\endgroup$ Commented Jan 18, 2023 at 21:59

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