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Is there such a thing as a (germ at the origin of) holomorphic mapping $f:\mathbb{C}^n\rightarrow\mathbb{C}^n$ which admits a differentiable inverse (of any class) which is not holomorphic?

The motivation for this question is that I'm seeing in some publications the expression "holomorphic diffeomorphism" as equivalent (or being used to refer to) biholomorphisms. My impression is that the word "diffeomorphism" is given emphasis due to the fact that it represents a new system of coordinates in $\mathbb{C}^n$.

This all matters only locally (at the origin) to what I'm studying, so I included the possibility for germs only, but I wondered if that was true in general as well.

Marra
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  • The inverse is holomorphic, this basically follows from the chain rule and the fact that you can characterize holomorphic functions as those for which $\partial / \partial \bar{z}$ vanishes. (This works in multiple variables as well, that partial means against all anti holomorphic coordinates) (I think. I'll write up an answer in which I check these details, but for now you can look at : http://www.jirka.org/scv/scv.pdf , and in particular theorem 1.3.6) – Elle Najt Jul 11 '17 at 18:00
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    Even more is true: An invertible holomorphic map between domains in $C^n$ is biholomorphic. https://math.stackexchange.com/questions/497548/why-holomorphic-injection-on-cnmust-be-biholomorphic – Moishe Kohan Jul 12 '17 at 14:41

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Here is a sketch, hopefully all correct. (I think I'm answering your question.) What do you think?

Prop: A smooth function $f : U \subset \mathbb{C}^n \to \mathbb{C}$ is holomorphic iff $\partial f / \partial \bar{z} = 0$ (constantly zero). Pf:This is equivalent to the CR equations. Here $\partial f/ \partial \bar{z}$ means the vector $(\partial f / \partial \bar{z_i})$, so this is equivalent to CR in each coordinate direction.

For $f : \mathbb{C}^n \to \mathbb{C}^m$, define $D(f) = (\frac{ \partial f_i}{\partial z_j})$ and $\bar{D}(f) = (\frac{ \partial f_i}{\partial \bar{z}_j})$.

Prop: Then $f$ is holomorphic iff $\bar{D}(f) = 0$. Pf: This follows from above.

Prop: (Chain rule) Suppose that $U \subset \mathbb{C}^n$, $V \subset \mathbb{C}^m$ are open sets, and $f : U \to V$ is smooth, and $g : V \to \mathbb{C}^k$ is smooth. Then $D(g \circ f) = D(g) \bar{D}(f) + \bar{D}(g) \bar{D}(\bar{f})$ Pf: Long computation in coordinates. (This is another way to write 1.3.1 in the linked notes - this is the spot I would check my writing carefully.)

Prop: $\bar{D}(\bar{f}) = \overline{ D(f)}$. Pf: The point is that $\overline{ (\partial_x + i \partial_y)(u + iv)} = \overline{ (\partial_x + i \partial_y)} \dot{} \overline{(u + iv)}$.

Prop: Suppose that $f$ is holomorphic, and $g \circ f$ is holomorphic. Suppose that $D(f)$ is everywhere invertible . Then $g$ is holomorphic.

$0 = \bar{D}(gf) = D(g) \bar{D}(f) + \bar{D}(g) \bar{D}({\bar{f}}) = \bar{D}(g) \overline{ D(f) }. $ Since the last term is an invertible matrix, this implies that $\bar{D}(g) = 0$. Hence $g$ is holomorphic.

Elle Najt
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  • Very well done! You indeed nailed it for me when you mentioned the chain rule. Gods smite me, I've forgotten about the implications of it. But thanks for your time! :) – Marra Jul 11 '17 at 18:22
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    @Marra Thanks :-) No problem -- I am always trying to find ways to force myself into working out certain details that I would usually recoil from, and social pressure is good for that. :) Yes, it is surprising how easily this conclusion follows! The chain rule is the tool, but it is all the magic of the CR equations. – Elle Najt Jul 11 '17 at 18:26