Let $U\subset\mathbb{C}^n$ and $V\subset\mathbb{C}^m$ be open subsets, and let $f : U\rightarrow V$ be a holomorphic map such that $df : TU\rightarrow TV$ has constant (complex) rank $k$. I'd like a reference for the statement:

There exists biholomorphic maps $\phi : U\rightarrow U'\subset\mathbb{C}^n$ and $\psi : V\rightarrow V'\subset\mathbb{C}^n$ such that $\psi\circ f\circ \phi^{-1}$ is given by the formula $(z_1,\ldots,z_n)\mapsto (z_1,\ldots,z_k,0,\ldots,0)$.

(I assume this is true? It seems like the proof of the real analog works just as well)

  • $\begingroup$ So presumably for the proof of the real case to carry over to the complex case, am I right in thinking that we just need the holomorphic version of the inverse function theorem? $\endgroup$ – Kenny Wong May 30 '18 at 22:30
  • $\begingroup$ @KennyWong I was imagining just using the usual inverse function theorem together with the fact that bijective holomorphic maps have holomorphic inverses (I think this is true?) $\endgroup$ – stupid_question_bot May 30 '18 at 22:37
  • 1
    $\begingroup$ Yeah, that's what I had in mind too! And yes, it's true that bijective holomorphic maps have holomorphic inverses - that's proved in Chapter 0.2 of Griffiths and Harris. $\endgroup$ – Kenny Wong May 30 '18 at 22:47

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