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Let $f:\mathbb{C}^n\rightarrow\mathbb{C}^n$ be a function such that $f=f(f_1,\ldots,f_n)$ and $f_i=f_i(x_1,\ldots,x_n)$.

Also, $f$ is bijective and its Jacobian matrix exists.

Does$f^{-1}\,$Jacobian matrix exist?

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    $\begingroup$ Consider $f:\mathbb{C}\to\mathbb{C}, z\mapsto z^3$. $\endgroup$ – Kevin Carlson Sep 17 '14 at 23:54
  • $\begingroup$ Did you mean $f=(f_1,\ldots,f_n)$ rather than $f=f(f_1,\ldots,f_n)$? ${}\qquad{}$ $\endgroup$ – Michael Hardy Sep 18 '14 at 0:01
  • $\begingroup$ Kevin Carlson has provided a counterexample, but I think next one should ask whether the conclusion holds when the Jacobian matrix is non-singular. $\endgroup$ – Michael Hardy Sep 18 '14 at 0:02
  • $\begingroup$ @KevinCarlson But the function you suggest isn't bijective since any element in the image corresponds to three elements in the domain. $\endgroup$ – Fujoyaki Sep 18 '14 at 0:34
  • $\begingroup$ @Fujoyaki Ick, yeah, over $\mathbb{C}$ I should say something like $f(x,y)=(x,y^3)$. Thanks. $\endgroup$ – Kevin Carlson Sep 18 '14 at 0:40
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It is a fact that an injective complex analytic map from an open subset of $\mathbb{C}^n$ to $\mathbb{C}^n$ has nonzero jacobians at all points and hence its image is an open subset of $\mathbb{C}^n$, and the inverse function is also analytic.

Yes, the Jacobian of the inverse function exists if we assume $f$ complex analytic.

If we only assume real analytic then the answer is no, see the example of Kevin Carlson above.

If we only assume the map to be injective and continuous we still get the image an open subset and the inverse map continuous, again a classic result.

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  • $\begingroup$ Who said $f$ was analytic? This is wrong if $f$ is just differentiable. $\endgroup$ – Kevin Carlson Sep 18 '14 at 0:40
  • $\begingroup$ Yep, see above edit. $\endgroup$ – Orest Bucicovschi Sep 18 '14 at 0:54

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