# Jacobian matrix of the inverse of a bijective function

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?

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

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.
• Who said $f$ was analytic? This is wrong if $f$ is just differentiable. – Kevin Carlson Sep 18 '14 at 0:40