2
$\begingroup$

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?

$\endgroup$
5
  • 2
    $\begingroup$ Consider $f:\mathbb{C}\to\mathbb{C}, z\mapsto z^3$. $\endgroup$ Sep 17, 2014 at 23:54
  • $\begingroup$ Did you mean $f=(f_1,\ldots,f_n)$ rather than $f=f(f_1,\ldots,f_n)$? ${}\qquad{}$ $\endgroup$ Sep 18, 2014 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. $\endgroup$ Sep 18, 2014 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, 2014 at 0:34
  • $\begingroup$ @Fujoyaki Ick, yeah, over $\mathbb{C}$ I should say something like $f(x,y)=(x,y^3)$. Thanks. $\endgroup$ Sep 18, 2014 at 0:40

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Who said $f$ was analytic? This is wrong if $f$ is just differentiable. $\endgroup$ Sep 18, 2014 at 0:40
  • $\begingroup$ Yep, see above edit. $\endgroup$
    – orangeskid
    Sep 18, 2014 at 0:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .