# Generalizations of the Inv. Function Thm.

1. Jan 11, 2008

### WWGD

Hi, everyone:

I was wondering if anyone knew of any extensions to the inverse function theorem
to this effect:

If f is a differentiable map, and df(x) is non-zero, then the IFT guarantees
there is a nhood (neighborhood) U_x containing x , such that $f|(U_x)$
a diffeom. $U_x-->f(U_x)$.

Now, under what conditions on f , can we be guaranteed to have that f
has a global differentiable inverse?, i.e, f has a global inverse $f^-1$ and
$f^-1$ is differentiable . I imagine df(x) not 0 for all x is necessary, but not
sure that it is sufficient.

For simple cases like f(x)=$x^2$ this is true in any interval [a,b]
not containing zero, and it may be relatively easy to do a proof of the claim
above for maps f:IR->IR . But I have no idea how well this would generalize
to maps f:$R^n\to\mathbb{R^n}$

Any Ideas?
Thanks.

2. Jan 12, 2008

### gel

If U is an open convex subset of $\mathbb{R}^n$ and $f:U\rightarrow\mathbb{R}^n$ is differentiable with positive definite Jacobian then it is not hard to see that f(U) is open and $f^{-1}:f(U)\rightarrow U$ exists.

edit: If J is the jacobian then I should have said that <x,Jx> > 0 for all non-zero vectors x. Or, equivalently, J+J^T is positive definite.

Last edited: Jan 13, 2008
3. Jan 13, 2008

### mathwonk

gel, what do you mean by positive definite jacobian?

does your statement cover the complex exponential map from C-->C? i.e. R^2-->R^2.

the jacobian is then e^x times a rotation matrix.

i.e. the term positive definite is usually applied to quadratic forms, not linear maps. or do you mean the quadratic form <x,Tx>, is a positive definite form where T is the linear jacobian.

Last edited: Jan 13, 2008
4. Jan 13, 2008

### gel

If f is the complex exponential then $f(x+iy)=e^x(\cos y + i\sin y)$. Then $df/dx=e^x (\cos y+ i\sin y)$ and $df/dy = e^x(-\sin y + i\cos y)$. As a map on $\mathbb{R}^2$ this has Jacobian
$$e^x\left(\begin{array}{ll} \cos y & \sin y\\ -\sin y & \cos y \end{array}\right)$$
which is e^x times a rotation matrix, as you say.
This is positive definite as long as cos y is positive, so the complex exponential restricted to complex numbers with imaginary part bounded by $\pm\pi/2$ has a smooth inverse.

5. Jan 13, 2008

### gel

That's true, no doubt you could restate the result in a different form avoiding the positive definiteness condition. I don't know if this is a standard result, it's just something that occurred to me when I was working through the basics of differentiable manifolds myself.

6. Jan 13, 2008

### Hurkyl

Staff Emeritus
(Assuming the domain is connected) I think that f has to induce a surjective map on the fundamental groups. I even find it plausible that that's a sufficient condition.

7. Jan 13, 2008

### gel

If f is invertible then it would have to be an isomorphism of fundamental groups, but it's not a sufficient condition. E.g., $f:[-1,1]\rightarrow [0,1]$ given by $f(x)=x^2$ is not invertible, but the fundamental groups of [-1,1] and [0,1] are trivial.

edit: maybe you meant that f has a local inverse and is surjective on the fundamental groups then it has a global inverse. Maybe that's true, I don't know, although in dimensions greater than 2 I expect you'd also need to require it to be surjective on the homology or homotopy groups.

Last edited: Jan 13, 2008
8. Jan 13, 2008

### Hurkyl

Staff Emeritus
I was implicitly assuming that f was already known to be locally invertible.

9. Jan 13, 2008

Positive definite for operators usually means all positive eigenvalues, doesn't it?

10. Jan 13, 2008

### gel

A real matrix is positive definite iff it is symmetric and has positive eigenvalues.

I was rather sloppy in my statement. A matrix M is positive definite if it is symmetric and <x,Mx> is positive for all non-zero vectors x. I wasn't requiring the Jacobian to be symmetric, just that <x,Mx> is positive, which is the same as requiring the symmetric part of M (i.e., (M+M^T)/2 ) to be positive definite. I should have been a bit clearer in my statement above.

11. Jan 17, 2008

### WWGD

Could you please elaborate on this?. I can't think of how properties of the fund.
group could affect the existence of a global smooth inverse, given that we have
the existence of local inverses.

12. Jan 17, 2008

### WWGD

To be more explicit: I understand the issue of homorphisms induced by maps, and
proofs that use this idea (I think there is a classic proof of Brower's fixed point thm.
that uses this), but I don't see how we could use induced homomorphisms , induced
locally by the Inv.Fn. Thm, to show the existence of a globally-defined inverse,
much less how these induced local homomorphisms would show any thing about
smoothness of f^-1, neither locally nor globally.

13. Jan 17, 2008

### Hurkyl

Staff Emeritus
Observe that if local inverses exist, a path in the codomain and a chosen preimage of any point on the path together uniquely determine a path in the domain. (which I will call its 'lift')

(Now, assume the codomain is path connected)

If we choose any point of the codomain, and choose a preimage for it, this gives a recipe for choosing preimages of every other point. I believe that this recipe is well-defined if and only if every loop in the codomain lifts to a loop in the domain. This is why I'm sure the fundamental group, or possibly the first homology or cohomology group, yield the condition we are looking for.

Since f^-1 is sewn together from the locally defined inverses...

14. Jan 17, 2008

### mathwonk

i think the essence of these arguments is that a proper local homeomorphism is a covering map. and a covering map is classified by what it does to the fundamental group.

in particular it is always injective on fundamental groups, hence is the unique homeomorphic covering if and only if it is surjective.

properness is essential however, since there exist local homeomorphisms which are surjective on fundamental groups that are not covering maps, and not surjective, such as the injection of the punctured plane into the full plane.