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I am curious as to what changes do we need to make to the hypotheses of the inverse function theorem in order to be able to find the global differentiable inverse to a differentiable function. We obviously need $f$ to be a bijection, and $f'$ to be non-zero. Is this sufficient for the existence of a global differentiable inverse?

For functions $f\colon\mathbb{R}\to\mathbb{R}$, we have


$f^{-1}(f(x))=x$, so $(f')^{-1}(f(x))f'(x)=1$

Then, we could define $(f')^{^-1}(f(x))$ to be $1/f'(x)$ ( this is the special case of the formula for the differentiable inverse -- when it exists -- in the IFT)

(and we are assumming $f'(x)\neq 0$)

In the case of $\mathbb{R}^2$, I guess we could think of all the branches of $\log z$ and $\exp z$, and we do have at least a branch-wise global inverse , i.e., if/when $\exp z$ is 1-1 (and it is , of course onto $\mathbb{C}-{0}$), then we have a differentiable inverse.

I guess my question would be: once the conditions of the IFT are satisfied: in how big of a neighborhood of $x$ can we define this local diffeomorphism, and, in which case would this neighborhood be the entire domain of definition of $f$?

I guess the case for manifolds would be a generalization of the case of $\mathbb{R}^n$, but it seems like we would need for the manifolds to have a single chart.

So, are the conditions of f being a bijective, differentiable map sufficient for the existence of a global differentiable inverse? And, if $f$ is differentiable, but not bijective, does the IFT hold in the largest subset of the domain of definition of $f$ where $f$ is a bijection?


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Please don't begin your lines with four or more spaces: instead of displaying, the renderer assumes you are typing code and posts it verbatim. I've edited your question for readability, adding mark-up. Please check that I didn't screw up anything. – Arturo Magidin Jan 30 '11 at 21:23

There is a theorem ("Introduction to Smooth Manifolds," Lee, Thm 7.15) for differentiable manifolds which says that:

If $F: M \to N$ is a differentiable bijective map of constant rank, then $F$ is a diffeomorphism -- so in particular $F^{-1}$ is differentiable.

Here, the rank of differentiable map $F\colon M \to N$ at a point $p \in M$ is defined to be the rank of its pushforward $F_*\colon T_pM \to T_{F(p)}N$. (Some authors use the word "differential" for "pushforward," and use the notation $d_pF$ for $F_*$.)

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So therefore, if $F\colon M \to N$ is differentiable, bijective, and $d_pF$ is invertible at every $p \in M$, then yes, there exists a global differentiable inverse. – Jesse Madnick Jan 31 '11 at 1:54
It might also be good to bear in mind that the main issue here is whether $d_pF$ is invertible, and not whether it is the zero matrix. – Jesse Madnick Jan 31 '11 at 1:59
As to how big one can make the domain of the local diffeomorphism... I don't know of a good answer to that. – Jesse Madnick Jan 31 '11 at 2:00

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