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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

Motivation:

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

Thanks.

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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.2011-01-30

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