Let $F: V \times W\rightarrow Z$, where $V,W,Z$ are finite-dimensional smooth (or analytic) manifolds and $F$ is smooth (or analytic). Assume that $\dim W=\dim Z$ and the usual inverse function theorem applies when we fix the first argument of $F$, so locally we have an inverse (w.r.t. the second argument of $F$) mapping $f: V \times Z \rightarrow W$ such that for any $v,w,z$ from suitably chosen small vicinities we have $F(v,f(v,z))\equiv z$ and $f$ is smooth (or analytic). Here comes the question:

Is there any sensible way to glue together these $f$'s so that we can speak of them globally, and under which conditions?

More precisely, is there a way to treat this situation "as if" $f$ were defined globally, without doing the "restrict ourselves to suitably chosen small vicinities" thing all the time, even if we allow the Jacobian of $F$ w.r.t. $w$ to vanish at some points or even on some submanifolds of $V\times W$ of nonzero codimension?

I do not quite expect that (under reasonably mild conditions) one can construct a global $f$ defined on the whole $Q=Z\times V/Sing$ where $Sing=\lbrace(v,z)\in V\times Z|f \quad \mbox{can't be reasonably defined}\rbrace$ but perhaps one can have a sheaf of such $f$'s on $Q$ or something of the sort? I just started with learning the sheaf theory, so I don't quite (yet) know how to make things work on my own.

While in my particular application I would like $V$ to be a manifold, if this makes things easier, I can quite well make do with $W$ and $Z$ being just open domains in $\mathbb{R}^n$ or $\mathbb{C}^n$ (or whole $\mathbb{R}^n$ or $\mathbb{C}^n$ for that matter); that's the main reason why I wrote the domain of $F$ as $V\times W$ rather than just some general manifold $M$.

In fact, I am quite convinced that things like that should have been treated in the literature but I wasn't able to google up anything reasonable so far (maybe I just don't know the right keywords), so pointing out any suitable references is most welcome.

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    $\begingroup$ Have you tried to work this out yourself? The inverse function theorem gives uniqueness as well as existence. What goes wrong when you try to extend the domain of the inverse map? $\endgroup$ – Deane Yang Nov 28 '10 at 22:21
  • $\begingroup$ @Deane: Thanks for your input; however, in the applications I have in mind I must allow the Jacobian of $F$ to vanish at some points (or rather on a submanifold of $V\times W$ of nonzero codimension in $V\times W$), so I can't genuinely globalize $f$, that's the whole point. $\endgroup$ – anonymous Nov 29 '10 at 5:41
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    $\begingroup$ If the Jacobian vanishes on a submanifold, then that's where all the action is. You have to study the behavior of $F$ on and near the singular submanifold in order to say anything more. There is not going to be a general theorem about this, since the behavior can be quite complicated. But perhaps your situation has some additional conditions (dimensional constraints, for example) that might reduce the possibilities. $\endgroup$ – Deane Yang Nov 29 '10 at 16:23

There exist several known global implicit function theorems. Those results tend to be tailored for specific applications. It seems to be rather difficult to state a universally useful one-size-fits-all version.

One result that I find particularly helpful goes back to Hadamard (see, e.g., Chapt. 6 of "The Implicit Function Theorem" by Krantz and Parks):

Theorem. Let $M$ and $N$ be smooth, connected manifolds of dimension $d$ and let $f:M\to N$ be a $C^1$ mapping. If

  • $f$ is proper (i.e. $f^{-1}(K)\subset M$ is compact whenever $K\subset N$ is compact),
  • the Jacobian of $f$ vanishes nowhere on $M$, and
  • $N$ is simply connected,

then $f$ is a homeomorphism.

You might be interested also in this paper by Rheinboldt. It contains some topological conditions on when the local solvability of the equation $$F(x,f(x,z))=z$$ leads to the global solvability.

  • $\begingroup$ Instead of assuming the target simply connected, you might assume the map has degree one in some sense, e.g. some fiber is one point, or if the manifolds are both compact, the map on top homology is isomorphic. $\endgroup$ – roy smith Nov 29 '10 at 0:41
  • $\begingroup$ Thank you, Andrey and Roy, that's interesting but it's going in a direction which is different from what I originally had in mind. The problem is that, in the notation you adopted, requiring the nonvanishing of Jacobian of $F$ everywhere on $M$ is too restrictive for the applications I need; that's why I used "globalize" (with the quotation marks!) in the title and threw in the remark on sheaves. $\endgroup$ – anonymous Nov 29 '10 at 4:39
  • $\begingroup$ I was very interested in this result by Hadamard. Do you know if I can take $M$ to be open ball of $\mathbb{R}^n$? I think open ball should be fine, but would a closed ball be fine also? Thank you $\endgroup$ – Johnny T. Mar 27 at 12:07

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