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Generally, local invertibility does not imply invertibility. However, for differentiable functions from $\mathbb{R}$ to $\mathbb{R}$ then surjectivity and local invertibility do imply invertibility.

As well as being the most obvious, it's also the only (non-contrived) case that I can think of. Are there any more?

Specifically, I'm looking for examples of spaces $X$ which are at least topological spaces (but may be more structured) and subsets of endomorphisms on $X$ for which local invertibility implies invertibility.

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    $\begingroup$ I'm having a hard time seeing why continuous (surjective) functions from R to R don't work. Is there a standard counterexample that forces you to restrict to differentiable ones? It seems like a continuous function is locally invertible iff locally monotone iff globally monotone iff globally invertible. And the set-theoretic inverse of an invertible continuous map R-->R is continuous. $\endgroup$
    – Hunter Brooks
    Commented Dec 31, 2009 at 2:25

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The "Jacobian Conjecture" is an example of such a statement. It reads :

"If a polynomial $P:\mathbb{C}^N\rightarrow \mathbb{C}^N$ has invertible differential everywhere, then it is globally invertible."

It is open, and considered as difficult.

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For a locally invertible holomorphic function on the unitary disk $f:\Delta\rightarrow \mathbb{C}$ we have a result of Becker (1972): if $(1-|z|^2)\left|\frac{zf''(z)}{f'(z)}\right|\leq 1$, then $f$ is invertible. We have also two criteria of Nehari involving the Schwarzian derivative $S(f)$ of $f$: if $|S(f)(z)|\leq\frac{2}{(1-|z|^2)^2}$ or $|S(f)(z)|\leq \frac{\pi^2}{2}$, then $f$ is invertible. A good reference is the book "geometric function theory in one and higher dimensions" by Graham-Kohr.

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Following Deane Yang, the answer is a definite yes: the map in question is a global diffeo, provided that (a) it is `locally invertible': i.e. its derivative is everywhere invertible, and (b) the domain and range are compact, simply connected, without boundary.

Proof: the map must be a covering map (``stack of records theorem'' -- see for example Guillemin and Pollack). But a covering map between simply connected spaces is an isomorphism -- here a diffeo.

To make this `non-contrived' take domain and range to be the two-sphere.

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  • $\begingroup$ This follows immediately from my comment to Deane's answer, since properness of the map f follows trivially from compactness of the domain X. The assumptions of compactness of the range Y, simple connectedness of X, boundarylessness of both are completely superfluous. $\endgroup$
    – Georges Elencwajg
    Commented Dec 31, 2009 at 9:26
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Another class of topological spaces of which this property holds are trees, and more generally R-trees. http://en.wikipedia.org/wiki/Real_tree

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There are a number of different ways to show that a local diffeomorphism $f: X \rightarrow Y$ is a global diffeomorphism:

For example, this follows if $X$ and $f(X)$ are both connected and simply connected. (this appears to be incorrect based on the comments below)

Or if $X$ and $Y$ are both simply connected compact manifolds without boundary.

I think but am not sure that the statement also follows if $X$ and $Y$ are both simply connected manifolds without boundary and the map $f$ is proper.

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    $\begingroup$ If f:X--->Y is a local homeomorphism between hausdorff topological spaces, then f is a finite covering if and only f is proper. Hence if moreover Y is simply connected and X connected, a proper local homeomorphism will be a homeomorphism. (No assumption of simple connectedness is needed for X) $\endgroup$
    – Georges Elencwajg
    Commented Dec 30, 2009 at 19:27
  • $\begingroup$ The first claim doesn't seem quite right; I think you need (at least) surjectivity of f as well. For instance there are local diffeomorphisms from the interval (0, 2) to itself, with image (0, 1). $\endgroup$
    – macbeth
    Commented Dec 30, 2009 at 19:54
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    $\begingroup$ In fact, I don't think the first claim is right at all. For instance, there's a surjective, not-injective local diffeomorphism from $\mathbb{R}^2$ to $S^2$: you can wind an infinitely long rubber ribbon around a ball so as to cover it completely. $\endgroup$
    – macbeth
    Commented Dec 30, 2009 at 20:28
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A local isometry between complete connected Riemannian manifolds must be a covering map. So a local isometry between complete connected Riemannian manifolds, with simply-connected range, should be a global isometry?

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