When does local invertibility imply invertibility? 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.
 A: 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.
A: 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.
A: 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. 
A: Another class of topological spaces of which this property holds are trees, and more generally R-trees. http://en.wikipedia.org/wiki/Real_tree
A: 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.
A: 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?
