If we have an entire function f that's locally invertible at every point z, then should the function be invertible on the entire complex plane C?
I am trying to solve this problem with the help of Inverse Function Theorem. Doing so makes me wonder if the converse is somehow true, i.e., whether local invertibility at some point implies non-vanishing derivative at that point? This is exactly where I have hit the wall.
Also, is there another way of solving this?
There are two different mathematical issues:
As noted in the comments by Conrad, a holomorphic function is locally bijective in some neighborhood of $z_{0}$ if and only if $f'(z_{0}) \neq 0$. (For example, there is no complex analogue of the real cubing function, which is bijective but has a critical point.)
An entire function turns out to be bijective (i.e., invertible; with a bit of work the inverse function is also seen to be entire) if and only if it is non-constant and affine, i.e., of the form $f(z) = az + b$ for some complex $a \neq 0$ and $b$.
And as noted in the comments by lhf, the second point may be irrelevant to the motivating goal, because there exists a familiar entire function having no critical points, and whose invertibility can be checked by inspection.