2
$\begingroup$

Let $f:\mathbb{D}=\{z\in\mathbb{C}\mid |z|<1\}\rightarrow\mathbb{C}$ be a local diffeomorphism (i.e. an immersion) from an open disk in the plane to the plane.

The only situation I can image where $f$ is not injective is that $f$ sends $\mathbb{D}$ to a "self-overlapping'' region, in which case $f$ can not have continuous injective boundary values. But it seems non-trivial to proof that such boundary values guarantee injectivity:

Question. Assume that $f$ extends to a continuous map $\overline{\mathbb{D}}\rightarrow\mathbb{C}$ such that $f|_{\partial\mathbb{D}}$ is injective and continuous, so that (by Jordan Curve Theorem) it maps $\partial\mathbb{D}$ homeomorphically to a Jordan curve which is the boundary of a simply connected domain $\Omega\subset\mathbb{C}$. Then it is true that $f$ is a homeomorphism from $\mathbb{D}$ to $\Omega$?

Essentially we can reduce the problem to the case where $f|_{\partial\mathbb{D}}$ maps $\partial\mathbb{D}$ identically to $\partial\mathbb{D}\subset\mathbb{C}$ and ask:

Question'. If $f:\overline{\mathbb{D}}\rightarrow\mathbb{C}$ is a continuous map such that $f|_{\partial\mathbb{D}}$ is the identity and $f|_{\mathbb{D}}$ is a local homeomorphism, then is $f$ always injective?

$\endgroup$

1 Answer 1

3
$\begingroup$

Yes, this sort of thing can be proved by degree theory. An outline of the argument is given here: https://math.stackexchange.com/questions/737358/locality-of-inverse-function-theorem

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .