# Invariance of domain in $\mathbb{R}^2$

Let $U \subseteq \mathbb{R}^2$ an open subset and let $f:U\rightarrow \mathbb {R}^2$ is be a continuous function.
I have the following version of Invariance of Domain Theorem (in $\mathbb{R}^2$):
If $f$ is injective then $f$ is homeomorphism.

I need to show:
a) If $f$ is locally injective then $f$ is open map.
b) If there is $a\in U$ such that $f^{-1}(a)$ is discrete subset, and $f$ is locally injective in $U\setminus {a}$, then $f$ is open map.

For (a) I tried: If $f(U)=\emptyset$ then there is nothing to prove. So I took some $y\in f(U)$, $y=f(x)$ for an $x\in U$. then there is an open subset $V\subseteq U$ such that $x\in V$ and $f|_V$ is injective. I tried to look at $\overline V$ and somehow use the invariance of domain theorem above, but I can't figure what to do.
For (b) - I'm lost.

can someone help me?

• The thesis of the IDM actually tells you: $f$ is a homeomorphism of $U$ with an open subset of $\mathbb{R}^2$. Does this answer your question? I guess in b you mean $f^{-1}(f(a))$. Oct 17 '14 at 19:51

The full statement of the Invariance of Domain Theorem: if $f$ is injective then $f$ is a homeomorphism between $U$ and the open set $f[U]$. (It also holds for all $n$, not just the planar case $\mathbb{R}^2$ with $n=2$.)
So now your start of the proof can be finished: let $O$ be non-empty open, let $y \in f[O]$, so $y = f(x)$ for some $x \in U$. Then $x$ has a neighbourhood $U$ such that $f$ restricted to $U$ is 1-1. Then $O \cap U$ is also a neighbourhood, $f$ is still 1-1 on it, so the Invariance of Domain (in my statement) shows us that $y \in f[O \cap U] \subset f[O]$, and $f[O \cap U]$ is open, showing that $y$ is an interior point of $f[O]$, as we needed to show.
As to the second part (where I'll assume you mean that $F = f^{-1}[\{f(a)\}]$ is discrete): suppose $O$ is again, non-empty open. If $y = f(x) \in f[O]$, and $y \neq f(a)$, then $x$ misses the discrete set $F$, so a small neighbourhood of $x$ can be chosen that misses $F$ and such that $f$ is injective on it. Apply the IoD theorem again, as above. And if $f(x) = f(a)$, we have a neighbourhood $O$ of $x$ such that $F \cap O = \{x\}$. Then I'd try to show that $f[O]$ is open. But $f$ need not (I think) be 1-1 on a small neighbourhood of $x$, a sone would hope. So some more thought needed there.