Proving a function is an open map, with limitations Following from a previous question I posted in here, given the same thing, i.e. an open set $U$ in $\mathbb{R}^2$ and a continuous function $f:U\rightarrow \mathbb{R}^2$ such that for each $u\in U$ exists a neighbourhood $V_u$ in $U$ such that $f\uparrow_{V_u}$ is one-to-one.
I'd like to show $f$ is an open map.
Now I was wondering what will happen if this last statement will be true except for a single point, i.e. exists a $u_0$ such that for each $u\in U$ if $u\neq u_0$ then there exists a neighbourhood $V_u$ in $U$ such that $f\uparrow_{V_u}$ is one-to-one.
Is it still true to say that $f$ is an open map? Or perhaps I need to add another assumption to make it correct?
I tried proving it by taking any open set $V\subseteq U$, and writing it as a union of open sets $\{V_i\}_{i\in I}$ (a solution suggested by @Joe Manlove ), by taking  $ \hat V_v $ to be a neighborhood with $ f\uparrow_{\hat{V}_v}$ that is $1:1$, for every $v\in V $, and then taking $ V_v := \hat V_v \cap V $, so that $V_v\subseteq V$, thus making their union be exactly $V$. Then I can use Invariance of Domain Theorem on each $V_i$, thus getting that $f(V)$ is a union of open sets, thus open as well. 
But the problem now is that I might not be able to find such cover, because of the possibility of $a$ being in $V$. What I tried to do was to take the union for all the other $V_v$ where $v\neq a$, and then proving that $a$ must be in this union, because $U$ is an open set. But then again I'm not so sure this is correct, and I thought perhaps there's another assumption (possibly a small one) I need to make before being able to prove this?
Other than this case, I was wondering, if this is correct, does it mean I can have more than a single "bad" point, i.e. $u_0,...,u_n$ and $f$ will still be an open set?
Edit: Does the following will help me - that the set $\hat{A}=\{a\in U \,\,\, | \,\,\,\, f(a)=f(u_0) \}$ is a discrete sub-space on $U$?
 A: The answer to your question is affirmative, that is, $f$ is still an open map. This follows from the result of your previous question, together with the following lemma: Suppose that $U\subset \mathbb R^n$ is an open set containing a point $u_0$ and that $f\colon U\to \mathbb R^m$, where $m>1$, is a continuous function such that $f(U-\{u_0\})$ is open. If there is an open and bounded set $D$ containing $u_0$ such that $\overline{D}\subset U$ and such that $f(D-\{u_0\})$ is open, then $f(U)$ is open.
The answer to your question follows from this lemma because in your scenario we can take $D$ to be a sufficiently small open disk containing $u_0$, the closure of which is contained in $U$; both $f(U-\{u_0\})$ and $f(D-\{u_0\})$ will be open by the result of your previous question, and then the lemma implies that $f(U)$ will be open as well. Because the hypotheses of your question hold for any open subset of $U$, it then follows that the map $f$ is open.
Let me now turn to the proof of the lemma. To begin with, we'll need the following fact: If $p$ is an isolated point of the boundary of an open set $V\subset \mathbb R^m$ (where $m>1$), then $p$ is interior to the closure of $V$. To prove this pick an open connected set $W$ containing $p$ such that $W\cap \partial V = \{p\}$ and put $W' = W-\{p\}$; then $W'\cap V$ is nonempty and $W'\cap V = W'\cap \overline{V}$, so by connectedness (which is what we need dimension $>1$ for) $W'\cap V = W'$. Thus $W\subset \overline V$ and $p$ is interior to $\overline V$ as claimed.
Going back to the situation of the lemma, write $U' = U-\{u_0\}$ and $D' = D-\{u_0\}$. That $f(D')$ is open in $\mathbb R^m$ implies that $\partial f(D')\subset f(\partial D') \subset f(\partial D) \cup \{f(u_0)\}$. Now $f(\partial D)$ is a compact subset of the open set $f(U')$, and so it has positive distance from the boundary $\partial f(U')$. Thus if $f(u_0)$ lies on the boundary of $f(U')$ (as it necessarily does if it is not already interior to $f(U')$), then $f(u_0)$ is an isolated point of $\partial f(D')$. By the fact just proved this means that $f(u_0)$ is interior to the closure of $f(D')$, and since the closure of $f(D')$ (which is equal to $f(\overline {D})$ by the compactness of $\overline{D}$) lies in $f(U)$ it follows at once that $f(u_0)$ is interior to $f(U)$. The conclusion now follows.
A: $f$ is an open map on the open $U_0=U\setminus\{a\}$ : if $V\subset U$ is open and $a\not\in V$ then $f(V)$ is open (https://math.stackexchange.com/a/1006024/14409).
If $V\subset U$ is an open neighborhood of $a$, $f(V)=f(V\setminus\{a\})\cup\{f(a)\}$ is not open unless $f(a)\in\overline{f(V\setminus\{a\})}$. In particular, in order to $f$ to be open it is necessary to have the following:
$$\forall r>0,\ f(a)\in\overline{f(B(a,r)\setminus\{a\})}$$
where $B(a,r)$ is the ball of radius $r$ centered in $a$.
Yes, $\hat{A}=\{x\in U,\ f(x)=f(a)\}$ is discrete (possibly equal to $\{a\}$), since $f$ cannot be constant in a neighborhood of any point $x\in\hat{A}$.
