Proving a function is an open map, with limitations

Solution 1:

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.

Solution 2:

$f$ 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 $x\in\hat{A}$.