In this MSE post A function $f:\mathbb{R}^2\to\mathbb{R}^2$ that is open and closed, but not continuous., user Moishe Kohan provides an example of a non-continuous open and closed ("clopen") function $\mathbb R^n \to \mathbb R^n$ for $n\geq 3$ by citing and using the following propositions of David Wilson:
Propositions 1 and 3 of
[Moishe Kohan's] Proposition [cooked up from Wilson's propositions]. Let $I^n$ be the closed $n$-dimensional cube, $n\ge 3$, and $J^n\subset int(I^n)$ is a closed subcube. Let $Q$ denote the interior of $J^n$. Then, there exists an open continuous map $g: I^n \setminus Q\to I^n$ which equals the identity on the boundary of $I^n$ and sends $\partial Q$ to the interior of $I^n$.
I am interested in the $2$-dimensional case, for which I propose the following conjecture:
Conjecture: there are no continuous open (w.r.t. subspace topologies) maps $f$ from the closed annulus $\mathbb A:= \{x\in \mathbb R^2: 1\leq |x|\leq 2\}$ to the closed disk $\mathbb D := \overline{B(0,2)}$, which restricts to the identity map on the outer boundary circle $\partial \mathbb D = C(0,2)\subseteq \mathbb A$, and sends the inner boundary circle $C(0,1)$ to the interior of $\mathbb D$.
I was not able to prove or disprove this conjecture for even the Simpler Case: where $f$ maps the inner boundary circle $C(0,1)$ to the single point $0\in \mathbb D$.
Some attempts I made on the Simpler Case.
Attempt 1: Because we want $f$ to be open, we in particular want any open neighborhood in $\mathbb A$ of any boundary point $b\in C(0,1)$ to map to an open set containing $0\in \mathbb D$, in particular containing some $B(0,\epsilon)$.
My idea was map circles $C(0,1+\eta)$ to circles $C(0,\eta)$, and to to "swirl"/"smear" the circles $C(0,1+\eta)$ more and more extremely as $\eta \searrow 0$, so that even a very small neighborhood $B(b,\delta)\cap \mathbb A$ of $b\in C(0,1)$ containing just a $\approx \frac{2\delta}{2\pi}$ fraction of the circles $C(0,1+\eta)$ for small enough $\eta>0$ would get "swirled"/"smeared" to contain the entire circle $C(0,\eta)$.
So something like $f$ maps $z:=re^{i\theta}\in \mathbb A$ (thinking of $\mathbb R^2$ as the complex plane) to $(r-1)e^{i\theta\cdot \frac{1}{r-1}}$, mapping an arc of the circle $C(0,1+\eta)$ of arclength $\ell$ to an arc of the circle $C(0,\eta)$ with arclength $\ell \cdot \frac{1}{r-1}$. Unfortunately, the corresponding formula for $f$ would be $$f(z)=\frac{|z|-1}{|z|^\frac{1}{|z|-1}} \cdot z^{\frac{1}{|z|-1}},$$ which maybe looks fine, until one remembers that raising a complex number to a non-integer power needs (non-continuous) branch cuts to define. :(
Attempt 2: One can make a continuous "swirling only" (no "spreading", i.e. no multiplicative factor in the argument variable) by mapping $r e^{i\theta} \mapsto (r-1) e^{i\theta + i\cdot \frac{1}{r-1}}$, i.e. $$f(z) = \frac{|z|-1}{|z|} \cdot z \cdot e^{\frac{1}{|z|-1}},$$ which I think is continuous, and does "swirl" a very small neighborhood $N_b:= B(b,\delta)\cap \mathbb A$ to something that does sort of "wrap around" $0 \in \mathbb D$, but in doing so has a lot of holes: this $f$ preserves that proportion of the arclength of $N_b \cap C(1+\eta)$, meaning $\text{arclength}(N_b \cap C(1+\eta)) = \text{arclength}(f(N_b) \cap C(0,\eta))$, so $f$ can't possibly map $N_b$ to something containing $B(0,\epsilon)$.
Attempt 3: Finally, because of this "monodromy problem" of defining this "swirling/smearing" map on the entire circle $C(0,1+\eta)$, I had an idea of cutting up $C(0,1+\eta)$ into $\frac 1\eta$ many pieces (restricting to $\eta \in \{\frac 1n: n=2, 3, 4, \ldots\}$), making $f$ "swirl/smear" each of those pieces into the entirety of $C(0,\eta)$, thus guaranteeing that $f(N_b)$ contains complete circles $C(0,\eta)$ arbitrarily close to $0\in \mathbb D$.
More precisely, I would partition $C(0,1+\frac 1n)$ into $2n$ equally sized, equally spaced pieces, and define $\{K_{n, i}\}_{i=1}^n$ to be the closures of the say odd-indexed pieces. I can map $K_{n,i}$ continuously to cover the entirety of $C(0,\frac 1n)$. Then for any $N_b:= B(b,\delta)\cap \mathbb A$, it does contain some $K_{n,i}$ for all $n$ sufficiently large, and hence $f(N_b)$ contains $C(0,\frac 1n)$ for all $n$ sufficiently large. I can extend this $f$ defined on $C(0,1) \cup \bigcup_{n=2}^\infty\bigcup_{i=1}^n K_{n,i}$ via the Tietze extension theorem to a continuous function $\mathbb A \to \mathbb D$, but again I really doubt it is a an open map.
(Cross posted to MO https://mathoverflow.net/questions/452140/does-there-exist-a-continuous-open-map-from-the-closed-annulus-to-the-closed-dis upon suggestion from comments)
