It is known that the notion of a function from spaces $R$ to space $D$ being open (set $x$ open in $R$ implies image $f(x)$ is open in $D$) is independent from the notion of a function being continuous (for any set $x$, if $f(x)$ is open in $D$, then $x$ is open in $R$). And this statement applies even if $R$ and $D$ are metric spaces.
Can you give me an understandable example of a function on metric spaces that is open but not continuous? (Obviously, then, the inverse is continuous but not open.)
I would prefer $f:\Bbb{R} \mapsto \Bbb{R}$ or if not, $f:\Bbb{R}^n \mapsto \Bbb{R}^m$ but iwll settle for any example if those don't exist.
It suffices to take the inverse of a continuous bijection (whose inverse fails to be continuous).
Here, then, is my go-to example: consider $S^1 \subset \Bbb R^2$, and the map $$ f:S^1 \subset \Bbb R^2 \to (-\pi,\pi]\\ (\cos( x), \sin(x)) \mapsto x $$ If we extend this example to the whole plane, we get something classic: the map $z \in \Bbb C \mapsto \ln(z)$ (with the usual branch cut) is open, but not continuous.
Or, if you prefer, take $z \in \Bbb C \mapsto \text{Arg}(z) \in \Bbb R$. This gives us an $\Bbb R^2 \to \Bbb R$ example, which I think is as low as we can go for $\Bbb R^n \to \Bbb R^m$.
It may be useful to note that for any open $f:X \to Y$: if $f$ is injective, $X$ is Hausdorff, and $f(X)$ is compact, then the map is necessarily continuous.