Continuous map from a connected open subset of the plane onto another one Let $\Omega_1, \Omega_2$ be two non empty open connected subsets of the plane. Is it always possible to find a continuous map $f$ from $\Omega_1$ onto $\Omega_2$?
Note: The origin of my question is this one: continuous-mapping-from-open-set-to-open-set.
 A: As I said, this is probably overkill, but if we use space-filling curves, we can take for granted that there exists a continuous surjection $[0,1]\rightarrow [0,1] \times [0,1]$.  
Now if $V \subset \mathbb{R}^2$ is open, connected, there exists a continuous surjection $V\rightarrow (0,1)$ (for instance, just project to the x-axis to get a disjoint union of open intervals, then choose one such interval $J$, take the identity on $J$ and send all the others into $J$) 
Let $U\subset \mathbb{R}^n$ be open, connected, and write $$U:=\cup_{i=1}^\infty C_i$$ where $C_i$ are a countable collection of closed squares (of varying sizes).  
We can define a surjective map $f_1:[1/2, 1)\rightarrow C_1$ and a path $\gamma :[1/3,1/2]\rightarrow  U$ with $\gamma(1/2)= f(1/2), \gamma (1/3)\in C_2$. Now define a surjective map $f_2: [1/4,1/3] \rightarrow C_2$. concatenating these maps gives a surjective map $[1/4,1)\rightarrow C_1\cup C_2$. Continuing in this way gives a surjection $(0,1) \rightarrow V$. 
....right?
Actually it seems silly to use space filling curves because we collapsed space just to fill it again. You could replace $(0,1)$ with an open strip, which you think of as a divided into squares, mapping the first square to $C_1$, the next square to $\gamma$, the next square to $C_2$... 
