Continuous mapping from open set to open set Suppose we have two open and bounded sets, $\Omega_1,\Omega_2 \in \mathbb{R}^2$. Is there a continuous function $\textbf{f}$ mapping $\Omega_1$ onto $\Omega_2$? 
\begin{align*}
\Omega_1 & = \{(x,y) \in \mathbb{R}^2 : 1 < x^2 + y^2 <2 \} \\
\Omega_2 & = \{(x,y) \in \mathbb{R}^2 : 3x^2 < x^2 + y^2 < 4\}
\end{align*}
I'm having trouble applying the definition of onto, i.e. for all $\textbf{y}$ in $\Omega_2$, there is AT LEAST ONE $\text{x} \in \Omega_1$ such that $\textbf{f}(\textbf{x}) = \textbf{y}$. Or I need to show that 
$$\textbf{f}(\Omega_1) = \Omega_2$$
 A: The set $\Omega_2 $ is not connected but the set $\Omega_1 $ is connected. Since the image by the continiuous function  of connected set is always connected thus such function does not exist.
A: My hint here is wrong.
I was having in mind that $\Omega_2$ was equal to $\{(x,y) \in \mathbb{R}^2 : x^2 + y^2 < 2 \text{ and } 3 x^2 \le y\}$ which is not the case. I however don't delete the answer as it is interesting to find a continuous mapping between $\Omega_1$ onto $\Omega_2^\prime=\{(x,y) \in \mathbb{R}^2 : x^2 + y^2 < 2 \text{ and } 3 x^2 \le y\}$.
You have $\Omega_1= \Omega_1^i \uplus \Omega_1^e$ where:
\begin{align*}
\Omega_1^i & = \{(x,y) \in \mathbb{R}^2 : 1 < x^2 + y^2 \le 3/2 \} \\
\Omega_1^e & = \{(x,y) \in \mathbb{R}^2 : 3/2 < x^2 + y^2 < 2 \}
\end{align*}
The disk $D=\{(x,y) \in \mathbb{R}^2 : x^2 + (y-1)^2 \le 1/4 \}$ is included in $\Omega_2$. To be verified precisely. If this is not true, a disk centered on $(0,1)$ with a radius small enough is included in $\Omega_1$ as $\Omega_1$ is open and $(0,1) \in \Omega_1$.
Now you can find


*

*$g$ continuous from $\Omega_1^i$ onto $D$.

*$h$ continuous from $\Omega_1^e$ onto $\Omega_2^\prime \setminus D$


in such a way that $f$ defined by
\begin{align*}
f|\Omega_1^i = g \\
f|\Omega_1^e = h
\end{align*}
is continuous.
