Proving that a function (involving the Cantor discontinuum ) is surjective.

Exercise. Let $\mathcal{D} = \{2\sum_{i=1}^{\infty} x_i3^{-i} : \forall i \in \Bbb N, x_i \in \{0,1\}\}$ be the Cantor discontinuum. Now, consider the function \begin{equation*} g : \mathcal{D} \rightarrow [0,1] \quad \text{ s.t. } \quad g\left(2\sum_{i=1}^{\infty} x_i3^{-i}\right) = \sum_{i=1}^{\infty} x_i2^{-i}. \end{equation*} Prove that $g$ is a surjective function.

My attempt. We want to prove that $g$ is surjective, i.e., that \begin{equation*} \forall x \in [0,1], \exists y \in \mathcal{D} \hspace{.15cm} | \hspace{.15cm} g(y) = x \end{equation*} Let $y \in \mathcal{D}$. Then, we might assume that $y = 2\sum_{i=1}^{\infty}x_i3^{-i}$, where $x_i \in \{0,1\}$. Besides this, we also see the following: \begin{equation*} g(y) = \sum_{i=1}^{\infty} x_i2^{-i} = \frac{1}{2}\left(2\sum_{i=1}^{\infty}x_i3^{-i}\left(\frac{2}{3}\right)^{-i}\right) = \frac{1}{2}y\sum_{i=1}^{\infty}\left(\frac{2}{3}\right)^{-i} \end{equation*} and thus, \begin{equation*} g(y) = x \Leftrightarrow \frac{y}{2}\sum_{i=1}^{\infty}\left(\frac{2}{3}\right)^{-i} = x \Leftrightarrow y = 2\sum_{i=1}^{\infty} (x2^i)3^{-i}. \end{equation*} And I believe this should be enough to prove the surjectivity. BUT $y$ doesn't seem to be a member of $\mathcal{D}$ since we can't say that $x_i = x2^{i} \in \{0,1\}$ (If $i$ is any integer and $x$ is any number on the unitary interval, then this obviously isn't true). This leads me to conclude that my resolution might not be right at all, and this is why I am posting this.

Thanks for any help in advance.

You have to start the proof by picking a $y \in [0,1]$. It has a binary expansion $0.x_1x_2x_3\ldots$, where all $x_i \in \{0,1\}$. This directly defines a sequence of $x_i$ to define a point $x$ in $\mathcal D$ such that $g(x)=y$. That’s all.

I think your approach is too complicated. You should know that each sequence $(x_i)$ with $x_i \in \{0,1\}$ determines a unique point $\phi((x_i)) = 2\sum_{i=1}^\infty x_i3^{-i} \in \mathcal{D}$. And should also know that each $x \in [0,1]$ has a dyadic representation $x = \sum_{i=1}^\infty x_i2^{-i}$ with $x_i \in \{0,1\}$. This is not unique, e.g. we have $1/2 = \sum_{i=1}^\infty x_i2^{-i}$ with $x_1 = 1$ and $x_i = 0$ for $i > 1$ and also $1/2 = \sum_{i=1}^\infty x_i2^{-i}$ with $x_1 = 0$ and $x_i = 1$ for $i > 1$.

But despite of non-uniqueness this shows that your function $g$ is surjective.