thus $ d_{\uu}(w) = s_0s_1s_2\cdots$. Since $r_0=\varphi_b(0)$ and $r_1=\varphi_b(1)$, we obtain also $\varphi_b(\uu') = \uu = \varphi_b(s_0)  \varphi_b(s_1) \varphi_b(s_2)  \cdots = \varphi_b(s_0s_1s_2\cdots)$. The statement in \[it:der\_of\_images\_fi\_b\_1\] now follows from injectivity of $\varphi_b$.

Now suppose that the prefix $w$ of $\uu$ is of length $>1$. As explained earlier, it suffices to consider right special prefixes. Since the letter $1$ is always followed by $0$, each right special factor must end in $0$. So the first and the last letter of $w$ is $0$, hence by Lemma \[lem:unique\_preimage\_under\_fi\_b\] there is a unique prefix $w'$ of $\uu'$ such that $\varphi_b(w')0 = w$. Let $r_0$ and $r_1$ be the two return words to $w$ and let $\uu = r_{s_0}r_{s_1}r_{s_2}\cdots$. Since the first letter of both $r_0$ and $r_1$ is $0$, there are uniquely given $r'_0$ and $r'_1$ such that $r_0=\varphi_b(r'_0)$ and $r_1=\varphi_b(r'_1)$ and $\uu' = r'_{s_0}r'_{s_1}r'_{s_2}\cdots$.

Clearly $w'$ is a prefix of $r'_{s_j}r'_{s_{j+1}}r'_{s_{j+2}}\cdots$ for all $j \in \N$ and so the number $|r'_{s_0}r'_{s_1}\cdots r'_{s_k}|$ is an occurrence of $w'$ in $\uu'$ for all $k \in \N$. Let $i > 0$ be an occurrence of $w'$ in $\uu'$. It follows that $\varphi_b\bigl(\uu'_{[0,i)}\bigr)w$ is a prefix of $\uu$ and $|\varphi_b\bigl(\uu'_{[0,i)}\bigr)|$ is an occurrence of $w$ in 