Let $X$ be a one-dimensional one-step irreducible shift of finite type and let $\pi$ be a one-block factor code from $X$ to a sofic $Y$. Suppose $y$ is a right transitive point of
$Y$ and $\pi(u)=y$ for some $u\in X$. Given $u\_0=a$ and a block $B$ of $X$, is there a point $x\in\pi^{-1}(y)$ such that $x\_0=a$ and $B$ occurs in $x\_{[0,\infty)}$? (Note that the stronger statement is true when $\pi$ is a finite-to-one code: a point is transitive if and only if it's image under $\pi$ is transitive, so $B$ actually occurs infinitely often in the right of $x$.)
Thanks to the helpers.



Relative irreducibility  
Sorry that this is slightly technical... It's uses some concepts that Mahsa showed me in answering a more simple question that I asked her.


Using results from <http://arxiv.org/abs/1001.5323v1>, we may assume that there exists a 'magic symbol' $k\in Y$ satsisfying that for any
word $y\_m\cdots y\_n\in Y$ with $y\_m=y\_n=k$, we have that any word
$x\_m\cdots x\_n\in X$ with $\pi(x\_m\cdots x\_n)=(y\_m\cdots y\_n)$ can be
extended to a sequence $x\in X$ with $\pi(x)=y$. This is not actually the definition of magic symbol but a theorem about them, see the above paper for a definition.


We assume that the desired word $B$ in $X$ starts and finishes with
elements of $\pi^{-1}(k)$, if not we just extend the word $B$ a
little bit. We let $B=b\_1\cdots b\_m$ and $C=\pi(B)$.


Let $A=\{a^1,\cdots,a^j\}$ be the set of possible values of $x\_1$ for
words $x\_1\cdots x\_m$ with $\pi(x\_1\cdots x\_m)=C$. Clearly $b\_1\in A$,
we let $a^1=b\_1$.


We begin with the following lemma:


\begin{lemma}
There exists a word $W=w\_1\cdots w\_n$ in
$Y$ with $w\_1=w\_n=k$ such that for each $a\in A$ there exists
a word $V=v\_1\cdots v\_n\in X$ with $\pi(V)=W$, $V$ contains $B$ and
$v\_1=a$.
\end{lemma}


To prove the lemma we define $W=w\_1\cdots w\_{i\_j}$ in chunks which
deal with the possible values of $v\_1\in A$.


We let $w\_1\cdots w\_{i\_1}$ be the word $C$. Then for $v\_1=a^1$ we are
done, since $B$ is a preimage of $C$ which starts with $a^1$. Since
$B$ starts and finishes with elements of $\pi^{-1}(k)$ it can be
extended to a sequence $x$ projecting to $y$.


Now we deal with $a^2$. By the definition of $A$, there exists a
preimage of $C$ starting with $a^2$. By the transitivity of $X$, this
preimage, which is a finite word in $X$, can be extended to be another
finite word in $x\_1\cdots x\_{i\_2}\in X$ which finishes with word $B$.
We let $w\_1\cdots w\_{i\_2}=\pi(x\_1\cdots x\_{i\_2})$, noting that
$i\_2>i\_1$ and $\pi(x\_1\cdots x\_{i\_1})=w\_1\cdots w\_{i\_1}$, so there is
no conflict with the values of $W$ which we already defined.


Now we do the same trick for $a^3$. There is a preimage of $w\_1\cdots
w\_{i\_1}$ $(=c\_1\cdots c\_m)$ starting with $a^3$, and hence there must
be a preimage of $w\_1\cdots w\_{i\_2}$ starting with $a^3$, because
$w\_1, w\_{i\_1}$ and $w\_{i\_2}$ are all equal to the magic symbol $k$.
Let $x\_1\cdots x\_{i\_2}$ be this word, and because $X$ is transitive we
can extend it to a finite word $x\_1\cdots x\_{i\_3}$ in $X$ finishing
with block $B$. Let $w\_1\cdots w\_{i\_3}=\pi(x\_1\cdots x\_{i\_3})$.


We continue this process until we have a word $W=w\_1\cdots w\_{i\_j}$,
this word $W$ satisfies the conditions of the lemma, so the lemma is
proved.


Now let $y\in Y$ be a transitive sequence. Then $y$ contains the word
$W$, say $y\_l\cdots y\_{l+i\_j}=W$. For our desired starting symbol $u$
there exists a word $x\_1\cdots x\_l$ with $x\_1=u$ and $\pi(x\_1\cdots
x\_l)=y\_1\cdots y\_l$, since there exists a preimage of $y$ starting
with $u$. $x\_l$ will be some member of $A$, and then using the lemma
we can extend $x\_1\cdots x\_l$ to a sequence which contains $B$ and
projects to $y$ as required.


