TITLE: Can we represent computable functions by r.e. sets ?
QUESTION [10 upvotes]: As we know, if $f$ is a computable function, then every pre-image of a r.e. set under $f$ is also a r.e. set, i.e. $f^{-1}(X)$ is a r.e.set if $X$ is a r.e. set. So I want to know that if a function satisfies that every pre-image of a r.e. set is also a r.e. set,  then can we conclude that $f$ is a computable function?

REPLY [5 votes]: Some related functions have the following application (D. Myers 2007, unpublished).
Define 
$$
A\le B\quad\Longleftrightarrow\quad A=f^{-1}(B) \quad\text{ for some $f$ that maps every $\Sigma^0_n$ set to a $\Sigma^0_n$ set, for each $n$}
$$
Theorem: For each $n\ge 1$, the properly $\Sigma^0_n$ sets (the sets in $\Sigma^0_n\backslash \Delta^0_n$) form a single equivalence class under the equivalence relation induced by $\le$. As do the properly $\Pi^0_n$ sets.
Thus $\le$ allows us to look at the arithmetical hierarchy as a degree structure. It is not clear whether this can be done with an arithmetical relation in place of $\le$. (Turing reducibility is $\Sigma^0_3$ hence arithmetical, but the relation $\le$ is on its face $\Sigma^1_1$ only.)