Does the direct image of a creative set also creative? This is the exercise 7.24. of Rotger's Theory of Recursive Functions and Effective Computability. (This is not a homework assignment. I solely prove exercises by my own.) He asks what happens if we take direct image or inverse image under a recursive one-to-one function to a productive set and a creative set. I can prove that:


*

*If $A$ is productive then $f[A]$ is also productive, and

*$f^{-1}[A]$ need not be neither creative nor productive even if $A$ does: for example, consider an one-to-one enumeration $f:\mathbb{N} \to K$ of a halting set. 


I have no idea whether $f[A]$ is creative if $A$ does, however. Thanks for any help.

Here are definitions for terms I have used in the question: 
$A\subset\mathbb{N}$ is productive if we have a partial recursive function $\psi$ such that if $W_e\subseteq A$ is not empty then $\psi(e)$ halts and $\psi(e)\in A-W_e$. $A$ is creative if $A$ is recursively enumerable and its complement is creative. For example, the complement of a halting set is creative under the function $\psi(x)=x$.
(Added in 26 April 2020: I swapped the definition of creativity and productiveness by confusion.
 Noah's answer is based on the my wrong previous definition. I corrected the definitions, but you need some care to read the answer.)
 A: The key fact is that we can pull back c.e. sets along computable maps (there's a topological analogy here, identifying "c.e." with "open" and "continuous" with "computable"). Specifically, suppose $f$ is computable and injective and $U$ is c.e. Then $f^{-1}[U]$ is also c.e., and this is uniform (= there is some computable $g$ such that $f^{-1}[W_e]=W_{g(e)}$ for all $e$).
With this in hand, here's how we can show that $f[C]$ is creative whenever $C$ is:
Suppose we're given that $W_e\subseteq f[C]$. Since $f$ is injective we always have $f^{-1}[f[A]]=A$, and so in particular we have $W_{g(e)}\subseteq C$. So $\psi(g(e))\in C\setminus W_{g(e)}$. This gives $f(\psi(g(e)))\in f[C]$, and so $f(\psi(g(e)))\not\in W_e$.
More snappily: if $\psi$ witnesses the creativity of $C$, then $f\circ \psi\circ g$ witnesses the creativity of $f[C]$.
(Incidentally, you don't need $W_e\not=\emptyset$ in the definition of creativity - we can always just nonuniformly pick some $a\in A$, and have $\psi$ start by moving from $W_e$ to $W_e\cup\{a\}$.)
