Find computable totally function $f: \mathbb{N} \rightarrow \mathbb{N} $ and solvable set $A$, when $f(A)$ will not be solvable.

Question

Find computable totally function $f: \mathbb{N} \rightarrow \mathbb{N} $ and recursive set $A$, when $f(A)$ will not be recursive.

I have an idea, that we should find enumerable but not recursive set.

Answer 1

Instead of $\Bbb N$, we can use any countable set (as long as the conversion between the two is computable), e.g. the set of turing macines, or even the set of tuples $(T,n,x)$ where $T$ encodes a Turing machine, $n$ is a natural number, and $x$ is a state of $T$.

Let $A$ be the set of all tuples $(T,n,x)$ for which $x$ is the state of $T$ after exactly $n$ steps; this is ecursive.

Let $$f(T,n,x)=\begin{cases}T&\text{if $x$ is a halting state of $T$}\\0&\text{otherwise.}\end{cases}$$ This is computable.

But $f(A)$ is the set of all halting Turing machines (union $\{0\}$). This is not recursive.