Many-one-reductions with finite image
As the name suggests, a many-one reduction can map many instances of&nbsp;$K$ onto a single instance of&nbsp;$A$.  If there are no restrictions on the power of&nbsp;$f$, then you can pick some $a\in A$ and $b\notin A$ and then define
$$f(x)=\begin{cases}
a &amp;\text{ if }x\in K\\
b &amp;\text{ if }x\notin K\,.
\end{cases}$$

However, if $f$&nbsp;is restricted to be computable then the image of&nbsp;$f$ is necessarily infinite. This is because every finite set is computable so, if the image of&nbsp;$f$ were finite, you'd have a computable reduction from the halting set to a computable set, which is impossible because that would mean the halting set would be computable too.

If $B$ is recursive and $f$ is computable then you can decide whether $x\in f^{-1}(B)$ just by checking whether $f(x)\in B$. If $B$&nbsp;isn't recursive or $f$&nbsp;isn't computable, all the bets are off.