Is decidability closed under the mapping f where f(a)=f(b)=0 and f(c)=1?
Let $f$ be any computable mapping such that $|f(x)| \geq c|x|$ for some $c &gt; 0$. For every language $L$, if $L$ is computable then so is $f(L)$.

To see this, let us be given an input $y$. We go over all strings $x$ of length at most $|y|/c$. For each of them, we check whether $x \in L$ and $f(x) = y$. If we found such a string $x$, then $y$ is in the language, otherwise it isn't.