Computability and continuity Say $L_1$ and $L_2$ are computable languages.  Let $f$ be a function $L_1 \rightarrow L_2$.  Let $C$ be the statement, &quot;if $l \subseteq L_2$ and $l$ is a computable language, the preimage $f^{-1}(l)$ is a computable language.&quot;  Does $C$ imply that $f$ is necessarily a computable function?
$C$ is similar to the topological concept of continuity; a function $g$ between topological spaces is continuous if, for each open set $S$ in the codomain of $g$, $g^{-1}(S)$ is an open set in the domain.  Here &quot;open sets&quot; have been replaced with &quot;computable languages.&quot;  (Computable languages can't actually be considered open sets here, because they aren't closed under infinite union.)
The converse, &quot;$f$ is computable implies $C$,&quot; is easily proved.  If $f$ is computable, and $l$ is a computable language that's a subset of $L_2$, then let $l' = f^{-1}(l)$.  You can computably tell whether something is in $l'$ by first mapping it through $f$ and asking if the result is in $l$.  So $l'$ must be a computable language, proving $C$.
Let $L_1 = L_2 = \mathbb{N}$ and let $M \subseteq \mathbb{N}$ be a maximal set and let $L = \mathbb{N} \setminus M$ be its complement. Recall that $L$ is infinite, and that every computably enumerable (c.e.) subset $S \subseteq \mathbb{N}$ contains either finitely many elements of $L$ or all but finitely many elements of $L$.
Let $f : \mathbb{N} \to \mathbb{N}$ be an enumeration of $L$ without repetitions. The map $f$ is not computable, or else $M$ would be a computable subset of $\mathbb{N}$, which a maximal set is not.
We claim that $f$ is a counter-example to your question. Consider any c.e. subset $S \subseteq \mathbb{N}$:

If $S \cap L$ is finite then $f^{-1}(S) = f^{-1}(S \cap L)$ is finite (as $f$ is injective) and therefore a computable set.
If $S \cap L$ contains all but finitely many elements of $L$ then $f^{-1}(S) = f^{-1}(S \cap L)$ contains all but finitely many elements of $\mathbb{N}$, therefore it is a computable set.

In your question you hint at the analogy between computability and continuity. Indeed there is one, but we need to be a bit more sensitive about computability.
Let $$\mathcal{E}  = \{S \subseteq \mathbb{N} \mid \text{$S$ is c.e.}\}$$ be the set of all c.e. subsets of $\mathbb{N}$, and let $W : \mathbb{N} \to \mathcal{E}$ be a standard enumeration of c.e. sets. We can think of $\mathcal{E}$ as the &quot;computably open&quot; subsets of $\mathbb{N}$. Indeed, they're closed under finite intersections and unions of computable families.
Say that a map $F : \mathcal{E} \to \mathcal{E}$ is computable if there exists a computable $r : \mathbb{N} \to \mathbb{N}$ such that $F(W_n) = W_{r(n)}$ for all $n \in \mathbb{N}$. (This is just the standard notion of computability for numbered sets.)
Your question can be rectified into a theorem like this (I just inserted the emphasized &quot;computable&quot;):
Theorem: A map $f : \mathbb{N} \to \mathbb{N}$ is computable if, and only if, its inverse image map $f^{-1} : \mathcal{P}(\mathbb{N}) \to \mathcal{P}(\mathbb{N})$ restricts to a computable map $f^{-1} : \mathcal{E} \to \mathcal{E}$.
I will leave the (easy) proof as an exercise. The above can easily be generalized from number-theoretic maps to languages.